If x+^2x=1, x(0,2), then [(^12x+^12x) + 3(^10x+^10x+^8x+^8x) + (^6x+^6x)] is equal to:

Question

If $\sin x+\sin^2x=1$, $x\in\left(0,\dfrac\pi2\right)$, then

$$(\cos^{12}x+\tan^{12}x) + 3(\cos^{10}x+\tan^{10}x+\cos^8x+\tan^8x) + (\cos^6x+\tan^6x)$$

is equal to:

Accepted Answer

From $\sin x+\sin^2x=1$: $\sin x = \cos^2x$, so $\tan x = \dfrac{\sin x}{\cos x} = \dfrac{\cos^2x}{\cos x} = \cos x$.

Substituting $\tan x=\cos x$ throughout, the expression becomes:

$$2\cos^{12}x + 6\cos^{10}x+6\cos^8x+2\cos^6x = 2\cos^6x\left(\cos^6x+3\cos^4x+3\cos^2x+1\right) = 2\cos^6x(\cos^2x+1)^3$$

Since $\cos^2x = \sin x$: $\cos^6x = \sin^3x$ and $\cos^2x+1 = \sin x+1$, so:

$$= 2\sin^3x(\sin x+1)^3 = 2[\sin x(\sin x+1)]^3 = 2[\sin x+\sin^2x]^3 = 2(1)^3 = 2$$

Solutions · 1