Search any question & find its solution
Question:
Answered & Verified by Expert
$\sin \left(\frac{\pi}{3}+x\right)-\cos \left(\frac{\pi}{6}+x\right)=$
Options:
Solution:
2887 Upvotes
Verified Answer
The correct answer is:
$\sin x$
$\sin \left(\frac{\pi}{3}+x\right)-\cos \left(\frac{\pi}{6}+x\right)$
$=\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x-\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x$
$=\sin x$
$=\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x-\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x$
$=\sin x$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.