3.Trigonometrical Ratios, Functions and Identities
medium

निम्नलिखित को सिद्ध कीजिए

$\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$

Option A
Option B
Option C
Option D

Solution

$\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)$

$=\frac{1}{2}\left[2 \cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)\right]+\frac{1}{2}\left[-2 \sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)\right]$

$=\frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}+\cos \left\{\left(\frac{\pi}{4}-x\right)-\left(\frac{\pi}{4}-y\right)\right\}\right]$

$+\frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}-\cos \left\{\frac{\pi}{4}-x\right\}-\left(\frac{\pi}{4}-y\right)\right]$

$\left[ \begin{gathered}
  \because 2\cos A\cos B = \cos (A + B) + \cos (A – B) \hfill \\
   – 2\sin A\sin B = \cos (A + B) – \cos (A – B) \hfill \\ 
\end{gathered}  \right]$

$=2 \times \frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}\right]$

$=\cos \left[\frac{\pi}{4}-(x+y)\right]$

$=\sin (x+y)$

$= R . H.S$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.