Question

How do you find the value of $\cos \left( \dfrac{3\pi }{8} \right)$ using a double or half angle formula?

Answer 1

Hint: We know that the formula for $\cos 2x$ is $2{{\cos }^{2}}x-1$ . So we can write ${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}$
Let’s assume 2x as y so the value of x will be $\dfrac{y}{2}$ . So by replacing these in the above formula we get
${{\cos }^{2}}\dfrac{y}{2}=\dfrac{1+\cos y}{2}$ we can use this formula to find the value of $\cos \left( \dfrac{3\pi }{8} \right)$

Complete step by step answer:
We have to find the value of $\cos \left( \dfrac{3\pi }{8} \right)$ , we will use the half angle formula to find the value.
We know the formula ${{\cos }^{2}}\dfrac{y}{2}=\dfrac{1+\cos y}{2}$
If we put the value of y equal to $\dfrac{3\pi }{4}$ then $\dfrac{y}{2}$ will be equal to $\dfrac{3\pi }{8}$ replacing these in above formula we get
${{\cos }^{2}}\dfrac{3\pi }{8}=\dfrac{1+\cos \dfrac{3\pi }{4}}{2}$ …eq1
We know the formula $\cos \left( \pi -x \right)=-\cos x$
Putting x equal to $\dfrac{\pi }{4}$ we get
$\cos \dfrac{3\pi }{4}=-\cos \dfrac{\pi }{4}$
We know the value of $\cos \dfrac{\pi }{4}$ is equal to $\dfrac{1}{\sqrt{2}}$
So the value of $\cos \dfrac{3\pi }{4}$ is equal to $-\dfrac{1}{\sqrt{2}}$ , putting the value eq1 we get
${{\cos }^{2}}\dfrac{3\pi }{8}=\dfrac{1-\dfrac{1}{\sqrt{2}}}{2}$
Further solving we get
${{\cos }^{2}}\dfrac{3\pi }{8}=\dfrac{\sqrt{2}-1}{2\sqrt{2}}$
 $\cos \left( \dfrac{3\pi }{8} \right)$ can be $\pm \sqrt{\dfrac{\sqrt{2}-1}{2\sqrt{2}}}$
$\dfrac{3\pi }{8}$ comes in the first quadrant so the value of $\cos \left( \dfrac{3\pi }{8} \right)$ will be positive so the value of $\cos \left( \dfrac{3\pi }{8} \right)$ is $\sqrt{\dfrac{\sqrt{2}-1}{2\sqrt{2}}}$

Note:
If the value of ${{\cos }^{2}}x$ is equal to y then the value of cos x can be $\pm \sqrt{y}$ , the value of cos x depends on the quadrant of x. If the quadrant of x is in first or fourth then the value of cos x will be $\sqrt{y}$, if the quadrant of x is in second or third quadrant then the value cos x will be equal to $-\sqrt{y}$. This cos is positive in the first and fourth quadrant, negative in the third and fourth quadrant.