double-angle and half-angle identities section 5.3
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Double-Angle and Half-Angle Identities
Section 5.3
Objectives• Apply the half-angle and/or double
angle formula to simplify an expression or evaluate an angle.
• Apply a power reducing formula to simplify an expression.
Double-Angle Identities
)cos()sin(2)2sin( aaa
)(sin)(cos)2cos( 22 aaa
Half-Angle Identities
2cos1
2sin
aa
2cos1
2cos
aa
Power-Reducing Identities
2
)2cos(1sin2 a
a
2
)2cos(1cos2 a
a
Use a half-angle identity to find the exact
value of
2cos1
2sin
aa
12
sin
We will use the half-angle formula for sine
We need to find out what a is in order to use this formula.
6122
212122
aa
a
continued on next slide
Use a half-angle identity to find the exact
value of
2
23
1
12sin
26
cos1
26sin
12
sin
We now replace a with in the formula to get6
continued on next slide
Use a half-angle identity to find the exact
value of
232
12sin
432
12sin
2
232
12sin
12
sin
continued on next slide
Now all that we have left to do is determine if the answer should be positive or negative.
Use a half-angle identity to find the exact
value of
232
12sin
12
sin
continued on next slide
We determine which to use based on what quadrant the original angle is in. In our case, we need to know what quadrant π/12 is in. This angle fall in quadrant I. Since the sine values in quadrant I are positive, we keep the positive answer.
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
)2cos( t
continued on next slide
For this we will need the double angle formula for cosine
)(sin)(cos)2cos( 22 aaa
In order to use this formula, we will need the cos(t) and sin(t). We can use either the Pythagorean identity or right triangles to find sin(t).
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
continued on next slide
Triangle for angle t
b
7
t
9
32
positiveislength
32
32
8149
97
2
2
222
b
b
b
b
b
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
932
)sin( t
continued on next slide
Triangle for angle t
7
t
932
Since angle t is in quadrant III, the sine value is negative.
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
)2cos( t
continued on next slide
Now we fill in the values for sine and cosine.
8117
)2cos(
8132
8149
)2cos(
932
97
)2cos(22
t
t
t
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
)2sin( t
continued on next slide
For this we will need the double angle formula for cosine
)cos()sin(2)2sin( aaa We know the values of both sin(t) and cos(t) since we found them for the first part of the problem.
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
)2sin( t
continued on next slide
813214
)2sin(
97
932
2)2sin(
t
t
Now we fill in the values for sine and cosine.
813214
)2sin( t
At this point, we can ask the question “What quadrant is the angle 2t in?”
This question can be answered by looking at the signs of the sin(2t) and cos(2t).
is positive
8117
)2cos( t is positive
and
The only quadrant where both the sine value and cosine value of an angle are positive is quadrant I.
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
2
sint
continued on next slide
2cos1
2sin
tt
We will use the half-angle formula for sine
Since we know the value of cos(t), we can just plug that into the formula.
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
2
sint
continued on next slide
297
1
2sin
2cos1
2sin
t
tt
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
2
sint
continued on next slide
2916
2sin
297
99
2sin
t
t
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
2
sint
continued on next slide
1816
2sin
t
Now all that we have left to do is determine if the answer should be positive or negative. In order to do this, we need to know which quadrant the angle t/2 falls in.
To do this we will need to use the information we have about the angle t.
What we need is information about t/2. If we divide each piece of the inequality, we will get t/2 in the middle of the inequality and bounds for the angle on the left and right sides.
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
23 t
2
sint
continued on next slide
1816
2sin
t
The information that we have about the angle t is
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
2
sint
continued on next slide
1816
2sin
t
43
22
22
3
22
23
t
t
t
Thus we see that the angle t/2 is in quadrant II.
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
2
sint
continued on next slide
1816
2sin
t
Since the angle t/2 is in quadrant II, the sine value must be positive.
1816
2sin
t
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
2
cost
continued on next slide
2cos1
2cos
tt
We will use the half-angle formula for sine
Since we know the value of cos(t), we can just plug that into the formula.
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
2
cost
continued on next slide
297
1
2cos
2cos1
2cos
t
tt
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
2
cost
continued on next slide
292
2cos
297
99
2cos
t
t
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
2
cost
continued on next slide
31
91
2cos
t
Now all that we have left to do is determine if the answer should be positive or negative. In order to do this, we need to know which quadrant the angle t/2 falls in.
To do this we will need to use the information we have about the angle t.
What we need is information about t/2. If we divide each piece of the inequality, we will get t/2 in the middle of the inequality and bounds for the angle on the left and right sides.
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
23 t
2
cost
continued on next slide
31
2cos
t
The information that we have about the angle t is
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
2
cost
continued on next slide
31
2cos
t
43
22
22
3
22
23
t
t
t
Thus we see that the angle t/2 is in quadrant II.
If
find the values of the following trigonometric functions.
23
where97
)cos( tt
2
cost
31
2cos
t
Since the angle t/2 is in quadrant II, the cosine value must be negative.
31
2cos
t
Use the power-reducing formula to simplify the expression
)7(sin)7(cos 44 xx
2
)2cos(1cos2 a
a
2
)2cos(1sin2 a
a
continued on next slide
We need to use the power-reducing identity for the cosine and sine functions to do this problem.
In our problem the angle a in the formula will be 7x in our problem. We also need to rewrite our problem.
2222 )7(sin)7(cos xx
Use the power-reducing formula to simplify the expression
)7(sin)7(cos 44 xx
continued on next slide
Now we just apply the identity to get:
4
)14(cos)14cos(21)14(cos)14cos(214
)14(cos)14cos(214
)14(cos)14cos(21
2)14cos(1
2)14cos(1
)7(sin)7(cos
22
22
22
2222
xxxx
xxxx
xx
xx
Use the power-reducing formula to simplify the expression
)7(sin)7(cos 44 xx
continued on next slide
)14cos(4
)14cos(44
)14(cos)14cos(21)14(cos)14cos(214
)14(cos)14cos(21)14(cos)14cos(21
22
22
x
x
xxxx
xxxx
This is much simpler than the original expression and the power (exponent) is clearly reduced.
Use the power-reducing formula to simplify the expression
)7(sin)7(cos)7(sin)7(cos 2222 xxxx
Is there another way to simplify this without using a power-reducing formula?
continued on next slide
The answer to this question is yes. The original expression is the difference of two squares and can be factoring into
)7(sin)7(cos 44 xx
Now you should notice that the expression in the second set of square brackets is the Pythagorean identity and thus is equal to 1.
Use the power-reducing formula to simplify the expression
1*)7(sin)7(cos 22 xx
)7(sin)7(cos 44 xx
)(sin)(cos)2cos( 22 aaa
Is there another way to simplify this without using a power-reducing formula?
Now you should notice that what is left is the right side of the double angle identity for cosine where the angle a is 7x.
This will allow us to rewrite the expression as
)14cos())7(2cos( xx