Question

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Answer 1

Step-by-step explanation:

Ans 2

is a Answer ok

solve it by

Answer 2

$\large\underline{\sf{To\:prove - }}$

$\rm \: {cos}^{2} \bigg(\dfrac{\pi}{8} \bigg) + {cos}^{2} \bigg(\dfrac{3\pi}{8} \bigg) + {cos}^{2} \bigg(\dfrac{5\pi}{8} \bigg) + {cos}^{2} \bigg(\dfrac{7\pi}{8} \bigg) = 2$

$\large\underline{\sf{Solution-}}$

Consider,

$\rm \: {cos}^{2} \bigg(\dfrac{\pi}{8} \bigg) + {cos}^{2} \bigg(\dfrac{3\pi}{8} \bigg) + {cos}^{2} \bigg(\dfrac{5\pi}{8} \bigg) + {cos}^{2} \bigg(\dfrac{7\pi}{8} \bigg)$

We know that,

$\rm :\longmapsto\:\dfrac{\pi}{8} + \dfrac{3\pi}{8} = \dfrac{\pi}{2}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{3\pi}{8} = \dfrac{\pi}{2} - \dfrac{\pi}{8}$

So,

$\rm :\longmapsto\:cos\dfrac{3\pi}{8} = cos \bigg(\dfrac{\pi}{2} - \dfrac{\pi}{8} \bigg)$

$\bf :\longmapsto\:cos\dfrac{3\pi}{8} = sin \dfrac{\pi}{8} - - -(1)$

Again,

$\rm :\longmapsto\:\dfrac{7\pi}{8} + \dfrac{5\pi}{8} = \dfrac{3\pi}{2}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{7\pi}{8} = \dfrac{3\pi}{2} - \dfrac{5\pi}{8}$

So,

$\rm :\longmapsto\:cos\dfrac{7\pi}{8} = cos \bigg(\dfrac{3\pi}{2} - \dfrac{5\pi}{8} \bigg)$

$\bf :\longmapsto\:cos\dfrac{7\pi}{8} = - \: sin \dfrac{5\pi}{8} - - -(2)$

So, given expression now

$\rm :\longmapsto\:{cos}^{2} \bigg(\dfrac{\pi}{8} \bigg) + {cos}^{2} \bigg(\dfrac{3\pi}{8} \bigg) + {cos}^{2} \bigg(\dfrac{5\pi}{8} \bigg) + {cos}^{2} \bigg(\dfrac{7\pi}{8} \bigg)$

can be rewritten as with the help of equation (1) and (2), as

$\rm = \: {cos}^{2} \bigg(\dfrac{\pi}{8} \bigg) + {sin}^{2} \bigg(\dfrac{\pi}{8} \bigg) + {cos}^{2} \bigg(\dfrac{5\pi}{8} \bigg) + {sin}^{2} \bigg(\dfrac{5\pi}{8} \bigg)$

$\rm \: = \: \:1 + 1$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {sin}^{2}x + {cos}^{2}x = 1 \bigg \}}$

$\rm \: = \: \:2$

$\large{\boxed{\bf{Hence, Proved}}}$

Additional Information :-

$\red{ \boxed{ \rm{ \: sin(A + B) = sinAcosB + sinBcosA}}}$

$\red{ \boxed{ \rm{ \: sin(A - B) = sinAcosB - sinBcosA}}}$

$\green{ \boxed{ \rm{ \: cos(A + B) = cosAcosB - sinAsinB}}}$

$\green{ \boxed{ \rm{ \: cos(A - B) = cosAcosB + sinAsinB}}}$

$\blue{ \boxed{ \rm{ \: tan(A + B) = \frac{tanA + tanB}{1 - tanAtanB}}}}$

$\blue{ \boxed{ \rm{ \: tan(A - B) = \frac{tanA - tanB}{1 + tanAtanB}}}}$