Question

Show that cos θ/2 (1-cosθ) = sin θ/2 sinθ

can anyone write the steps please? many Thanks! ​

Answer 1

`\cos (\theta)/(2)(1-\cos \theta)=\sin (\theta)/(2) \sin \theta`

Explanation:

`\textsf{To show that}\;\;\cos (\theta)/(2)(1-\cos \theta)=\sin (\theta)/(2) \sin \theta:`

`\textsf{Let}\;\;u=(\theta)/(2) \implies 2u=\theta`

Therefore:

`\implies \cos (\theta)/(2)(1-\cos \theta)=\cos u(1-\cos 2u)`

Cosine Double Angle Identities

cos(A±B) = cosA cosB ∓ sinA sinB

cos(2θ) = cos²θ - sin²θ

cos(2θ) = 2cos²θ - 1

cos(2θ) = 1 - 2sin²θ

Use the cos double angle identity cos(2θ) = 1 - 2sin²θ to rewrite cos(2u) in terms of sin:

`\begin{aligned}\implies \cos (\theta)/(2)(1-\cos \theta)&=\cos u(1-\cos 2u)\\&=\cos u(1-(1-2 \sin^2 u))\end{aligned}`

Simplify:

`\begin{aligned}\implies \cos (\theta)/(2)(1- \cos \theta)&=\cos u(1- \cos 2u)\\&=\cos u(1-(1-2 \sin^2 u))\\&=\cos u(1-1+2 \sin^2 u)\\&= \cos u(2 \sin^2 u)\\&=2 \sin u \cos u \sin u\end{aligned}`

Sine Double Angle Identities

sin(A±B) = sinAcosB ± cosAsinB

sin(2θ) = 2sinθcosθ

Using the sine double angle identity to rewrite 2sin(u)cos(u) as sin(2u):

`\begin{aligned}\implies \cos (\theta)/(2)(1-\cos \theta)&=\cos u(1- \cos 2u)\\&=\cos u(1-(1-2 \sin^2 u))\\&= \cos u(1-1+2 \sin^2 u)\\&= \cos u(2 \sin^2 u)\\&=2 \sin u \cos u \sin u\\&= \sin(2u) \sin u\end{aligned}`

Finally, substitute u = θ/2 back in:

`\begin{aligned}\implies \cos (\theta)/(2)(1-\cos \theta)&=\cos u(1-\cos 2u)\\&=\cos u(1-(1-2 \sin^2 u))\\&=\cos u(1-1+2 \sin^2 u)\\&= \cos u(2\sin^2 u)\\&=2\sin u\cos u \sin u\\&=\sin(2u) \sin u\\&=\sin\left(2 \cdot (\theta)/(2)\right) \sin (\theta)/(2)\\&=\sin \theta \sin (\theta)/(2)\\&=\sin (\theta)/(2) \sin \theta \end{aligned}`

If you don't want to substitute u = θ/2, the full calculation using the same double angle identities is:

`\begin{aligned}\implies\cos(\theta)/(2)(1-\cos\theta)&=\cos(\theta)/(2)\left(1-\left(1-2\sin^2(\theta)/(2)\right)\right)\\\\&=\cos(\theta)/(2)\left(1-1+2\sin^2(\theta)/(2)\right)\\\\&=\cos(\theta)/(2)\left(2\sin^2(\theta)/(2)\right)\\\\&=2\sin (\theta)/(2)\cos (\theta)/(2) \sin (\theta)/(2)\\\\&=\sin\left(2\cdot(\theta)/(2)\right)\sin(\theta)/(2)\\\\&=\sin\theta\sin(\theta)/(2)\\\\&=\sin(\theta)/(2)\sin\theta \end{aligned}`