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Question

Mathematics Question on Trigonometric Functions

If sin4x2+cos4x3=15\frac{sin^4 x}{2}+\frac{cos^4 x}{3}=\frac{1}{5} then

A

tan2x=23tan^2 x=\frac{2}{3}

B

sin8x8+cos8x27=1125\frac{sin^8 x}{8}+\frac{cos^8 x}{27}=\frac{1}{125}

C

tan2x=13tan^2 x=\frac{1}{3}

D

sin8x8+cos8x27=2125\frac{sin^8 x}{8}+\frac{cos^8 x}{27}=\frac{2}{125}

Answer

sin8x8+cos8x27=1125\frac{sin^8 x}{8}+\frac{cos^8 x}{27}=\frac{1}{125}

Explanation

Solution

sin4x2+cos43=15\frac{sin^4 x}{2}+\frac{cos^4}{3}=\frac{1}{5}
\Rightarrow \hspace25mm \frac{sin^4 x}{2}+\frac{(1-sin^2 x)^2}{3}=\frac{1}{5}
\Rightarrow \hspace25mm \frac{sin^4}{2}+\frac{1+sin^4 x-2sin^2 x}{3}=\frac{1}{5}
\Rightarrow \hspace25mm 5 sin^4 x -4sin^2 x+2=\frac{6}{5}
\Rightarrow \hspace25mm 25 sin^4 x-20sin^2 x+4=0
\Rightarrow \hspace35mm (5 sin^2 x-2)^2=0
\Rightarrow \hspace40mm sin^2 x=\frac{2}{5}
\hspace40mm sin^2 x=\frac{3}{5}, tan^2 x=\frac{2}{3}
sin8x8+cos8x27=1125\frac{sin^8 x}{8}+\frac{cos^8 x}{27}=\frac{1}{125}