Question

prove it .please answer fastly and please don't answer if u don't know the answer​

Answer 1

Since $\sf{\sin(90^o-x)=\cos x,}$

$\longrightarrow\sf{\sin78^o=\sin(90^o-12^o)}$

$\longrightarrow\sf{\sin78^o=\cos12^o}$

Since $\sf{\cos(180^o-x)=-\cos x,}$

$\longrightarrow\sf{\cos132^o=\cos(180^o-48^o)}$

$\longrightarrow\sf{\cos132^o=-\cos48^o}$

Therefore,

$\longrightarrow\sf{\sin78^o+\cos132^o=\cos12^o-\cos48^o}$

We know the sum - to - product identity,

$\longrightarrow\sf{\cos A-\cos B=-2\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right)}$

So,

$\longrightarrow\sf{\sin78^o+\cos132^o=-2\sin\left(\dfrac{12^o+48^o}{2}\right)\sin\left(\dfrac{12^o-48^o}{2}\right)}$

$\longrightarrow\sf{\sin78^o+\cos132^o=-2\sin\left(\dfrac{60^o}{2}\right)\sin\left(\dfrac{-36^o}{2}\right)}$

$\longrightarrow\sf{\sin78^o+\cos132^o=-2\sin30^o\sin(-18^o)}$

Since $\sf{\sin(-x)=-\sin x,}$

$\longrightarrow\sf{\sin78^o+\cos132^o=2\sin30^o\sin18^o}$

$\longrightarrow\sf{\sin78^o+\cos132^o=2\times\dfrac{1}{2}\times\dfrac{1}{\sqrt5+1}}$

$\longrightarrow\sf{\sin78^o+\cos132^o=\dfrac{1}{\sqrt5+1}}$

$\longrightarrow\sf{\sin78^o+\cos132^o=\dfrac{\sqrt5-1}{(\sqrt5+1)(\sqrt5-1)}}$

$\longrightarrow\sf{\sin78^o+\cos132^o=\dfrac{\sqrt5-1}{5-1}}$

$\longrightarrow\underline{\underline{\sf{\sin78^o+\cos132^o=\dfrac{\sqrt5-1}{4}}}}$

Hence Proved!