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ยง6-3 Double-Angle and Half-Angle Identities
In this section, we will learn following identities:
Double-angle and half-angle identities Double-Angle Identities sin 2x = 2 sin x cos x cos 2x = cos^2 x โ sin^2 x = 1 โ 2 sin^2 x = 2 cos^2 x โ 1 tan 2x = 2 tan^ x 1 โ tan^2 x
= 2 cot^ x cot^2 x โ 1
cot x โ tan x Half-Angle Identities
sin x 2
1 โ cos x 2
cos x 2
1 + cos x 2
tan
x 2 =^ ยฑ
1 โ cos x 1 + cos x =^
sin x 1 + cos x =
1 โ cos x sin x where the sign is determined by the quadrant in which x/2 lies.
Verify the identities
Double-Angle Identities: We start with the sum identities: sin(x + y) = sin x cos y + cos x sin y (1) cos(x + y) = cos x cos y โ sin x sin y (2) tan(x + y) =
tan x + tan y 1 โ tan x tan y (3) Double-angle identity for sine:
sin 2 x = sin(x + x) = sin x cos x + cos x sin x = 2 sin x cos x Double-angle identity for cosine: cos 2 x = cos(x + x) = cos x cos x โ sin x sin x = cos^2 x โ sin^2 x First double โ angle identity for cosine = (1 โ sin^2 x) โ sin^2 x = 1 โ 2 sin^2 x Second double โ angle identity for cosine = cos^2 x โ (1 โ cos^2 x) = 2 cos^2 x โ 1 Third double โ angle identity for cosine 1
Double-angle identity for tangent:
tan 2 x = tan(x + x) = tan^ x^ + tan^ x 1 โ (tan x) (tan x) = 2 tan^ x 1 โ tan^2 x
First double โ angle identity for tangent
=
(^2) cot^1 x 1 โ ( (^) cot^1 x )^2
=
(^2) cot^1 x cot^2 x cot^2 x(1 โ ( (^) cot^1 x )^2 )
=
2 cot x cot^2 x โ 1 Second double^ โ^ angle identity for tangent =
2 cot x tan x (cot^2 x โ 1) tan x =
cot x โ tan x Third double^ โ^ angle identity for tangent
Half-Angle Identities
Half-angle identities are simply double-angle identities stated in an alter- nate form. We start from the double-angle identity for cosine:
cos 2 m = 1 โ 2 sin^2 m = 2 cos^2 m โ 1
Replace m with x/2 and solve for sin(x/2) or cos(x/2):
cos x = 1 โ 2 sin^2 x 2 sin^2
x 2 =^
1 โ cos x 2 sin x 2
1 โ cos x 2
Half โ angleidentityforsine
where the sign is determined by the quadrant in which x/2 lies.
cos x = 2 cos^2 x 2
cos^2 x 2
= 1 + cos^ x 2 cos x 2
1 + cos x 2
Half โ angleidentityforcosine
where the sign is determined by the quadrant in which x/2 lies.
Solution. (A).
1 โ tan^2 x 1 + tan^2 x =^
1 โ sin
(^2) x cos^2 x 1 + sin cos^22 xx
cos^2 x
1 โ sin
(^2) x cos^2 x
cos^2 x
1 + (^) cossin^22 xx
= cos
(^2) x โ sin (^2) x cos^2 x + sin^2 x = cos^2 โ sin^2 x = cos 2x (B). Start from the left side, we have
sin x 2
1 โ cos x 2 sin^2 x 2
= 1 โ^ cos^ x 2
Start from the right side, we have
tan x โ sin x 2 tan x =
sin x cos x โ^ sin^ x (^2) cossin^ x x
=
cos x ( sin cos^ xx โ sin x) (cos x) 2 (^) cossin^ xx
=
sin x โ cos x sin x 2 sin x = sin^ x2 sin(1^ โ^ cosx^ x)
= 1 โ^ cos^ x 2 Hence, we have
sin^2
x 2 =
1 โ cos x 2 =
tan x โ sin x 2 tan x. ยง
Example Find exact values (A) Find the exact values of sin 2x and cos 2x if tan x = โ 34 and x is a quadrant IV angle. (B) Compute the exact values of sin 165o^ using a half-angle identity. (C) Find the exact values of cos(x/2) and sin(x/2) is sin x = โ 35 , ฯ < x < 3 ฯ/2.
Solution. (A). From the reference triangle for x (Figure 2), we have
r =
(โ3)^2 + 4^2 = 5
and sin x = โ 3 5
cos x =^4 5 Now use double-angle identities for sine and cosine:
6
r โ^3
Figure 1
sin 2x = 2 sin x cos x = 2(โ 3 5
)(^4
cos 2x = 2 cos^2 x โ 1 = 2(
(B).
sin 165o^ = sin
330 o 2 =
1 โ cos 330o 2
=
2 ร 2
(C). From the reference triangle for x (Figure ??), we have a = โ
52 โ (โ3)^2 = โ 4
and cos x = โ
If ฯ < x < 3 ฯ/2, then
