Test Index

AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 21 of 160

Marks: +1, -0

\sin^4 \frac{\pi}{8} + \sin^4 \frac{2\pi}{8}

\sin^4 \frac{3\pi}{8} + \sin^4 \frac{4\pi}{8}

\sin^4 \frac{5\pi}{8} + \sin^4 \frac{6\pi}{8}

\sin^4 \frac{7\pi}{8}

Solution:

Let's consider the expression as,

x = \sin^4 \frac{\pi}{8} + \sin^4 \frac{2\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{4\pi}{8} + \sin^4 \frac{5\pi}{8} + \sin^4 \frac{6\pi}{8} + \sin^4 \frac{7\pi}{8}

x = \left[ \sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{5\pi}{8} + \sin^4 \frac{7\pi}{8} \right] + \left[ \begin{array}{c} \sin^4 \frac{2\pi}{8} + \sin^4 \frac{4\pi}{8} \\ + \sin^4 \frac{6\pi}{8} \end{array} \right]

= \left[ \begin{array}{c} \sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \\ \cos^4\left(\frac{\pi}{2} - \frac{5\pi}{8}\right) + \cos^4\left(\frac{\pi}{2} - \frac{7\pi}{8}\right) \end{array} \right] + \left[ \begin{array}{c} \sin^4 \frac{\pi}{4} + \sin^4 \frac{\pi}{2} \\ + \sin^4\left(\frac{\pi}{2} + \frac{\pi}{4}\right) \end{array} \right]

= \left[ \sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} \right] + \left[ \begin{array}{c} \sin^4 \frac{\pi}{4} + \sin^4 \frac{\pi}{2} \\ + \cos^4 \frac{\pi}{4} + \cos^4 \frac{3\pi}{8} \end{array} \right]

= \left[ \left(\sin^4 \frac{\pi}{8} + \cos^4 \frac{\pi}{8}\right) + \left(\sin^4 \frac{3\pi}{8} + \cos^4 \frac{3\pi}{8}\right) \right] + \left[ \begin{array}{c} \left(\frac{1}{\sqrt{2}}\right)^4 + 1^4 \\ + \left(\frac{1}{\sqrt{2}}\right)^4 \end{array} \right]

Further simplify the above,

x = \left[ \left(1 - 2 \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}\right) + \left(1 - 2 \sin^2 \frac{3\pi}{8} \cos^2 \frac{3\pi}{8}\right) \right] + \left[ \frac{1}{4} + 1 + \frac{1}{4} \right]

= \left[ 2 - \frac{1}{2} 4 \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8} - \frac{1}{2} 4 \sin^2 \frac{3\pi}{8} \cos^2 \frac{3\pi}{8} \right] + \left[ \frac{1}{2} + 1 \right]

= 3

Go to Question:

Prev Question

Next Question