10-3 Double Angles and Half Angles

To find the what the double of sin or cos or whatever of an angle or half of sin or cos or whatever of an angle, you DO NOT multiply by two or divide by two. The sin(2a) or sin(a/2) is deceiving.

The formulas you'll need to know are:
sin(2a) = 2sin(a)cos(a)

cos(2a) = cos²(a)-sin²(a)

or = 1-2sin²(a)

or = 2cos²

tan(2a) = 2tan²(a)/1-tan²(a)

sin(a/2) = ±√(1-cos[a]/2)

cos(a/2) = ± √(1+cos[a]/2)

tan(a/2) = ± √(1-cos[a]/1+cos[a])

or = sin(a)/1+cos(a)

or = 1-cos(a)/sin(a)

Example:

If cosA = 1/3, find:

a. sin2A

b. cos2A

c. tan2A

d. sinA/2

e. cosA/2

f. tanA/2

First, we need to make a triangle to find the other trig functions.

cosA = 1/3, so 1 is x and 3 is the radius. We use the Pythagorean theorem to find y.
1²+y² = 3²

1+y² = 9

y² = 8

y = √(8) which can be simplified to 2√(2)

cosA= 1/3

sinA = 2√(2)/3

tanA = 2√(2)

a.sin2A = 2sinAcosA

= 2[2√(2)/3][1/3]

=4√(2)/9

b.cos2A = cos²A-sin²A
= (1/3)² - (2√(2)/3)²

= (1/9)-(8/9)

= -7/9

You could have used any of the three formulas for cos2A.

c. tan2A = 2tanA/1-tan²A

= 2[2√(2)]/1-[2√(2)]²

= 4√(2)/(1-8)

= -4√(2)/7

d.sinA/2 = ± √(1-cos[A]/2)

= ± √(1-(1/3)/2)

= ± √([2/3]/2)

= ± √(2/6)

= ± √(1/3)

e.cosA/2 = ± √(1+cosA/2)

= ± √(1+[1/3]/2)

= ± √([4/3]/2)

= ± √(4/6)

= ± √(2/3)

f.tanA/2 = sin(A)/1+cos(A)

= [2√(2)/3]/1+[1/3]

= [2√(2)/3]/[4/3]

= 6√(2)/12

=√(2)/2

Again, you could have used any of the three formulas.