Question

How do you find the value of $\sin \left( \dfrac{7\pi }{8} \right)$ using the double or double angle formula?

Answer 1

To solve the problem we have to find the value of $\sin \left( \dfrac{7\pi }{8} \right)$ . For that we have to convert it as $\left( \pi -\dfrac{\pi }{8} \right)$ . After that by using the double angle formula we can find the value of $\sin \left( \dfrac{\pi }{8} \right)$ . By using the double angle formula we can easily find the value of $\sin \left( \dfrac{\pi }{8} \right)$ .

Complete step by step solution:
For the given problem we have to find the value of $\sin \left( \dfrac{7\pi }{8} \right)$ using half or double angle formula.
Let us find the value of $\sin \left( \dfrac{7\pi }{8} \right)$ using double angle formula $\sin \left( \dfrac{\pi }{8} \right)$
Let us consider the given equation as equation (1).
$a=\sin \left( \dfrac{7\pi }{8} \right).........\left( 1 \right)$
By the trigonometric table of special arcs and unit circle, we can write the equation (1) as

& a=\sin \left( \pi -\dfrac{\pi }{8} \right) \\\ & \Rightarrow a=\sin \left( \dfrac{\pi }{8} \right) \\\ \end{aligned}$$ Let us consider the equation as equation (2). $$a=\sin \left( \dfrac{\pi }{8} \right).........\left( 2 \right)$$ Finding $$\sin \left( \dfrac{\pi }{8} \right)$$ by using trigonometry identity. As we know the trigonometry identity $$2{{\sin }^{2}}a=1-\cos 2a$$ Let us consider the above formula as formula (f1). $$2{{\sin }^{2}}a=1-\cos 2a..............\left( f1 \right)$$ For finding the value of $$\sin \left( \dfrac{\pi }{8} \right)$$ . Let us apply formula (f1), we get $$\Rightarrow 2{{\sin }^{2}}\left( \dfrac{\pi }{8} \right)=1-\cos \left( 2.\dfrac{\pi }{8} \right)$$ By simplifying it we get $$\Rightarrow 2{{\sin }^{2}}\left( \dfrac{\pi }{8} \right)=1-\cos \left( \dfrac{\pi }{4} \right)$$ Let us consider the above equation as equation (3). $$2{{\sin }^{2}}\left( \dfrac{\pi }{8} \right)=1-\cos \left( \dfrac{\pi }{4} \right)..........\left( 3 \right)$$ Substituting the value of $$\cos \left( \dfrac{\pi }{4} \right)$$ in equation (3). $$\Rightarrow 2{{\sin }^{2}}\left( \dfrac{\pi }{8} \right)=1-\dfrac{\sqrt{2}}{2}$$ Let us consider the above equation as equation (4) $$2{{\sin }^{2}}\left( \dfrac{\pi }{8} \right)=1-\dfrac{\sqrt{2}}{2}........\left( 4 \right)$$ Dividing the equation (4) with 2 we get $$\Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{8} \right)=\dfrac{1}{2}-\dfrac{\sqrt{2}}{4}$$ Cleaning a bit, we get $$\Rightarrow {{\sin }^{2}}\left( \dfrac{\pi }{8} \right)=\dfrac{2-\sqrt{2}}{4}$$ Let us consider the above equation as equation (5) $${{\sin }^{2}}\left( \dfrac{\pi }{8} \right)=\dfrac{2-\sqrt{2}}{4}.......\left( 5 \right)$$ After squaring equation (5) we get $$\Rightarrow \sin \left( \dfrac{\pi }{8} \right)=\sqrt{\dfrac{2-\sqrt{2}}{4}}$$ Cleaning a bit , we get $$\Rightarrow \sin \left( \dfrac{\pi }{8} \right)=\dfrac{\sqrt{2-\sqrt{2}}}{2}$$ Let us consider the above equation as equation (6). $$\sin \left( \dfrac{\pi }{8} \right)=\dfrac{\sqrt{2-\sqrt{2}}}{2}............\left( 6 \right)$$ Substituting the value of $$\sin \left( \dfrac{\pi }{8} \right)$$ from the equation (6) to the equation (2). Therefore, $$\begin{aligned} & a=\sin \left( \dfrac{\pi }{8} \right) \\\ & a=\dfrac{\sqrt{2-\sqrt{2}}}{2} \\\ \end{aligned}$$ Let us consider $$a=\dfrac{\sqrt{2-\sqrt{2}}}{2}......\left( 7 \right)$$ So, therefore value of $$\sin \left( \dfrac{7\pi }{8} \right)$$ is$$\dfrac{\sqrt{2-\sqrt{2}}}{2}$$. **Note:** Students must be aware of all trigonometric table and trigonometric formulas. While squaring the equation $${{\sin }^{2}}\left( \dfrac{\pi }{8} \right)=\dfrac{2-\sqrt{2}}{4}$$ we will get two case i.e. positive and negative case but negative answer is rejected because $$\sin \left( \dfrac{\pi }{8} \right)$$ is positive. This problem can be solved using sine half angle formula too.