Question

How do you use the half-angle formula to find $\sin \left( {\dfrac{{9\pi }}{8}} \right)$?

Answer 1

We start solving the problem by recalling the half-angle formula for the sine function as $\sin \dfrac{x}{2} = \sqrt {\dfrac{{1 - \cos x}}{2}}$. We then find the value of x so that $\dfrac{x}{2}$ is equal to the value $\dfrac{{9\pi }}{8}$. We then make use of the results $\cos \left( {2\pi + \theta } \right) = \cos \theta$ to proceed through the problem. We then make the necessary arrangements inside the square root and make use of the fact that $\dfrac{{9\pi }}{8}$ lies in the third quadrant and sin function is negative in the third quadrant to get the required answer.

Complete step-by-step solution:
According to the problem, we are asked to find the value of $\sin \dfrac{{9\pi }}{8}$ using the half-angle formula. Let us recall the half-angle formula for the sine function.
We know that the half-angle formula for sine function is defined as,
$\sin \dfrac{x}{2} = \sqrt {\dfrac{{1 - \cos x}}{2}}$.............….. (1)
Now, we need to find the value of $\sin \dfrac{{9\pi }}{8}$. So, we have
$\Rightarrow \dfrac{x}{2} = \dfrac{{9\pi }}{8}$
Multiply both sides by 2,
$\Rightarrow x = \dfrac{{9\pi }}{4}$
Let us substitute the value in equation (1)
$\Rightarrow \sin \dfrac{{9\pi }}{8} = \sqrt {\dfrac{{1 - \cos \dfrac{{9\pi }}{4}}}{2}}$
As we know $\cos \left( {2\pi + \theta } \right) = \cos \theta$. Then,
$\Rightarrow \sin \dfrac{{9\pi }}{8} = \sqrt {\dfrac{{1 - \cos \left( {2\pi + \dfrac{\pi }{4}} \right)}}{2}}$
Simplify the terms,
$\Rightarrow \sin \dfrac{{9\pi }}{8} = \sqrt {\dfrac{{1 - \cos \left( {\dfrac{\pi }{4}} \right)}}{2}}$
We know that $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$. Then,
$\Rightarrow \sin \dfrac{{9\pi }}{8} = \sqrt {\dfrac{{1 - \dfrac{1}{{\sqrt 2 }}}}{2}}$
Take LCM and move $\sqrt 2$ in the denominator,
$\Rightarrow \sin \dfrac{{9\pi }}{8} = \sqrt {\dfrac{{\sqrt 2 - 1}}{{2\sqrt 2 }}}$
Multiply numerator and denominator by $\sqrt 2$,
$\Rightarrow \sin \dfrac{{9\pi }}{8} = \sqrt {\dfrac{{2 - \sqrt 2 }}{4}}$
Simplify the terms,
$\Rightarrow \sin \dfrac{{9\pi }}{8} = \dfrac{{\sqrt {2 - \sqrt 2 } }}{2}$
Since $\dfrac{{9\pi }}{8}$ lies in the third quadrant. So, the sine function will be negative.
$\therefore \sin \dfrac{{9\pi }}{8} = - \dfrac{{\sqrt {2 - \sqrt 2 } }}{2}$

Hence, the value of $\sin \dfrac{{9\pi }}{8}$ is $- \dfrac{{\sqrt {2 - \sqrt 2 } }}{2}$

Note: We should perform each step carefully in order to avoid calculation mistakes and confusion. We should keep in mind the nature of the values of trigonometric functions in different quadrants while solving this type of problem. Similarly, we can expect the formulas to find the value of $\sin \dfrac{{9\pi }}{8}$ and $\cos \dfrac{{9\pi }}{8}$ using the formula of $\cos 2\theta$.