If $ \sin x=\frac{6 \sin 30^{\circ}-8 \cos 60^{\circ}+2 \tan 45^{\circ}}{2\left(\sin ^{2} 30^{\circ}+\cos ^{2} 60^{\circ}\right)} $, then $ x $ is equal to

Given:

\( \sin x=\frac{6 \sin 30^{\circ}-8 \cos 60^{\circ}+2 \tan 45^{\circ}}{2\left(\sin ^{2} 30^{\circ}+\cos ^{2} 60^{\circ}\right)} \)

To do:

We have to find the value of $x$.

Solution:  

We know that,

$\sin 30^{\circ}=\frac{1}{2}$

$\cos 60^{\circ}=\frac{1}{2}$

$\tan 45^{\circ}=1$

RHS $=\frac{6 \sin 30^{\circ}-8 \cos 60^{\circ}+2 \tan 45^{\circ}}{2(\sin ^{2} 30^{\circ}+\cos ^{2} 60^{\circ})}$

$ \begin{array}{l}
=\frac{6\times \frac{1}{2} -8\times \frac{1}{2} +2\times 1}{2\left[\left(\frac{1}{2}\right)^{2} +\left(\frac{1}{2}\right)^{2}\right]}\\
=\frac{3-4+2}{2\left(\frac{1}{4} +\frac{1}{4}\right)}\\
=\frac{5-4}{2\times \frac{2}{4}}\\
=1
\end{array}$

$=sin 90^{\circ}$         [Since $sin 90^{\circ}=1$]

Hence, $x=90^o$.

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Updated on: 10-Oct-2022

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