Download 492 18–3 Functions of Double Angles Sine of Twice an Angle

492
Chapter 18
◆
Trigonometric Identities and Equations
19. sin cos(30 ) sin(60 )
20. cos(x y) cos(x y) 2 cos x cos y
21. cos x sin p x q p sin x q
6
6
22. tan(360 ) tan 23. cos(60 ) sin(330 ) 0
sin 4x cos 4x
24. sin 5x sec x csc x sin x
cos x
sin(x y) tan x tan y
25. sin(x y) tan x tan y
26. cos(x 60) cos(60 x) cos x
cos(x y)
27. tan y cot x
sin x cos y
28. cotp x q tanp x q 0
4
4
cos sin 29. tan( 45) cos sin cot cot 1
30. cot( )
cot cot 1 tan x
31. tan p x q
1 tan x
4
Express as a single sine function.
32. y 47.2 sin t 64.9 cos t
33. y 8470 sin t 7360 cos t
34. y 1.83 sin t 2.74 cos t
35. y 84.2 sin t 74.2 cos t
18–3
Functions of Double Angles
Sine of 2
An equation for the sine of 2 is easily derived by setting in Eq. 167.
sin( ) sin cos cos sin which is rewritten in the following form:
Sine of Twice an Angle
sin 2 2 sin cos 170
Cosine of 2
Similarly, setting in Eq. 168, we have
cos( ) cos cos sin sin which is also given as follows:
Cosine of Twice an Angle
cos 2 cos2 sin2 171a
Section 18–3
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493
Functions of Double Angles
There are two alternative forms for Eq. 171a. From Eq. 164,
cos2 1 sin2 Substituting into Eq. 171a yields
cos 2 1 sin2 sin2 cos 2 1 sin2 Cosine of Twice an Angle
171b
We can similarly use Eq. 164 to eliminate the sin2 term from Eq. 171a.
cos 2 cos2 (1 cos2 )
Thus:
cos 2 2 cos2 1
Cosine of Twice an Angle
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171c
Example 17: Express sin 3x in terms of the single angle x.
Solution: We can consider 3x to be x 2x, and, using Eq. 167, we have
sin 3x sin(x 2x) sin x cos 2x cos x sin 2x
We now replace cos 2x by cos2 x sin2 x, and sin 2x by 2 sin x cos x.
sin 3x sin x(cos2 x sin2 x) cos x(2 sin x cos x)
sin x cos2 x sin3 x 2 sin x cos2 x
3 sin x cos2 x sin3 x
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Example 18: Prove that
cos 2A sin(A B) 0
where A and B are the two acute angles of a right triangle.
Solution: By Eqs. 171a and 167,
cos 2A sin(A B) cos2 A sin2 A sin A cos B cos A sin B
But, by the cofunctions of Eq. 154,
cos B sin A
and
sin B cos A
So
cos 2A sin(A B) cos2 A sin2 A sin A sin A cos A cos A
cos2 A sin2 A sin2 A cos2 A 0
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Example 19: Simplify the expression
sin 2x
1 cos 2x
Solution: By Eqs. 170 and 171c,
sin 2x
1 cos 2x
2 sin x cos x
1 2 cos2 x 1
2 sin x cos x
2 cos2 x
sin x
tan x
cos x
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494
Chapter 18
◆
Trigonometric Identities and Equations
Tangent of 2␣
Setting in Eq. 169 gives
tan tan tan ( ) 1 tan tan Therefore:
Tangent of
Twice an Angle
◆◆◆
2 tan x
tan 2 = 1 tan2 x
172
Example 20: Prove that
2 cot x
csc2 x 2
tan 2x
Solution: By Eq. 172,
2 tan tan 2x 1 tan2 and, by Eq. 152c,
2
cot x
tan 2x 1
1 cot2 x
Multiply numerator and denominator by cot2 x.
2 cot x
tan 2x cot2 x 1
Then, by Eq. 166,
2 cot x
2 cot x
tan 2x 2
(csc x 1) 1 csc2 x 2
Common
Error
Exercise 3
◆
The sine of twice an angle is not twice the sine of the angle.
Nor is the cosine (or tangent) of twice an angle equal to
twice the cosine (or tangent) of that angle.
Remember to use the formulas from this section for all
of the trig functions of double angles.
Functions of Double Angles
Simplify.
1. 2 sin2 x cos 2x
2. 2 sin 2 cos 2
2 tan x
3. 1 tan2 x
2 sec2 x
4. sec2 x
sin 6x
cos 6x
5. sin 2x
cos 2x
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Section 18–4
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495
Functions of Half-Angles
Prove.
2 tan 6. tan 2
1 tan2 1 tan2x
7. cos 2x
1 tan2x
8. 2 csc 2 tan cot 9. 2 cot 2x cot x tan x
1 cot2 x
10. sec 2x cot2 x 1
1 tan x
cos 2x
11. tan x 1 sin 2x 1
sin 2 sin 12. tan 1 cos cos 2
2 cos 2x
13. 1 cot x sin 2x 2 sin2 x
14. 4 cos3 x 3 cos x cos 3x
sin 2 1
15. sin cos cos sin cot2 x 1
16. cot 2x
2 cot x
If A and B are the two acute angles in a right triangle, show that:
17. sin 2A sin 2B.
18. tan(A B) cot 2A.
19. sin 2A cos(A B).
18–4 Functions of Half-Angles
Sine of ␣冒2
The double-angle formulas derived in Sec. 18–3 can also be regarded as half-angle formulas,
because if one angle is double another, the second angle must be half the first.
Starting with Eq. 171b, we obtain
cos 2 1 2 sin2 We solve for sin .
2 sin2 1 cos 2
sin 兹
1 cos 2
2
For emphasis, we replace by 冒2.
Sine of Half
an Angle
sin 2
兹
1 cos 2
173
The sign in Eq. 173 (and in Eqs. 174 and 175c as well) is to be read as plus or minus, but not
both. This sign is different from the sign in the quadratic formula, for example, where we
took both the positive and the negative values.
The reason for this difference is clear from Fig. 18–4, which shows a graph of sin 冒2 and
a graph of 兹 (1 cos )冒2 . Note that the two curves are the same only when sin 冒2 is
positive. When sin 冒2 is negative, it is necessary to use the negative of 兹 (1 cos )冒2 .
This occurs when 冒2 is in the third or fourth quadrant. Thus we choose the plus or the minus
according to the quadrant in which 冒2 is located.