Question

Sin10° + sin 50° - sin 80° + sin 140°=√2. Sin 25°

Answer 1

Hope this will help you

Answer 2

Answer: Proved.

Step-by-step explanation:  We are given to prove that -

$\sin 10^\circ+\sin 50^\circ -\sin 80^\circ +\sin 140^\circ=\sqrt2~\sin 25^\circ.$.

To arrive at our result, we need to remember the following identities from Trigonometry.

$\sin A+\sin B=2\sin(\dfrac{A+B}{2})\cos(\dfrac{A-B}{2}),\\\\\sin A-\sin B=2\cos(\dfrac{A+B}{2})\sin(\dfrac{A-B}{2}).$

Also, we recall the following trigonometric values-

$\sin 30^\circ=\dfrac{1}{2},~~\sin 45^\circ=\dfrac{1}{\sqrt2}.$

In addition to these, we should remember

$\cos(-A)=\cos A.$

Let us start the proof.

$L.H.S\\\\=\sin 10^\circ+\sin 50^\circ -\sin 80^\circ +\sin 140^\circ\\\\=2\sin(\dfrac{10^\circ+50^\circ}{2})\cos(\dfrac{50^\circ-10^\circ}{2})+2\cos(\dfrac{140^\circ+80^\circ}{2})\sin(\dfrac{140^\circ-80^\circ}{2})\\\\=2\sin 30^\circ\cos 20^\circ+2\cos 110^\circ\sin 30^\circ\\\\=2\times \dfrac{1}{2}\cos 20^\circ+2\times \dfrac{1}{2}\cos 110^\circ\\\\=\cos 20^\circ+\cos 110^\circ\\\\=\cos 20^\circ +\cos(90^\circ+20^\circ)\\\\=\cos 20^\circ-\sin 20^\circ\\\\=\sqrt2(\sin 45^\circ\cos 20^\circ-\cos 45^\circ\sin 20^\circ)\\\\=\sqrt2\sin(45^\circ-20^\circ)\\\\=\sqrt2\sin 25^\circ\\\\R.H.S.$