Q119 / 180 · COMEDK 2013 · Mathematics

If $\sin x+\sin ^{2} x=1$, then $\cos ^{12} x+3 \cos ^{10} x+3 \cos ^{8} x+\cos ^{6} x$ is equal to

  • Correct answer
    A
    1
  • B
    2
  • C
    3
  • D
    0

Solution

We have, $$ \begin{array}{ll} & \sin x+\sin ^{2} x=1 \\ \Rightarrow \quad & \sin x=1-\sin ^{2} x \\ \Rightarrow \quad & \sin x=\cos ^{2} x \\ \text { Now, } \cos ^{12} x & +3 \cos ^{10} x+3 \cos ^{8} x+\cos ^{6} x \\ & =\sin ^{6} x+3 \sin ^{5} x+3 \sin ^{4} x+\sin ^{3} x \\ & =\sin ^{3} x\left[\sin ^{3} x+3 \sin ^{2} x+3 \sin x+1\right] \\ & =\sin ^{3} x(\sin x+1)^{3} \\ & =\left(\sin ^{2} x+\sin x\right)^{3} \\ & =(1)^{3}=1 \end{array} $$

Hence, the correct answer is option: 'A'.

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