Question

sin78-sin18+cos132=0​

Answer 1

SOLUTION:-

Take L.H.S.

sin78° - sin18° + cos132°

$2cos( \frac{78 \degree + 18 \degree}{2} ).sin( \frac{78 \degree- 18 \degree}{2} ) + cos132 \degree \\ \\ = > 2cos( \frac{96}{2} ).sin( \frac{70 \degree - 18 \degree}{2} ) + cos132 \degree \\ \\ = > 2cos48 \degree.sin30 \degree + cos132 \degree \\ \\ = > 2cos48 \degree.( \frac{1}{2} ) + cos132 \degree \\ \\ = > cos48 \degree + cos132 \degree \\ \\ = > 2cos( \frac{48 \degree + 132 \degree}{2} ).cos( \frac{48 \degree - 132 \degree}{2} ) \\ \\ = > 2.cos( \frac{180 \degree}{2} ).cos( \frac{ - 84}{2} ) \\ \\ = > 2.cos90 \degree.cos( - 42 \degree) \\ \\ = > 2.(0).cos( - 42) \\ \\ = > 0.cos ( - 42) \\ \\ = > 0 \: \: \: \: \: \: \: \: \: [R.H.S.]$

Proved.

Hope it helps ☺️

Answer 2

To prove : $\sin 78^\circ -\sin 18^\circ +\cos 132^\circ=0$

Step-by-step explanation:

Now as according to question

Consider L.H.S.

$\sin 78^\circ -\sin 18^\circ +\cos 132^\circ$

As we know

$\sin x - \sin y = 2 \cos \dfrac{x+y}{2} \sin\dfrac{x-y}{2}$

we have

$\sin 78^\circ -\sin 18^\circ +\cos 132^\circ\\\\\Rightarrow 2 \cos \dfrac{78^\circ+18^\circ}{2} \sin \dfrac{78^\circ-18^\circ}{2}+ \cos 132^\circ\\\\\Rightarrow 2 \cos 48^\circ \sin 30^\circ+ \cos 132^\circ\\\\\Rightarrow 2 \cos 48^\circ\times \dfrac{1}{2} + \cos 132^\circ\\\\\Rightarrow \cos48^\circ +\cos132^\circ$

Also

$\cos x+ \cos y = 2 \cos \dfrac{x+y}{2} \cos \dfrac{x-y}{2}$

So we get

$\Rightarrow 2 \cos \dfrac{48+132}{2} \cos \dfrac{48-132}{2}\\\\\Rightarrow 2 \cos 90^\circ \cos (-84^\circ)\\\\\Rightarrow 2\times 0 \times \cos (-84^\circ) =0$

which is equal to R.H.S.

Hence proved the required result i.e. $\sin 78^\circ -\sin 18^\circ +\cos 132^\circ=0$