Question

The value of ${\cos ^4}\left( {\dfrac{\pi }{8}} \right) + {\cos ^4}\left( {\dfrac{{3\pi }}{8}} \right) + {\cos ^4}\left( {\dfrac{{5\pi }}{8}} \right) + {\cos ^4}\left( {\dfrac{{7\pi }}{8}} \right)$ is
A. $0$
B. $\dfrac{1}{2}$
C. $\dfrac{3}{2}$
D. $1$

Answer 1

Hint: In the given problem, we need to find the value of ${\cos ^4}\left( {\dfrac{\pi }{8}} \right) + {\cos ^4}\left( {\dfrac{{3\pi }}{8}} \right) + {\cos ^4}\left( {\dfrac{{5\pi }}{8}} \right) + {\cos ^4}\left( {\dfrac{{7\pi }}{8}} \right)$. At first, we will try to simplify the given expression in terms of $\cos $ function. After this, we will use algebraic identity and trigonometric formulae to simplify the given expression and get our required answer.

Formula Used:
$\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta $
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
$\sin 2x = 2\sin x\cos x$

Complete step by step solution:
We have, ${\cos ^4}\left( {\dfrac{\pi }{8}} \right) + {\cos ^4}\left( {\dfrac{{3\pi }}{8}} \right) + {\cos ^4}\left( {\dfrac{{5\pi }}{8}} \right) + {\cos ^4}\left( {\dfrac{{7\pi }}{8}} \right)$
$ = {\left( {\cos \dfrac{\pi }{8}} \right)^4} + {\left( {\cos \dfrac{{3\pi }}{8}} \right)^4} + {\left( {\cos \dfrac{{5\pi }}{8}} \right)^4} + {\left( {\cos \dfrac{{7\pi }}{8}} \right)^4}$
We can write the above written expression as,
$ = {\left( {\cos \left( {\dfrac{\pi }{8}} \right)} \right)^4} + {\left( {\cos \left( {\dfrac{{3\pi }}{8}} \right)} \right)^4} + {\left( {\cos \left( {\pi - \dfrac{{3\pi }}{8}} \right)} \right)^4} + {\left( {\cos \left( {\pi - \dfrac{\pi }{8}} \right)} \right)^4}$
We know that $\cos \left( {\pi - \theta } \right) = - \cos \theta $ (Here, $\cos $ is negative because $\pi - \theta $ lies in the second quadrant, and in second quadrant $\cos $ is negative). Therefore, we get
$ = {\left( {\cos \dfrac{\pi }{8}} \right)^4} + {\left( {\cos \dfrac{{3\pi }}{8}} \right)^4} + {\left( { - \cos \dfrac{{3\pi }}{8}} \right)^4} + {\left( { - \cos \dfrac{\pi }{8}} \right)^4}$
$ = {\cos ^4}\left( {\dfrac{\pi }{8}} \right) + {\cos ^4}\left( {\dfrac{{3\pi }}{8}} \right) + {\cos ^4}\left( {\dfrac{{3\pi }}{8}} \right) + {\cos ^4}\left( {\dfrac{\pi }{8}} \right)$
On addition, we get
$ = 2{\cos ^4}\left( {\dfrac{\pi }{8}} \right) + 2{\cos ^4}\left( {\dfrac{{3\pi }}{8}} \right)$
$ = 2\left[ {{{\cos }^4}\left( {\dfrac{\pi }{8}} \right) + {{\cos }^4}\left( {\dfrac{{3\pi }}{8}} \right)} \right]$
The above written expression can also be written as
$ = 2\left[ {{{\cos }^4}\left( {\dfrac{\pi }{8}} \right) + {{\cos }^4}\left( {\dfrac{\pi }{2} - \dfrac{\pi }{8}} \right)} \right]$
We know that $\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta $. Therefore, we get
$ = 2\left[ {{{\cos }^4}\dfrac{\pi }{8} + {{\sin }^4}\dfrac{\pi }{8}} \right]$
Now, we will add $2{\sin ^2}\dfrac{\pi }{8}{\cos ^2}\dfrac{\pi }{8}$ to ${\cos ^4}\dfrac{\pi }{8} + {\sin ^4}\dfrac{\pi }{8}$ and subtract $2{\sin ^2}\dfrac{\pi }{8}{\cos ^2}\dfrac{\pi }{8}$ from ${\cos ^4}\dfrac{\pi }{8} + {\sin ^4}\dfrac{\pi }{8}$.
$ = 2\left[ {\left( {{{\cos }^4}\dfrac{\pi }{8} + {{\sin }^4}\dfrac{\pi }{8} + 2{{\sin }^2}\dfrac{\pi }{8}{{\cos }^2}\dfrac{\pi }{8}} \right) - 2{{\sin }^2}\dfrac{\pi }{8}{{\cos }^2}\dfrac{\pi }{8}} \right]$
We know that ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. Therefore, we get $ = 2\left[ {{{\left( {{{\cos }^2}\dfrac{\pi }{8} + {{\sin }^2}\dfrac{\pi }{8}} \right)}^2} - 2{{\sin }^2}\dfrac{\pi }{8}{{\cos }^2}\dfrac{\pi }{8}} \right]$
We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$. Therefore, we get
$ = 2\left[ {1 - 2{{\sin }^2}\dfrac{\pi }{8}{{\cos }^2}\dfrac{\pi }{8}} \right]$
Multiply and divide $2{\sin ^2}\dfrac{\pi }{8}{\cos ^2}\dfrac{\pi }{8}$ by $2$
$ = 2\left[ {1 - \dfrac{4}{2} \times {{\sin }^2}\dfrac{\pi }{8}{{\cos }^2}\dfrac{\pi }{8}} \right]$
$ = 2\left[ {1 - \dfrac{1}{2} \times {{\left( {2\sin \dfrac{\pi }{8}\cos \dfrac{\pi }{8}} \right)}^2}} \right]$
We know that $\sin 2x = 2\sin x\cos x$. Therefore, we get
$ = 2\left[ {1 - \dfrac{1}{2} \times {{\left( {\sin 2.\dfrac{\pi }{8}} \right)}^2}} \right]$
$ = 2\left[ {1 - \dfrac{1}{2} \times {{\left( {\sin \dfrac{\pi }{4}} \right)}^2}} \right]$
We know that $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$. Therefore, we get
$ = 2\left[ {1 - \dfrac{1}{2} \times {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}} \right]$
$ = 2\left[ {1 - \dfrac{1}{2} \times \dfrac{1}{2}} \right]$
On further simplification, we get
$ = 2 \times \dfrac{3}{4}$
$ = \dfrac{3}{2}$
Hence, the value of ${\cos ^4}\left( {\dfrac{\pi }{8}} \right) + {\cos ^4}\left( {\dfrac{{3\pi }}{8}} \right) + {\cos ^4}\left( {\dfrac{{5\pi }}{8}} \right) + {\cos ^4}\left( {\dfrac{{7\pi }}{8}} \right)$ is $\dfrac{3}{2}$.

Option ‘C’ is correct

Note: To solve these types of questions, one must remember all the standard formulas of trigonometric functions. Most of the trigonometric functions questions are just based on substitutions, we can only solve the problem if we know the formulas and sign convention. Always try to convert the given expression in the form of an identity or formula so that you can simplify the expression easily.
Some trigonometric formulas are written below:
$\sin 2A = 2\sin A\cos A$
$\cos 2A = {\cos ^2}A - {\sin ^2}A$
$\cos 2A = 2{\cos ^2}A - 1$
$\cos 2A = 1 - 2{\sin ^2}A$
$\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$
$\sin 2A = \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}$
$\cos 2A = \dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}$
We use them accordingly to the given problem