Question

Prove that ${\sin ^4}\dfrac{\pi }{8} + {\sin ^4}\dfrac{{3\pi }}{8} + {\sin ^4}\dfrac{{5\pi }}{8}{\sin ^4}\dfrac{{7\pi }}{8} = \dfrac{3}{2}$

Answer 1

Here, we will use the basic identities of the trigonometric functions. Firstly we will take the LHS of the equation and solve it. So we will apply the properties of the trigonometric function for the simplification of the LHS of the equation and by solving the simplified equation we will get the LHS equals to RHS.

Complete step by step solution:
Firstly we will take the LHS of the equation.
${\sin ^4}\dfrac{\pi }{8} + {\sin ^4}\dfrac{{3\pi }}{8} + {\sin ^4}\dfrac{{5\pi }}{8}{\sin ^4}\dfrac{{7\pi }}{8}$
We will use the trigonometry property i.e. $\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta$. But we will apply this property for the last two terms only. Therefore, we get
${\sin ^4}\dfrac{\pi }{8} + {\sin ^4}\dfrac{{3\pi }}{8} + {\cos ^4}\left( {\dfrac{\pi }{2} - \dfrac{{5\pi }}{8}} \right) + {\cos ^4}\left( {\dfrac{\pi }{2} - \dfrac{{7\pi }}{8}} \right)$
${\sin ^4}\dfrac{\pi }{8} + {\sin ^4}\dfrac{{3\pi }}{8} + {\cos ^4}\left( { - \dfrac{\pi }{2}} \right) + {\cos ^4}\left( { - \dfrac{{3\pi }}{8}} \right)$
We know that $\cos \left( { - \theta } \right) = \sin \theta$. Therefore, we get
${\sin ^4}\dfrac{\pi }{8} + {\sin ^4}\dfrac{{3\pi }}{8} + {\cos ^4}\left( {\dfrac{\pi }{2}} \right) + {\cos ^4}\left( {\dfrac{{3\pi }}{8}} \right)$
We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
Now by squaring both side of the equation we get ${\sin ^4}\theta + {\cos ^4}\theta + 2{\sin ^2}\theta {\cos ^2}\theta = 1$.
Therefore ${\sin ^4}\theta + {\cos ^4}\theta = 1 - 2{\sin ^2}\theta {\cos ^2}\theta$. So, by applying this we get
$\Rightarrow \left( {1 - 2{{\sin }^2}\dfrac{\pi }{8}{{\cos }^2}\dfrac{\pi }{8}} \right) + \left( {1 - 2{{\sin }^2}\dfrac{{3\pi }}{8}{{\cos }^2}\dfrac{{3\pi }}{8}} \right)$
$\Rightarrow 2 - 2{\sin ^2}\dfrac{\pi }{8}{\cos ^2}\dfrac{\pi }{8} - 2{\sin ^2}\dfrac{{3\pi }}{8}{\cos ^2}\dfrac{{3\pi }}{8}$
$\Rightarrow 2 - \dfrac{1}{2} \times 4{\sin ^2}\dfrac{\pi }{8}{\cos ^2}\dfrac{\pi }{8} - \dfrac{1}{2} \times 4{\sin ^2}\dfrac{{3\pi }}{8}{\cos ^2}\dfrac{{3\pi }}{8}$
We know that $2\sin \theta {\rm{ }}\cos \theta = \sin {\rm{ }}2\theta$. Therefore by applying this property in the equation we get
$\Rightarrow 2 - \dfrac{1}{2}{\sin ^2}\dfrac{{2\pi }}{8} - \dfrac{1}{2}{\sin ^2}\dfrac{{2 \times 3\pi }}{8}$
$\Rightarrow 2 - \dfrac{1}{2}{\sin ^2}\dfrac{\pi }{4} - \dfrac{1}{2}{\sin ^2}\dfrac{{3\pi }}{4}$
We know that the value of ${\sin ^2}\dfrac{\pi }{4} = \dfrac{1}{2}$ and ${\sin ^2}\dfrac{{3\pi }}{4} = \dfrac{1}{2}$. Therefore, we get
$\Rightarrow 2 - \dfrac{1}{2} \times \dfrac{1}{2} - \dfrac{1}{2} \times \dfrac{1}{2} = 2 - \dfrac{1}{4} - \dfrac{1}{4} = 2 - \dfrac{1}{2} = \dfrac{{4 - 1}}{2} = \dfrac{3}{2}$
Therefore we get the value of LHS of the equation as $\dfrac{3}{2}$.
Hence LHS is equal to the RHS of the equation.
Hence proved.

Note:
We must remember the different properties of the trigonometric function and also in which quadrant which function is positive or negative as in the first quadrant all the functions i.e. sin, cos, tan, cot, sec, cosec is positive. In the second quadrant, only the sine and cosecant function are positive and all the other functions are negative. In the third quadrant, only tan and cot function is positive and in the fourth quadrant, only cos and sec function is positive. Also, we keep in mind the basic properties of the trigonometric functions and with the help of this concept, this question can be easily solved.
Properties used in the question: $2\sin \theta {\rm{ }}\cos \theta = \sin {\rm{ }}2\theta$and${\cos ^2}\theta - {\sin ^2}\theta = \cos {\rm{ }}2\theta$