SSC CGL Model Paper 15

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Question : 72 of 100

Marks: +1, -0

If

\sin x+\cos x =c,

then

\sin^6x+\cos^6x

is equal to.

${1+6c^2-3c^4}/16$
${1+6c^2-3c^4}/4$
${1+6c^2+3c^4}/16$
${1+6c^2+3c^4}/4$

Solution:

\sin x+\cos x =c

Squaring both sides

\sin^2x+\cos^2x+2\sin x\cos x=c^2

\sin x\cos x={c^2-1}/2

We know that,

\sin ^6x+\cos^6x

=1-2\sin^2x\cos^2x

=1-3({c^2-1}/2)^2

={1+6c^2-3c^4}/4

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