Question

Prove following expression:
${{\sin }^{2}}\dfrac{\pi }{8}+{{\sin }^{2}}\dfrac{3\pi }{8}+{{\sin }^{2}}\dfrac{5\pi }{8}+{{\sin }^{2}}\dfrac{7\pi }{8}=2$.

Answer 1

Hint: In this question, we first need to express the given expression in terms of some standard angles by using trigonometric ratios of multiple angles formula. Then apply the transformation formula to simplify it further. Now, on further simplification we get the result.
$1-\cos 2A=2{{\sin }^{2}}A$
$\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$

Complete step by step answer:
As we already know that form the trigonometric ratios of multiple angles
$1-\cos 2A=2{{\sin }^{2}}A$
Here, from the transformation formula we also have
$\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
Now, by considering the given equation we get,
$\Rightarrow {{\sin }^{2}}\dfrac{\pi }{8}+{{\sin }^{2}}\dfrac{3\pi }{8}+{{\sin }^{2}}\dfrac{5\pi }{8}+{{\sin }^{2}}\dfrac{7\pi }{8}$
Now, on applying trigonometric ratios of multiple angles we get,

& \Rightarrow \dfrac{1}{2}\left( 1-\cos \dfrac{\pi }{4} \right)+\dfrac{1}{2}\left( 1-\cos \dfrac{3\pi }{4} \right)+\dfrac{1}{2}\left( 1-\cos \dfrac{5\pi }{4} \right)+\dfrac{1}{2}\left( 1-\cos \dfrac{7\pi }{4} \right) \\\ & \left[ \because 1-\cos 2A=2{{\sin }^{2}}A \right] \\\ & \\\ \end{aligned}$$ By rearranging the terms in the above equation we get, $$\begin{aligned} & \Rightarrow \dfrac{1}{2}\left( 4-\cos \dfrac{\pi }{4}-\cos \dfrac{3\pi }{4}-\cos \dfrac{5\pi }{4}-\cos \dfrac{7\pi }{4} \right) \\\ & \Rightarrow 2-\dfrac{1}{2}\left( \cos \dfrac{\pi }{4}+\cos \dfrac{3\pi }{4}+\cos \dfrac{5\pi }{4}+\cos \dfrac{7\pi }{4} \right) \\\ \end{aligned}$$ Now, by applying transformation formula to the above equation we get, $$\begin{aligned} & \Rightarrow 2-\dfrac{1}{2}\left( 2\cos \left[ \dfrac{\frac{\pi }{4}+\dfrac{3\pi }{4}}{2} \right]\cos \left[ \dfrac{\dfrac{\pi }{4}-\dfrac{3\pi }{4}}{2} \right]+2\cos \left[ \dfrac{\dfrac{5\pi }{4}+\dfrac{7\pi }{4}}{2} \right]\cos \left[ \dfrac{\dfrac{5\pi }{4}-\dfrac{7\pi }{4}}{2} \right] \right) \\\ & \left[ \because \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \right] \\\ \end{aligned}$$ Now, on further simplifying the above expression we get, $$\Rightarrow 2-\left( \cos \dfrac{\pi }{2}\cos \dfrac{-\pi }{4}+\cos \dfrac{3\pi }{2}\cos \dfrac{-\pi }{4} \right)$$ Let us now take the common terms out to simplify it further $$\Rightarrow 2-\cos \dfrac{\pi }{4}\left( \cos \dfrac{\pi }{2}+\cos \dfrac{3\pi }{2} \right)\text{ }\left[ \because \cos \left( -\theta \right)=\cos \theta \right]$$ Now, let us substitute the respective values to simplify it further $$\Rightarrow 2-\cos \dfrac{\pi }{4}\times 0\text{ }\left[ \because \cos \dfrac{\pi }{2}=\cos \dfrac{3\pi }{2}=0 \right]$$ Now, this can be rewritten as $$\begin{aligned} & \Rightarrow 2 \\\ & \therefore {{\sin }^{2}}\dfrac{\pi }{8}+{{\sin }^{2}}\dfrac{3\pi }{8}+{{\sin }^{2}}\dfrac{5\pi }{8}+{{\sin }^{2}}\dfrac{7\pi }{8}=2 \\\ \end{aligned}$$ Note: Instead of directly writing the $${{\sin }^{2}}A$$ terms in terms of $$\cos 2A$$ we can convert those terms into $${{\cos }^{2}}A$$ using the identity $${{\sin }^{2}}A+{{\cos }^{2}}A=1$$ and then convert it in terms of $$\cos 2A$$ is one of the alternate methods. We can use the trigonometric ratios of compound angles formula in place of the transformation formula in order to expand the equation and then simplify it accordingly. $$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$$ While simplifying the equation we need to be careful and apply the corresponding values accordingly because neglecting any of the terms changes the result of the equation completely.