Question

$find \: the \: value \: of \cos( \frac{\pi}{4} + x ) + \cos( \frac{\pi}{4} - x)$
​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore cos(\frac{\pi}{4}+x)+cos(\frac{\pi}{4}-x)=\sqrt{2}cos\:x}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline \bold{To \: find: } \\ \implies cos (\frac{\pi}{4} + x) + cos( \frac{\pi}{4} - x) = ?$

• According to given question :

$\bold{Using \: identity : } \\ \bold{\implies cos (a + b) = cos \: a \times cos \: b - sin \:a \times sin \: b} \\ \\ \bold{\implies cos(a - b) = cos \: a \times cos \: b + sin \:a \times sin \: b} \\ \\ \bold{For \: finding \: value : } \\ \implies cos (\frac{\pi}{4} + x) + cos (\frac{\pi}{4} - x) \\ \\ \implies cos \frac{\pi}{4} \times cos \: x - sin \: \frac{\pi}{4} \times sin \: x + cos \frac{\pi}{4} \times cos \: x + sin \: \frac{\pi}{4} \times sin \: x \\ \\ \bold{cos \frac{\pi}{4} = \frac{1}{ \sqrt{2} } = sin \frac{\pi}{4} } \\ \\ \implies \frac{cos \: x}{ \sqrt{2} } - \cancel{\frac{sin \: x}{ \sqrt{2} }} + \frac{cos \:x}{ \sqrt{2} } + \cancel{\frac{sin \: x}{\sqrt{2} }} \\ \\ \implies \frac{2cos \: x}{ \sqrt{2} } \\ \\ \implies \frac{2cos \:x \sqrt{2} }{2} \\ \\ \bold{\implies \sqrt{2} cos \: x}$

Answer 2

$\boxed{\sf \red{ Answer}}$

$\mathsf{\sqrt{2} cosx}</p><p>$

$\boxed{\sf \red{ Explanation}}$

To Find:

Value of,

$\mathsf{\pink{cos(\dfrac{\pi}{4}+x) + cos (\dfrac{\pi}{4}-x) }}$

Solution:

$\mathsf{ cos(\dfrac{\pi}{4}+x) + cos (\dfrac{\pi}{4}-x) }</p><p>$

W.k.t

$\boxed{\bold\blue{ cos(A+B) = cosA.cosB - sinA.SinB}}$

$\boxed{\bold\blue{ cos(A-B) = cosA.cosB + sinA.SinB}}$

By using the identities

$\mathsf{\implies cos(\dfrac{\pi}{4}+x) + cos (\dfrac{\pi}{4}-x) }</p><p>$

$\mathsf{\implies cos\dfrac{\pi}{4} cosx - sin \dfrac{\pi}{4}. sinx + cos\dfrac{\pi}{4} cosx +sin \dfrac{\pi}{4}. sinx }</p><p>$

$\mathsf{\implies cos\dfrac{\pi}{4} cosx - \cancel{sin \dfrac{\pi}{4}. sinx }+ cos\dfrac{\pi}{4} cosx +\cancel{sin \dfrac{\pi}{4}. sinx} }</p><p>$

$\mathsf{\implies cos\dfrac{\pi}{4} cosx + cos\dfrac{\pi}{4} cosx }</p><p>$

$\mathsf{\implies 2(cos\dfrac{\pi}{4} cosx ) }</p><p>$

$\mathsf{\implies 2( \dfrac{1}{\sqrt{2}} cosx ) }</p><p>$

$\mathsf{\implies{\sqrt{2} \times \sqrt{2} ( \dfrac{1}{\sqrt{2}} cosx ) }}</p><p>$

$\mathsf{\implies{\sqrt{2} \times \cancel{\sqrt{2}} ( \dfrac{1}{\cancel{\sqrt{2}}} cosx ) }}</p><p>$

$\mathsf{\implies \sqrt{2} cosx }</p><p>$

#$\texttt{\red{ Aravind} \: \blue{Reddy}...!}$