Question

sin^2(180/4-theta)=1/2(1-Sin2theta)​

Answer 1

Given Question :-

Prove that

$\rm \: {sin}^{2}\bigg(\dfrac{180\degree }{4} - \theta \bigg) = \dfrac{1}{2}\bigg(1 - sin2\theta \bigg)$

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

Consider LHS

$\rm \: {sin}^{2}\bigg(\dfrac{180\degree }{4} - \theta \bigg)$

can be rewritten as

$\rm \: = \: {sin}^{2}\bigg(45\degree - \theta \bigg)$

$\rm \: = \: \dfrac{1}{2} \times 2 {sin}^{2}\bigg(45\degree - \theta \bigg)$

We know

$\boxed{ \rm{ \:cos2x = 1 - {2sin}^{2}x \: }}$

So, using this result, we have

$\rm \: = \: \dfrac{1}{2} \bigg(1 - cos2(45\degree - \theta )\bigg)$

$\rm \: = \: \dfrac{1}{2} \bigg(1 - cos(90\degree - 2\theta )\bigg)$

$\rm \: = \: \dfrac{1}{2} \bigg(1 - sin2\theta \bigg)$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: {sin}^{2}\bigg(\dfrac{180\degree }{4} - \theta \bigg) = \dfrac{1}{2}\bigg(1 - sin2\theta \bigg) \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered} {\boxed{ \begin{array}{c} \underline{\underline{ \text{Additional \: lnformation}}} \\ & \rm \: sin2x \: = 2 \: sinx \: cosx\:\\ & \rm \: cos2x = 1 - {2sin}^{2}x \\ & \rm \: cos2x = {2cos}^{2}x - 1 \\ & \rm \: cos2x = {cos}^{2}x - {sin}^{2}x \\ & \rm \:tan2x = \dfrac{2tanx}{1 - {tan}^{2} x} \\ & \rm \: sin2x = \frac{2tanx}{1 + {tan}^{2}x } \\ & \rm \:sin3x = 3sinx - {4sin}^{3}x \\ & \rm \: cos3x = {4cos}^{3}x - 3cosx \\ & \rm \: tan3x = \dfrac{3tanx - {tan}^{3} x}{1 - {3tan}^{2}x} \end{array}}}\end{gathered}$