Question

# If $$\cos x + \cos^{2}x = 1$$, then the value of $$\sin^{4}x + \sin^{6}x$$ is

A
1+5
B
152
C
152
D
1+52
E
15

Solution

## The correct option is E $$\dfrac {-1 + \sqrt {5}}{2}$$Given, $$\cos x + \cos^{2}x = 1$$     .... (i)$$\Rightarrow \cos^{2}x + \cos x - 1 = 0$$Therefore, $$\cos x = \dfrac {-1\pm \sqrt {1 + 4}}{2} (\because$$ quadratic in $$\cos x)$$$$= \dfrac {-1\pm \sqrt {5}}{2} = \dfrac {-1 + \sqrt {5}}{2} \left (\because \cos x \neq \dfrac {-1 - \sqrt {5}}{2}\right )$$Also from Eq. (i),$$\cos x = 1 - \cos^{2} x$$$$\Rightarrow \cos x = \sin^{2}x$$     .... (ii)Now, $$\sin^{4}x + \sin^{6}x$$$$= \cos^{2} x + \cos^{3}x$$$$= \cos^{2} x (1 + \cos x)$$$$= \left (\dfrac {-1 + \sqrt {5}}{2}\right )^{2} \left (1 + \dfrac {-1 + \sqrt {5}}{2}\right )$$$$= \left (\dfrac {-1 + \sqrt {5}}{2}\right )^{2} \left (\dfrac {1 + \sqrt {5}}{2}\right )$$$$= \left (\dfrac {-1 + \sqrt {5}}{2}\right ) \left (\dfrac {5 - 1}{4}\right ) = \left (\dfrac {-1 + \sqrt {5}}{2}\right ) \times 1$$$$= \dfrac {-1 + \sqrt {5}}{2}$$Maths

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