algebra of the trigonometric functions.€¦ · some trig identities. let’s recall the things we...
TRANSCRIPT
Algebra of the trigonometric functions.
Algebra of the trigonometric functions. 1 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity)
cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 =
1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) =
sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,
sec(θ) = 1cos(θ) , csc(θ) = 1
sin(θ) , cot(θ) = cos(θ)sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) =
1cos(θ) , csc(θ) = 1
sin(θ) , cot(θ) = cos(θ)sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) ,
csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) =
1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) ,
cot(θ) = cos(θ)sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) =
cos(θ)sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) =
cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ),
sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) =
− sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),
cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) =
− cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ),
sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) =
sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),
cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) =
− cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ),
sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) =
− sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),
cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) =
sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ),
sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) =
cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) =
cos(θ)
Algebra of the trigonometric functions. 2 / 7
Some trig identities.
Let’s recall the things we know about the trig functions
Theorem
(The most important identity) cos(θ)2 + sin(θ)2 = 1
(The obvious identities)
tan(θ) = sin(θ)cos(θ) ,sec(θ) = 1
cos(θ) , csc(θ) = 1sin(θ) , cot(θ) = cos(θ)
sin(θ) ,
(Previously)
cos(−θ) = cos(θ), sin(−θ) = − sin(θ),cos(π − θ) = − cos(θ), sin(π − θ) = sin(θ),cos(π + θ) = − cos(θ), sin(π + θ) = − sin(θ),cos(π/2 − θ) = sin(θ), sin(π/2 − θ) = cos(θ).
Can you get a formula for cos(π/2 + θ) in terms of sin(θ) and cos(θ)?
Proposition
For every angle θ, cos(π/2 + θ) = − sin(θ), sin(π/2 + θ) = cos(θ)
Algebra of the trigonometric functions. 2 / 7
Getting identities by putting everything in terms of sin andcos.
1 (I’ll do this one) Simplify tan(γ) + cot(γ).
tan(γ) + cot(γ) =
sin(γ)cos(γ) + cos(γ)
sin(γ)
=
sin(γ)2+cos(γ)2
cos(γ)·(sin(γ)
Use the most important identity
=
1cos(γ)·(sin(γ)
=
sec(γ) · csc(γ)
2 (group work) Simplify tan(γ) · cot(γ).
Algebra of the trigonometric functions. 3 / 7
Getting identities by putting everything in terms of sin andcos.
1 (I’ll do this one) Simplify tan(γ) + cot(γ).
tan(γ) + cot(γ) =
sin(γ)cos(γ) + cos(γ)
sin(γ)
=
sin(γ)2+cos(γ)2
cos(γ)·(sin(γ)
Use the most important identity
=
1cos(γ)·(sin(γ)
=
sec(γ) · csc(γ)
2 (group work) Simplify tan(γ) · cot(γ).
Algebra of the trigonometric functions. 3 / 7
Getting identities by putting everything in terms of sin andcos.
1 (I’ll do this one) Simplify tan(γ) + cot(γ).
tan(γ) + cot(γ) = sin(γ)cos(γ) + cos(γ)
sin(γ)
=
sin(γ)2+cos(γ)2
cos(γ)·(sin(γ)
Use the most important identity
=
1cos(γ)·(sin(γ)
=
sec(γ) · csc(γ)
2 (group work) Simplify tan(γ) · cot(γ).
Algebra of the trigonometric functions. 3 / 7
Getting identities by putting everything in terms of sin andcos.
1 (I’ll do this one) Simplify tan(γ) + cot(γ).
tan(γ) + cot(γ) = sin(γ)cos(γ) + cos(γ)
sin(γ)
= sin(γ)2+cos(γ)2
cos(γ)·(sin(γ)
Use the most important identity
=
1cos(γ)·(sin(γ)
=
sec(γ) · csc(γ)
2 (group work) Simplify tan(γ) · cot(γ).
Algebra of the trigonometric functions. 3 / 7
Getting identities by putting everything in terms of sin andcos.
1 (I’ll do this one) Simplify tan(γ) + cot(γ).
tan(γ) + cot(γ) = sin(γ)cos(γ) + cos(γ)
sin(γ)
= sin(γ)2+cos(γ)2
cos(γ)·(sin(γ) Use the most important identity
=
1cos(γ)·(sin(γ)
=
sec(γ) · csc(γ)
2 (group work) Simplify tan(γ) · cot(γ).
Algebra of the trigonometric functions. 3 / 7
Getting identities by putting everything in terms of sin andcos.
1 (I’ll do this one) Simplify tan(γ) + cot(γ).
tan(γ) + cot(γ) = sin(γ)cos(γ) + cos(γ)
sin(γ)
= sin(γ)2+cos(γ)2
cos(γ)·(sin(γ) Use the most important identity
= 1cos(γ)·(sin(γ)
=
sec(γ) · csc(γ)
2 (group work) Simplify tan(γ) · cot(γ).
Algebra of the trigonometric functions. 3 / 7
Getting identities by putting everything in terms of sin andcos.
1 (I’ll do this one) Simplify tan(γ) + cot(γ).
tan(γ) + cot(γ) = sin(γ)cos(γ) + cos(γ)
sin(γ)
= sin(γ)2+cos(γ)2
cos(γ)·(sin(γ) Use the most important identity
= 1cos(γ)·(sin(γ)
= sec(γ) · csc(γ)
2 (group work) Simplify tan(γ) · cot(γ).
Algebra of the trigonometric functions. 3 / 7
Proving something is an identity
1 (I’ll do this one) Prove that sec(α)2 + csc(α)2 = (sec(α) · csc(α))2
Expand sec(α)2 + csc(α)2
sec(α)2 + csc(α)2 =
1cos(α)2
+ 1sin(α)2
=
sin(α)2+cos(α)2
cos(α)2·sin(α)2
The important identity
=
1cos(α)2·sin(α)2
=
(sec(α) · csc(α))2
This is what we set out to prove!
2 (Group work) Prove that 1 + tan(θ)2 = sec(θ)2 is an identity bysimplifying one side of the equation.
Algebra of the trigonometric functions. 4 / 7
Proving something is an identity
1 (I’ll do this one) Prove that sec(α)2 + csc(α)2 = (sec(α) · csc(α))2
Expand sec(α)2 + csc(α)2
sec(α)2 + csc(α)2 =
1cos(α)2
+ 1sin(α)2
=
sin(α)2+cos(α)2
cos(α)2·sin(α)2
The important identity
=
1cos(α)2·sin(α)2
=
(sec(α) · csc(α))2
This is what we set out to prove!
2 (Group work) Prove that 1 + tan(θ)2 = sec(θ)2 is an identity bysimplifying one side of the equation.
Algebra of the trigonometric functions. 4 / 7
Proving something is an identity
1 (I’ll do this one) Prove that sec(α)2 + csc(α)2 = (sec(α) · csc(α))2
Expand sec(α)2 + csc(α)2
sec(α)2 + csc(α)2 = 1cos(α)2
+ 1sin(α)2
=
sin(α)2+cos(α)2
cos(α)2·sin(α)2
The important identity
=
1cos(α)2·sin(α)2
=
(sec(α) · csc(α))2
This is what we set out to prove!
2 (Group work) Prove that 1 + tan(θ)2 = sec(θ)2 is an identity bysimplifying one side of the equation.
Algebra of the trigonometric functions. 4 / 7
Proving something is an identity
1 (I’ll do this one) Prove that sec(α)2 + csc(α)2 = (sec(α) · csc(α))2
Expand sec(α)2 + csc(α)2
sec(α)2 + csc(α)2 = 1cos(α)2
+ 1sin(α)2
= sin(α)2+cos(α)2
cos(α)2·sin(α)2
The important identity
=
1cos(α)2·sin(α)2
=
(sec(α) · csc(α))2
This is what we set out to prove!
2 (Group work) Prove that 1 + tan(θ)2 = sec(θ)2 is an identity bysimplifying one side of the equation.
Algebra of the trigonometric functions. 4 / 7
Proving something is an identity
1 (I’ll do this one) Prove that sec(α)2 + csc(α)2 = (sec(α) · csc(α))2
Expand sec(α)2 + csc(α)2
sec(α)2 + csc(α)2 = 1cos(α)2
+ 1sin(α)2
= sin(α)2+cos(α)2
cos(α)2·sin(α)2 The important identity
=
1cos(α)2·sin(α)2
=
(sec(α) · csc(α))2
This is what we set out to prove!
2 (Group work) Prove that 1 + tan(θ)2 = sec(θ)2 is an identity bysimplifying one side of the equation.
Algebra of the trigonometric functions. 4 / 7
Proving something is an identity
1 (I’ll do this one) Prove that sec(α)2 + csc(α)2 = (sec(α) · csc(α))2
Expand sec(α)2 + csc(α)2
sec(α)2 + csc(α)2 = 1cos(α)2
+ 1sin(α)2
= sin(α)2+cos(α)2
cos(α)2·sin(α)2 The important identity
= 1cos(α)2·sin(α)2
=
(sec(α) · csc(α))2
This is what we set out to prove!
2 (Group work) Prove that 1 + tan(θ)2 = sec(θ)2 is an identity bysimplifying one side of the equation.
Algebra of the trigonometric functions. 4 / 7
Proving something is an identity
1 (I’ll do this one) Prove that sec(α)2 + csc(α)2 = (sec(α) · csc(α))2
Expand sec(α)2 + csc(α)2
sec(α)2 + csc(α)2 = 1cos(α)2
+ 1sin(α)2
= sin(α)2+cos(α)2
cos(α)2·sin(α)2 The important identity
= 1cos(α)2·sin(α)2
= (sec(α) · csc(α))2
This is what we set out to prove!
2 (Group work) Prove that 1 + tan(θ)2 = sec(θ)2 is an identity bysimplifying one side of the equation.
Algebra of the trigonometric functions. 4 / 7
Proving something is an identity
1 (I’ll do this one) Prove that sec(α)2 + csc(α)2 = (sec(α) · csc(α))2
Expand sec(α)2 + csc(α)2
sec(α)2 + csc(α)2 = 1cos(α)2
+ 1sin(α)2
= sin(α)2+cos(α)2
cos(α)2·sin(α)2 The important identity
= 1cos(α)2·sin(α)2
= (sec(α) · csc(α))2
This is what we set out to prove!
2 (Group work) Prove that 1 + tan(θ)2 = sec(θ)2 is an identity bysimplifying one side of the equation.
Algebra of the trigonometric functions. 4 / 7
Proving something is NOT an identity
If you suspect that something might be an identity, but you are not sure,then try to check if for some easy to compute inputsExample: Is sin(α + β) = sin(α) + sin(β) an identity?
Does it check out for α = 0, β = 0?
sin(0 + 0) = sin(0) = 0 and sin(0) + sin(0) = 0 + 0 = 0sin(0) + sin(0) = 0 + 0 = 0Passes this test
Does it check out for α = π/2, β = π/2?
sin(π/2 + π/2) = sin(π) = 0sin(π/2) + sin(π/2) = 1 + 1 = 2This is not an identity!
In your groups, and in your notes:Prove that cos(α · β) = cos(α) · cos(β) is not an identity.
Algebra of the trigonometric functions. 5 / 7
Proving something is NOT an identity
If you suspect that something might be an identity, but you are not sure,then try to check if for some easy to compute inputsExample: Is sin(α + β) = sin(α) + sin(β) an identity?
Does it check out for α = 0, β = 0?sin(0 + 0) = sin(0) = 0 and sin(0) + sin(0) = 0 + 0 = 0
sin(0) + sin(0) = 0 + 0 = 0Passes this test
Does it check out for α = π/2, β = π/2?
sin(π/2 + π/2) = sin(π) = 0sin(π/2) + sin(π/2) = 1 + 1 = 2This is not an identity!
In your groups, and in your notes:Prove that cos(α · β) = cos(α) · cos(β) is not an identity.
Algebra of the trigonometric functions. 5 / 7
Proving something is NOT an identity
If you suspect that something might be an identity, but you are not sure,then try to check if for some easy to compute inputsExample: Is sin(α + β) = sin(α) + sin(β) an identity?
Does it check out for α = 0, β = 0?sin(0 + 0) = sin(0) = 0 and sin(0) + sin(0) = 0 + 0 = 0sin(0) + sin(0) = 0 + 0 = 0
Passes this test
Does it check out for α = π/2, β = π/2?
sin(π/2 + π/2) = sin(π) = 0sin(π/2) + sin(π/2) = 1 + 1 = 2This is not an identity!
In your groups, and in your notes:Prove that cos(α · β) = cos(α) · cos(β) is not an identity.
Algebra of the trigonometric functions. 5 / 7
Proving something is NOT an identity
If you suspect that something might be an identity, but you are not sure,then try to check if for some easy to compute inputsExample: Is sin(α + β) = sin(α) + sin(β) an identity?
Does it check out for α = 0, β = 0?sin(0 + 0) = sin(0) = 0 and sin(0) + sin(0) = 0 + 0 = 0sin(0) + sin(0) = 0 + 0 = 0Passes this test
Does it check out for α = π/2, β = π/2?
sin(π/2 + π/2) = sin(π) = 0sin(π/2) + sin(π/2) = 1 + 1 = 2This is not an identity!
In your groups, and in your notes:Prove that cos(α · β) = cos(α) · cos(β) is not an identity.
Algebra of the trigonometric functions. 5 / 7
Proving something is NOT an identity
If you suspect that something might be an identity, but you are not sure,then try to check if for some easy to compute inputsExample: Is sin(α + β) = sin(α) + sin(β) an identity?
Does it check out for α = 0, β = 0?sin(0 + 0) = sin(0) = 0 and sin(0) + sin(0) = 0 + 0 = 0sin(0) + sin(0) = 0 + 0 = 0Passes this test
Does it check out for α = π/2, β = π/2?sin(π/2 + π/2) = sin(π) = 0
sin(π/2) + sin(π/2) = 1 + 1 = 2This is not an identity!
In your groups, and in your notes:Prove that cos(α · β) = cos(α) · cos(β) is not an identity.
Algebra of the trigonometric functions. 5 / 7
Proving something is NOT an identity
If you suspect that something might be an identity, but you are not sure,then try to check if for some easy to compute inputsExample: Is sin(α + β) = sin(α) + sin(β) an identity?
Does it check out for α = 0, β = 0?sin(0 + 0) = sin(0) = 0 and sin(0) + sin(0) = 0 + 0 = 0sin(0) + sin(0) = 0 + 0 = 0Passes this test
Does it check out for α = π/2, β = π/2?sin(π/2 + π/2) = sin(π) = 0sin(π/2) + sin(π/2) = 1 + 1 = 2
This is not an identity!
In your groups, and in your notes:Prove that cos(α · β) = cos(α) · cos(β) is not an identity.
Algebra of the trigonometric functions. 5 / 7
Proving something is NOT an identity
If you suspect that something might be an identity, but you are not sure,then try to check if for some easy to compute inputsExample: Is sin(α + β) = sin(α) + sin(β) an identity?
Does it check out for α = 0, β = 0?sin(0 + 0) = sin(0) = 0 and sin(0) + sin(0) = 0 + 0 = 0sin(0) + sin(0) = 0 + 0 = 0Passes this test
Does it check out for α = π/2, β = π/2?sin(π/2 + π/2) = sin(π) = 0sin(π/2) + sin(π/2) = 1 + 1 = 2This is not an identity!
In your groups, and in your notes:Prove that cos(α · β) = cos(α) · cos(β) is not an identity.
Algebra of the trigonometric functions. 5 / 7
Computing all the trig functions when you know only one.
1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)
2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).
Express tan(θ) = 3 in terms of sin and cos.
3 = sin(θ)cos(θ) so
3 · cos(θ) = sin(θ)
Now use the best identity:
1 = sin(θ)2 + cos(θ)2 =
(3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1
cos(θ) = ±√1010
In which quadrant are we? Quadrant 3. cos(θ) and sin(θ) are negative.
cos(θ) = −√1010
sin(θ) = 3 cos(θ) =
−√1010
Algebra of the trigonometric functions. 6 / 7
Computing all the trig functions when you know only one.
1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)
2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).
Express tan(θ) = 3 in terms of sin and cos.
3 = sin(θ)cos(θ) so
3 · cos(θ) = sin(θ)
Now use the best identity:
1 = sin(θ)2 + cos(θ)2 =
(3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1
cos(θ) = ±√1010
In which quadrant are we? Quadrant 3. cos(θ) and sin(θ) are negative.
cos(θ) = −√1010
sin(θ) = 3 cos(θ) =
−√1010
Algebra of the trigonometric functions. 6 / 7
Computing all the trig functions when you know only one.
1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)
2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).
Express tan(θ) = 3 in terms of sin and cos.
3 = sin(θ)cos(θ) so
3 · cos(θ) = sin(θ)
Now use the best identity:
1 = sin(θ)2 + cos(θ)2 =
(3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1
cos(θ) = ±√1010
In which quadrant are we? Quadrant 3. cos(θ) and sin(θ) are negative.
cos(θ) = −√1010
sin(θ) = 3 cos(θ) =
−√1010
Algebra of the trigonometric functions. 6 / 7
Computing all the trig functions when you know only one.
1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)
2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).
Express tan(θ) = 3 in terms of sin and cos.
3 = sin(θ)cos(θ) so 3 · cos(θ) = sin(θ)
Now use the best identity:
1 = sin(θ)2 + cos(θ)2 =
(3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1
cos(θ) = ±√1010
In which quadrant are we? Quadrant 3. cos(θ) and sin(θ) are negative.
cos(θ) = −√1010
sin(θ) = 3 cos(θ) =
−√1010
Algebra of the trigonometric functions. 6 / 7
Computing all the trig functions when you know only one.
1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)
2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).
Express tan(θ) = 3 in terms of sin and cos.
3 = sin(θ)cos(θ) so 3 · cos(θ) = sin(θ)
Now use the best identity:
1 = sin(θ)2 + cos(θ)2 =
(3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1
cos(θ) = ±√1010
In which quadrant are we? Quadrant 3. cos(θ) and sin(θ) are negative.
cos(θ) = −√1010
sin(θ) = 3 cos(θ) =
−√1010
Algebra of the trigonometric functions. 6 / 7
Computing all the trig functions when you know only one.
1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)
2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).
Express tan(θ) = 3 in terms of sin and cos.
3 = sin(θ)cos(θ) so 3 · cos(θ) = sin(θ)
Now use the best identity:
1 = sin(θ)2 + cos(θ)2 = (3 cos(θ)2) + cos(θ)2 so
10 cos(θ)2 = 1
cos(θ) = ±√1010
In which quadrant are we? Quadrant 3. cos(θ) and sin(θ) are negative.
cos(θ) = −√1010
sin(θ) = 3 cos(θ) =
−√1010
Algebra of the trigonometric functions. 6 / 7
Computing all the trig functions when you know only one.
1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)
2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).
Express tan(θ) = 3 in terms of sin and cos.
3 = sin(θ)cos(θ) so 3 · cos(θ) = sin(θ)
Now use the best identity:
1 = sin(θ)2 + cos(θ)2 = (3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1
cos(θ) = ±√1010
In which quadrant are we? Quadrant 3. cos(θ) and sin(θ) are negative.
cos(θ) = −√1010
sin(θ) = 3 cos(θ) =
−√1010
Algebra of the trigonometric functions. 6 / 7
Computing all the trig functions when you know only one.
1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)
2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).
Express tan(θ) = 3 in terms of sin and cos.
3 = sin(θ)cos(θ) so 3 · cos(θ) = sin(θ)
Now use the best identity:
1 = sin(θ)2 + cos(θ)2 = (3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1
cos(θ) = ±√1010
In which quadrant are we? Quadrant 3. cos(θ) and sin(θ) are negative.
cos(θ) = −√1010
sin(θ) = 3 cos(θ) =
−√1010
Algebra of the trigonometric functions. 6 / 7
Computing all the trig functions when you know only one.
1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)
2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).
Express tan(θ) = 3 in terms of sin and cos.
3 = sin(θ)cos(θ) so 3 · cos(θ) = sin(θ)
Now use the best identity:
1 = sin(θ)2 + cos(θ)2 = (3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1
cos(θ) = ±√1010
In which quadrant are we?
Quadrant 3. cos(θ) and sin(θ) are negative.
cos(θ) = −√1010
sin(θ) = 3 cos(θ) =
−√1010
Algebra of the trigonometric functions. 6 / 7
Computing all the trig functions when you know only one.
1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)
2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).
Express tan(θ) = 3 in terms of sin and cos.
3 = sin(θ)cos(θ) so 3 · cos(θ) = sin(θ)
Now use the best identity:
1 = sin(θ)2 + cos(θ)2 = (3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1
cos(θ) = ±√1010
In which quadrant are we? Quadrant 3. cos(θ) and sin(θ) are negative.
cos(θ) = −√1010
sin(θ) = 3 cos(θ) =
−√1010
Algebra of the trigonometric functions. 6 / 7
Computing all the trig functions when you know only one.
1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)
2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).
Express tan(θ) = 3 in terms of sin and cos.
3 = sin(θ)cos(θ) so 3 · cos(θ) = sin(θ)
Now use the best identity:
1 = sin(θ)2 + cos(θ)2 = (3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1
cos(θ) = ±√1010
In which quadrant are we? Quadrant 3. cos(θ) and sin(θ) are negative.
cos(θ) = −√1010
sin(θ) = 3 cos(θ) =
−√1010
Algebra of the trigonometric functions. 6 / 7
Computing all the trig functions when you know only one.
1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)
2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).
Express tan(θ) = 3 in terms of sin and cos.
3 = sin(θ)cos(θ) so 3 · cos(θ) = sin(θ)
Now use the best identity:
1 = sin(θ)2 + cos(θ)2 = (3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1
cos(θ) = ±√1010
In which quadrant are we? Quadrant 3. cos(θ) and sin(θ) are negative.
cos(θ) = −√1010
sin(θ) = 3 cos(θ) =
−√1010
Algebra of the trigonometric functions. 6 / 7
Computing all the trig functions when you know only one.
1 (I’ll do this one) If tan(θ) = 3 and π < θ < 3π/3 then computecos(θ) and sin(θ)
2 (Groupwork) If sec(θ) = 2 and 0 < θ < π/2, compute cos(θ) andsin(θ).
Express tan(θ) = 3 in terms of sin and cos.
3 = sin(θ)cos(θ) so 3 · cos(θ) = sin(θ)
Now use the best identity:
1 = sin(θ)2 + cos(θ)2 = (3 cos(θ)2) + cos(θ)2 so 10 cos(θ)2 = 1
cos(θ) = ±√1010
In which quadrant are we? Quadrant 3. cos(θ) and sin(θ) are negative.
cos(θ) = −√1010
sin(θ) = 3 cos(θ) = −√1010
Algebra of the trigonometric functions. 6 / 7
The algebra tricks you know and love are still good.
1 (I’ll do this one) Prove that cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ) .
2 (For you) Simplify each side to prove that1
1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.
Start with the right hand side and multiply and divide by the correct thingto make the denominator a difference of squares
cos(θ)
1 + sin(θ)=
cos(θ) · (1 − sin(θ))
(1 + sin(θ)) · (1 − sin(θ))=
cos(θ) · (1 − sin(θ))
1 − sin(θ)2
Use the best identity cos(θ)1+sin(θ) = cos(θ)·(1−sin(θ))
cos(θ)2
Cancellation: cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ)But that’s what we wanted to prove!
Algebra of the trigonometric functions. 7 / 7
The algebra tricks you know and love are still good.
1 (I’ll do this one) Prove that cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ) .
2 (For you) Simplify each side to prove that1
1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.
Start with the right hand side and multiply and divide by the correct thingto make the denominator a difference of squares
cos(θ)
1 + sin(θ)=
cos(θ) · (1 − sin(θ))
(1 + sin(θ)) · (1 − sin(θ))=
cos(θ) · (1 − sin(θ))
1 − sin(θ)2
Use the best identity cos(θ)1+sin(θ) = cos(θ)·(1−sin(θ))
cos(θ)2
Cancellation: cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ)But that’s what we wanted to prove!
Algebra of the trigonometric functions. 7 / 7
The algebra tricks you know and love are still good.
1 (I’ll do this one) Prove that cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ) .
2 (For you) Simplify each side to prove that1
1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.
Start with the right hand side and multiply and divide by the correct thingto make the denominator a difference of squares
cos(θ)
1 + sin(θ)=
cos(θ) · (1 − sin(θ))
(1 + sin(θ)) · (1 − sin(θ))=
cos(θ) · (1 − sin(θ))
1 − sin(θ)2
Use the best identity cos(θ)1+sin(θ) = cos(θ)·(1−sin(θ))
cos(θ)2
Cancellation: cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ)But that’s what we wanted to prove!
Algebra of the trigonometric functions. 7 / 7
The algebra tricks you know and love are still good.
1 (I’ll do this one) Prove that cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ) .
2 (For you) Simplify each side to prove that1
1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.
Start with the right hand side and multiply and divide by the correct thingto make the denominator a difference of squares
cos(θ)
1 + sin(θ)=
cos(θ) · (1 − sin(θ))
(1 + sin(θ)) · (1 − sin(θ))=
cos(θ) · (1 − sin(θ))
1 − sin(θ)2
Use the best identity cos(θ)1+sin(θ) = cos(θ)·(1−sin(θ))
cos(θ)2
Cancellation: cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ)But that’s what we wanted to prove!
Algebra of the trigonometric functions. 7 / 7
The algebra tricks you know and love are still good.
1 (I’ll do this one) Prove that cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ) .
2 (For you) Simplify each side to prove that1
1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.
Start with the right hand side and multiply and divide by the correct thingto make the denominator a difference of squares
cos(θ)
1 + sin(θ)=
cos(θ) · (1 − sin(θ))
(1 + sin(θ)) · (1 − sin(θ))=
cos(θ) · (1 − sin(θ))
1 − sin(θ)2
Use the best identity cos(θ)1+sin(θ) = cos(θ)·(1−sin(θ))
cos(θ)2
Cancellation: cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ)But that’s what we wanted to prove!
Algebra of the trigonometric functions. 7 / 7
The algebra tricks you know and love are still good.
1 (I’ll do this one) Prove that cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ) .
2 (For you) Simplify each side to prove that1
1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.
Start with the right hand side and multiply and divide by the correct thingto make the denominator a difference of squares
cos(θ)
1 + sin(θ)=
cos(θ) · (1 − sin(θ))
(1 + sin(θ)) · (1 − sin(θ))=
cos(θ) · (1 − sin(θ))
1 − sin(θ)2
Use the best identity
cos(θ)1+sin(θ) = cos(θ)·(1−sin(θ))
cos(θ)2
Cancellation: cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ)But that’s what we wanted to prove!
Algebra of the trigonometric functions. 7 / 7
The algebra tricks you know and love are still good.
1 (I’ll do this one) Prove that cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ) .
2 (For you) Simplify each side to prove that1
1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.
Start with the right hand side and multiply and divide by the correct thingto make the denominator a difference of squares
cos(θ)
1 + sin(θ)=
cos(θ) · (1 − sin(θ))
(1 + sin(θ)) · (1 − sin(θ))=
cos(θ) · (1 − sin(θ))
1 − sin(θ)2
Use the best identity cos(θ)1+sin(θ) =
cos(θ)·(1−sin(θ))cos(θ)2
Cancellation: cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ)But that’s what we wanted to prove!
Algebra of the trigonometric functions. 7 / 7
The algebra tricks you know and love are still good.
1 (I’ll do this one) Prove that cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ) .
2 (For you) Simplify each side to prove that1
1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.
Start with the right hand side and multiply and divide by the correct thingto make the denominator a difference of squares
cos(θ)
1 + sin(θ)=
cos(θ) · (1 − sin(θ))
(1 + sin(θ)) · (1 − sin(θ))=
cos(θ) · (1 − sin(θ))
1 − sin(θ)2
Use the best identity cos(θ)1+sin(θ) = cos(θ)·(1−sin(θ))
cos(θ)2
Cancellation: cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ)But that’s what we wanted to prove!
Algebra of the trigonometric functions. 7 / 7
The algebra tricks you know and love are still good.
1 (I’ll do this one) Prove that cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ) .
2 (For you) Simplify each side to prove that1
1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.
Start with the right hand side and multiply and divide by the correct thingto make the denominator a difference of squares
cos(θ)
1 + sin(θ)=
cos(θ) · (1 − sin(θ))
(1 + sin(θ)) · (1 − sin(θ))=
cos(θ) · (1 − sin(θ))
1 − sin(θ)2
Use the best identity cos(θ)1+sin(θ) = cos(θ)·(1−sin(θ))
cos(θ)2
Cancellation:
cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ)But that’s what we wanted to prove!
Algebra of the trigonometric functions. 7 / 7
The algebra tricks you know and love are still good.
1 (I’ll do this one) Prove that cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ) .
2 (For you) Simplify each side to prove that1
1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.
Start with the right hand side and multiply and divide by the correct thingto make the denominator a difference of squares
cos(θ)
1 + sin(θ)=
cos(θ) · (1 − sin(θ))
(1 + sin(θ)) · (1 − sin(θ))=
cos(θ) · (1 − sin(θ))
1 − sin(θ)2
Use the best identity cos(θ)1+sin(θ) = cos(θ)·(1−sin(θ))
cos(θ)2
Cancellation: cos(θ)1+sin(θ) =
1−sin(θ)cos(θ)
But that’s what we wanted to prove!
Algebra of the trigonometric functions. 7 / 7
The algebra tricks you know and love are still good.
1 (I’ll do this one) Prove that cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ) .
2 (For you) Simplify each side to prove that1
1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.
Start with the right hand side and multiply and divide by the correct thingto make the denominator a difference of squares
cos(θ)
1 + sin(θ)=
cos(θ) · (1 − sin(θ))
(1 + sin(θ)) · (1 − sin(θ))=
cos(θ) · (1 − sin(θ))
1 − sin(θ)2
Use the best identity cos(θ)1+sin(θ) = cos(θ)·(1−sin(θ))
cos(θ)2
Cancellation: cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ)
But that’s what we wanted to prove!
Algebra of the trigonometric functions. 7 / 7
The algebra tricks you know and love are still good.
1 (I’ll do this one) Prove that cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ) .
2 (For you) Simplify each side to prove that1
1+cos(θ) + 11−cos(θ) = 2 + 2 · cot(θ)2.
Start with the right hand side and multiply and divide by the correct thingto make the denominator a difference of squares
cos(θ)
1 + sin(θ)=
cos(θ) · (1 − sin(θ))
(1 + sin(θ)) · (1 − sin(θ))=
cos(θ) · (1 − sin(θ))
1 − sin(θ)2
Use the best identity cos(θ)1+sin(θ) = cos(θ)·(1−sin(θ))
cos(θ)2
Cancellation: cos(θ)1+sin(θ) = 1−sin(θ)
cos(θ)But that’s what we wanted to prove!
Algebra of the trigonometric functions. 7 / 7
One of the homework problems
Prove the identity:
tan(θ) + sec(θ) − 1
tan(θ) − sec(θ) + 1=
1 + sin(θ)
cos(θ)
Algebra of the trigonometric functions. 8 / 7