Download 495 18–4 Functions of Half-Angles Sine of Half an Angle

Section 18–4
◆
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.
496
Chapter 18
◆
Trigonometric Identities and Equations
y
1
y = sin α
2
0.5
0
180˚
360˚
540˚
720˚
α (deg)
y
0.5
0
y = + 1 − cos α
2
α (deg)
FIGURE 18–4
◆◆◆
Example 21: Given that cos 200 0.9397, find sin 100.
Solution: From Eq. 173,
1 (0.9397)
200
1 cos 200
sin 0.9848
2
2
2
Since 100 is in the second quadrant, the sine is positive, so we drop the minus sign and get
兹
兹
sin 100 0.9848
◆◆◆
Cosine of 冒2
Similarly, starting with Eq. 171c, we have
cos 2 2 cos2 1
Then we solve for cos .
2 cos2 1 cos 2
cos 兹
1 cos 2
2
We next replace by 冒2 and obtain the following:
Here also, we choose the plus
or the minus according to the
quadrant in which /2 is located.
Cosine of
Half an Angle
cos 2
兹
1 cos 2
174
Section 18–4
◆
497
Functions of Half-Angles
Tangent of 冒2
There are three formulas for the tangent of a half-angle; we will show the derivation of each in
turn. First, from Eq. 162,
sin 2
tan 2 cos 2
Multiply numerator and denominator by 2 sin(冒2).
2 sin2 2
tan 2 2 sin cos 2
2
Then, by Eqs. 171b and 170, we have the following:
1 cos tan 2
sin Tangent of
Half an Angle
175a
Another form of this identity is obtained by multiplying numerator and denominator by
1 cos .
1 cos 1 cos tan 2
1 cos sin 1 cos2 sin2 sin (1 cos ) sin (1 cos )
also written in the following form:
Tangent of
Half an Angle
sin tan 2 1 cos 175b
We can obtain a third formula for the tangent by dividing Eq. 173 by Eq. 174.
1 cos sin 2
2
tan 2 cos 1 cos 2
2
兹
兹
Tangent of
Half an Angle
◆◆◆
tan 2
兹
1 cos 1 cos Example 22: Prove that
1 sin2 3 cos 2
3 cos 1 cos2 2
175c
498
Chapter 18
◆
Trigonometric Identities and Equations
Solution: By Eqs. 173 and 174,
1 cos 1 sin2 1 2
2
1
cos 1 cos2 1 2
2
Multiply numerator and denominator by 2.
2 1 cos 2 1 cos 3 cos 3 cos Exercise 4
◆
◆◆◆
Functions of Half-Angles
Prove each identity.
1. 2 sin2 cos 1
2
x
x
2. 4 cos2 sin2 1 cos2 x
2
2
x sec x 1
3. 2 cos2 sec x
2
4. cot csc cot 2
sin 5. sin sec cos 1
2
2
cos2 cos 2
6. 1
sin2 2
7. sec tan tan 1
2
x
x 2
8. sin x 1 p cos sin q
2
2
9. 2 sin sin 2 4 sin cos2 2
1 tan2 2
11. cos 1 tan2 2
1 cos x
x
10. tan sin x
2
In right triangle ABC, show the following:
a
A
12. tan 2 bc
18–5
A
13. sin 2
cb
兹 2c
Trigonometric Equations
Solving Trigonometric Equations
One use for the trigonometric identities we have just studied is in the solution of trigonometric
equations.
A trigonometric equation will usually have an infinite number of roots. However, it is
customary to list only nonnegative values less than 360 that satisfy the equation.