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Question Number 126562 by mey3nipaba last updated on 21/Dec/20

P r o v e t h a t 2 \sin \frac{θ + ϕ}{2} \cos \frac{θ - ϕ}{2} = \sin θ + \sin \emptyset

I n e e d h e l p i m m e d i a t e l y p l e a s e

Commented by mr W last updated on 21/Dec/20

θ = \frac{θ + φ}{2} + \frac{θ - φ}{2}

φ = \frac{θ + φ}{2} - \frac{θ - φ}{2}

\sin θ = \sin (\frac{θ + φ}{2} + \frac{θ - φ}{2}) = \sin \frac{θ + φ}{2} \cos \frac{θ - φ}{2} + \cos \frac{θ + φ}{2} \sin \frac{θ - φ}{2}

\sin φ = \sin (\frac{θ + φ}{2} - \frac{θ - φ}{2}) = \sin \frac{θ + φ}{2} \cos \frac{θ - φ}{2} - \cos \frac{θ + φ}{2} \sin \frac{θ - φ}{2}

\Rightarrow \sin θ + \sin φ = 2 \sin \frac{θ + φ}{2} \cos \frac{θ - φ}{2}

Answered by Olaf last updated on 21/Dec/20

Many ways to prove that .

Example :

Let f_{ϕ} (θ) = 2 \sin \frac{θ + ϕ}{2} \cos \frac{θ - ϕ}{2}

and g_{ϕ} (θ) = \sin θ + \sin ϕ

(ϕ is a parameter)

f_{ϕ}^{'} (θ) = 2 [\frac{1}{2} \cos \frac{θ + ϕ}{2} \cos \frac{θ - ϕ}{2} - \frac{1}{2} \sin \frac{θ + ϕ}{2} \cos \frac{θ + ϕ}{2}]

f_{ϕ}^{'} (θ) = \cos (\frac{θ + ϕ}{2} + \frac{θ - ϕ}{2}) = \cos θ = g_{ϕ}^{'} (θ)

\Rightarrow f_{ϕ} (θ) = g_{ϕ} (θ) + C

C = f_{ϕ} (ϕ) - g_{ϕ} (ϕ) = 2 \sin ϕ - 2 \sin ϕ = 0

f_{ϕ} (θ) = g_{ϕ} (θ)

Answered by physicstutes last updated on 21/Dec/20

lets begin with

\sin (A + B) = \sin A \cos B + \sin B \cos A \dots \dots (x)

\sin (A - B) = \sin A \cos B - \sin B \cos A \dots \dots . (y)

let A + B = θ \dots . . (i)

and A - B = \emptyset \dots . . (i i)

(i i) + (i) \Rightarrow A = \frac{θ + \emptyset}{2} similarly B = \frac{θ - \emptyset}{2}

(x) + (y) \Rightarrow \sin \emptyset + \sin θ = 2 \sin A \cos B

\Rightarrow \sin θ + \sin \emptyset = 2 \sin (\frac{θ + \emptyset}{2}) \cos (\frac{θ - \emptyset}{2})