Question

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

Answer 1

Hint: 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
\[\begin{align}
  & a=\sin \left( \pi -\dfrac{\pi }{8} \right) \\
 & \Rightarrow a=\sin \left( \dfrac{\pi }{8} \right) \\
\end{align}\]
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,
\[\begin{align}
  & a=\sin \left( \dfrac{\pi }{8} \right) \\
 & a=\dfrac{\sqrt{2-\sqrt{2}}}{2} \\
\end{align}\]
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.