# can someone help me with my math homework?

Prove, using the trigonometric identities,

that

cos((1/2)*θ)= sqrt((cos(θ )+1)/2)

for any 0 ≤ θ ≤ π. Remember to separate the left and right sides in

your proof. Start from the RIGHT side! Moreover: θ =(1/2)*θ+ (1/2)*θ

### 3 Answers

- MathmomLv 72 years agoFavorite Answer
Angle sum identity:

cos(x+y) = cos x cos y − sin x sin y

cosθ = cos(θ/2 + θ/2)

........ = cos(θ/2) cos(θ/2) − sin(θ/2) sin(θ/2)

........ = cos²(θ/2) − sin²(θ/2)

........ = cos²(θ/2) − (1 − cos²(θ/2))

........ = 2cos²(θ/2)−1

RHS = √[(cosθ + 1) / 2]

= √[((2cos²(θ/2)−1) + 1) / 2]

= √(2cos²(θ/2) / 2)

= √(cos²(θ/2))

= |cos(θ/2)|

= cos(θ/2) ..... since cos(θ/2) > 0 for 0 ≤ θ ≤ π

= LHS

- IndikosLv 72 years ago
We have cos(a+b) = cos(a)cos(b) - sin(a)sin(b)

set a = b = x

then cos(2x)

= cos(x + x)

= cos(x)cos(x) - sin(x)sin(x)

= cos^2(x) - sin^2(x)

= cos^2(x) - ( 1 - cos^2(x)

= cos^2(x) - 1 + cos^2(x)

= 2cos^2(x) - 1

so cos(2x) = 2cos^2(x) - 1

so 2cos^2(x) = cos(2x) + 1

cos^2(x) = (cos(2x)+1)/2

or cos(x) = sqrt( (cos(2x)+1)/2)

Now set x = θ/2

so cos(θ/2) = sqrt( (cos(2*θ/2)+1)/2)

= sqrt( (cos(θ)+1)/2)