double angle formulas
DESCRIPTION
Double Angle Formulas. Let sinA=1/5 with A in QI. Find sin(2A). Let sinA=1/5 with A in QI. Find sin(2A). Let sinA=1/5 with A in QI. Find sin(2A). Find sin(90 °) using a double angle formula. Simplify 2sin30 °cos30°. Simplify cos 2 15 °–sin 2 15°. - PowerPoint PPT PresentationTRANSCRIPT
Double Angle Formulas
)90sin( ° )452sin( °⋅= )45sin(2?
°=22
2⋅=
)90sin( °22
2?
⋅=
1 2?=
21 ≠AA sin2)2sin( ≠
)2sin( A)sin( AA+=
⋅= Asin Acos + Acos Asin
AAcossin= AAcossin+
AAcossin2=
)2sin( A AAcossin2=
Let sinA=1/5 with A in QI. Find sin(2A).
)2sin( A AAcossin2=
1cossin 22 =+ AA
AA 22 sin1cos −=
AA 2sin1cos −±=
2
51
1cos ⎟⎠⎞
⎜⎝⎛−+=A
251
1cos −=A
251
2525
cos −=A
Let sinA=1/5 with A in QI. Find sin(2A).
)2sin( A AAcossin2= 2
51
1cos ⎟⎠⎞
⎜⎝⎛−+=A
251
1cos −=A
251
2525
cos −=A
2524
cos =A
524
cos =A5
62=
Let sinA=1/5 with A in QI. Find sin(2A).
)2sin( A AAcossin2=
524
cos =A5
62=
2=51⋅
562⋅
)2sin( A25
64=
Find sin(90°) using a double angle formula.
)90sin( °)452sin( °⋅=
)45cos()45sin(2 °⋅°⋅=
⋅=2 ⋅22
22
2)2( 2
=22= 1=
Simplify 2sin30°cos30°.
°° 30cos30sin2
)302sin( °⋅=
)60sin( °=
23=
.8
cos8
sin Simplify ⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛ ππ
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛
8cos
8sin
ππ
( )8
cos8
sin221
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛⋅=
ππ
⎟⎠⎞
⎜⎝⎛ ⋅⋅=
82sin
21 π
⎟⎠⎞
⎜⎝⎛⋅=
4sin
21 π
22
21⋅=
42=
cos(2A). Find)2cos( A
)cos( AA+=
⋅= Acos Acos − ⋅Asin Asin
A2cos= − A2sin
)2cos( A A2cos= − A2sin
Simplify cos215°–sin215°.
°−° 15sin15cos 22
)152cos( °⋅=
)30cos( °=
23=
Derive an alternative form of the identity cos(2A)=cos2A–sin2A.
)2cos( A A2cos= − A2sin)sin1( 2 A−= − A2sin
1= A2sin2−)2cos( A 1= A2sin2−
Derive an alternative form of the identity cos(2A)=cos2A–sin2A.
)2cos( A A2cos= − A2sinA2cos= − )cos1( 2 A−
)2cos( A
A2cos= 1− A2cos+A2cos2= 1−
A2cos2= 1−
AAA 22 sincos)2cos( −=AA 2sin21)2cos( −=1cos2)2cos( 2 −= AA
The identity with its alternative forms.
Suppose that cos x =1/ 10 with xQ IV, find sin(2x) and cos(2x).
)2sin( x xxcossin2=
)2cos( x xx 22 sincos −=)2cos( x x2sin21−=)2cos( x 1cos2 2 −= x ⋅=2
2
101⎟⎠⎞
⎜⎝⎛ 1−
101
2⋅= 1−51= 1−
54−=
Suppose that cos x =1/ 10 with xQ IV, find sin(2x) and cos(2x).
)2sin( x xxcossin2=
1cossin 22 =+ xx
xx 22 cos1sin −=
xx 2cos1sin −±=
2
101
1sin ⎟⎠⎞
⎜⎝⎛−−=x
101
1sin −−=x
109
sin −=x103−=
Suppose that cos x =1/ 10 with xQ IV, find sin(2x) and cos(2x).
)2sin( x xxcossin2=
109
sin −=x103−=
⋅=2103−
101
2)10(
6−=106−=
53−=
)2cos( x54−=
Suppose that cos x =1/ 10 with xQ IV, find sin(2x) and cos(2x).
)2sin( x53−=
)2cos( x54−=
1)2(cos)2(sin 22 =+ xx)2(cos)2(sin 22 xx +
2
53⎟⎠⎞
⎜⎝⎛−= +
2
54⎟⎠⎞
⎜⎝⎛−
259= +
2516
2525=
1=
Simplify 2cos2105°–1.
1105cos2 2 −°)1052cos( °⋅=
)210cos( °=
−= cos−= °−= 30cos
23−=
Prove that(cos x – sin x)(cos x + sin x)=cos(2x)
proof
)sin)(cossin(cos xxxx +−
xx 22 sincos −=
)2cos( x=
.2
2cos1sinthat Prove 2 θθ −=
proof
2)2cos(1 θ−
2)sin(cos1 22 θθ −−=
2sincos1 22 θθ +−=
2sinsin 22 θθ +=
2sin2 2θ=
θ2sin=
.cos3cos4)3cos(that Prove 3 θθθ −=proof
)3cos( θ)2cos( θθ +=
⋅= θ2cos θcos − ⋅θ2sin θsin)1cos2( 2 −= θ θcos − )cossin2( θθ θsin
θ3cos2= θcos− θθ cossin2 2−θ3cos2= θcos− 2− )cos1( 2θ− θcosθ3cos2= θcos− 2− θcos )cos1( 2θ−
.cos3cos4)3cos(that Prove 3 θθθ −=proof
θ3cos2= θcos− 2− θcos )cos1( 2θ−θ3cos2= θcos− θcos2− θ3cos2+θ3cos4= θcos3−
BB
BBB
22
22
seccsc
cscsec)2sec( Prove
−=
proof
BB
BB22
22
seccsc
cscsec
−
BB
BB
22
22
cos
1
sin
1sin
1
cos
1
−
⋅=
BB
BBB
22
22
seccsc
cscsec)2sec( Prove
−=
proof
BB
BB22
22
seccsc
cscsec
−
BB
BB
22
22
cos
1
sin
1sin
1
cos
1
−
⋅=
BB
BBB
22
22
seccsc
cscsec)2sec( Prove
−=
proof
BB
BB22
22
seccsc
cscsec
−
BB
BB
22
22
cos
1
sin
1sin
1
cos
1
−
⋅=
BB
BBB
22
22
seccsc
cscsec)2sec( Prove
−=
proof
BB
BB22
22
seccsc
cscsec
−
BB
BB
22
22
cos
1
sin
1sin
1
cos
1
−
⋅=
BB
BBB
22
22
seccsc
cscsec)2sec( Prove
−=
proof
BB
BB22
22
seccsc
cscsec
−
BB
BB
22
22
cos
1
sin
1sin
1
cos
1
−
⋅=
BB
BBB
22
22
seccsc
cscsec)2sec( Prove
−=
proof
BB
BB22
22
seccsc
cscsec
−
BB
BB
22
22
cos
1
sin
1sin
1
cos
1
−
⋅=
BB
BBB
22
22
seccsc
cscsec)2sec( Prove
−=
proof
BB
BB22
22
seccsc
cscsec
−
BB
BB
22
22
cos
1
sin
1sin
1
cos
1
−
⋅=
BB
BB
22
22
cos
1
sin
1sin
1
cos
1
−
⋅=
BB
BBBB
22
22
22
cossin
sincoscossin
1
−=
BB
BBB
22
22
seccsc
cscsec)2sec( Prove
−=
proof
BB
BB
22
22
cos
1
sin
1sin
1
cos
1
−
⋅=
BB
BBBB
22
22
22
cossin
sincoscossin
1
−=
BB
BBB
22
22
seccsc
cscsec)2sec( Prove
−=
proof
BB
BB
22
22
cos
1
sin
1sin
1
cos
1
−
⋅=
BB
BBBB
22
22
22
cossin
sincoscossin
1
−=
BB
BBB
22
22
seccsc
cscsec)2sec( Prove
−=
proof
BB
BB
22
22
cos
1
sin
1sin
1
cos
1
−
⋅=
BB
BBBB
22
22
22
cossin
sincoscossin
1
−=
BB
BBB
22
22
seccsc
cscsec)2sec( Prove
−=
proof
BB
BBB
22
22
seccsc
cscsec)2sec( Prove
−=
proof
BB
BBBB
22
22
22
cossin
sincoscossin
1
−=
BB 22 sincos
1
−=
BB
BBB
22
22
seccsc
cscsec)2sec( Prove
−=
proof
BB 22 sincos
1
−=
)2cos(1
B=
( )B2sec=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−=
Derive a double angle formula for the tangent function.
)2tan( A)tan( AA+=
AAAA
tantan1tantan
−+=
A
A2tan1
tan2
−= )2tan( A
A
A2tan1
tan2
−=
Given cos θ =1/10 and x QIV, find tan(2θ).
)2tan( θθ
θ2tan1
tan2
−=
θcos101=
rx=
222 ryx =+222 xry −=
22 xry −±=
110 −−=y9−=y 3−=
θtanxy=
13−= 3−=
( ) 22 110 −−=y
Given cos θ =1/10 and x QIV, find tan(2θ).
)2tan( θθ
θ2tan1
tan2
−=
θcos101=
rx=
222 ryx =+222 xry −=
22 xry −±=
( ) 22 110 −−=y110 −−=y
9−=y 3−=θtan
xy=
13−= 3−=
2)3(1
)3(2
−−
−⋅=91
6−−=
Given cos θ =1/10 and x QIV, find tan(2θ).
)2tan( θθ
θ2tan1
tan2
−=
222 ryx =+222 xry −=
22 xry −±=
( ) 22 110 −−=y110 −−=y
9−=y 3−=θtan
xy=
13−= 3−=
2)3(1
)3(2
−−
−⋅=91
6−−=
86
−−=
43=
)(21 −⋅= )tan(21 −⋅= )45tan(21 °−⋅=
⎟⎠⎞
⎜⎝⎛−
⎟⎠⎞
⎜⎝⎛
π
π
83
tan1
83
tan Simplify
2
⎟⎠⎞
⎜⎝⎛−
⎟⎠⎞
⎜⎝⎛
π
π
83
tan1
83
tan
2⎟⎠⎞
⎜⎝⎛−
⎟⎠⎞
⎜⎝⎛
⋅=π
π
83
tan1
83
tan2
21
2⎟⎠⎞
⎜⎝⎛ ⋅⋅= π
83
2tan21
⎟⎠⎞
⎜⎝⎛⋅= π
43
tan21 )135tan(
21 °⋅= ⋅=
21
)1(21 −=
21−=
Given csc t = 7 with t in QII, find sin(2t) and cos(2t).
tsintcsc
1=7
1=
)2sin( t tt cossin2=
1cossin 22 =+ tttt 22 sin1cos −=
tt 2sin1cos −±=
2
71
1cos ⎟⎠⎞
⎜⎝⎛−−=t
71
1cos −−=t
76
cos −=t76−=
Given csc t = 7 with t in QII, find sin(2t) and cos(2t).
tsintcsc
1=7
1=
)2sin( t tt cossin2=
2
71
1cos ⎟⎠⎞
⎜⎝⎛−−=t
76
cos −=t76−=
⋅=2 ⋅7
1⎟⎠⎞
⎜⎝⎛−
76
( )27
62−=
762−=
Given csc t = 7 with t in QII, find sin(2t) and cos(2t).
tsintcsc
1=7
1=
76
cos −=t76−=
)2cos( t tt 22 sincos −=2
76⎟⎠⎞
⎜⎝⎛−= −
2
71⎟⎠⎞
⎜⎝⎛
76= −
71
75=