If f(x)=x3−x and g(x)=sin⁡2x, thenfgπ12=

g(π/12)=sin⁡(2×π/12)=sin⁡(π/6)=1/2f(g(π/12))=f(1/2)=(1/2)3−(1/2)=1/8−1/2=1/8−4/8=−3/8

If f(x)=x3−x and g(x)=sin⁡2x, thenfgπ12=

g(π/12)=sin⁡(2×π/12)=sin⁡(π/6)=1/2f(g(π/12))=f(1/2)=(1/2)3−(1/2)=1/8−1/2=1/8−4/8=−3/8

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If $f (x) = x^{3} - x$ and $g (x) = \sin 2 x,$ then $f (g (\frac{π}{12})) =$

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$- \frac{3}{8}$

answer is C.

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$\begin{aligned} g (π / 12) = \sin (2 \times π / 12) = \sin (π / 6) = 1 / 2 \\ f (g (π / 12)) = f (1 / 2) = (1 / 2)^{3} - (1 / 2) = 1 / 8 - 1 / 2 = 1 / 8 - 4 / 8 = - 3 / 8 \end{aligned}$

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If $f (x) = x^{3} - x$ and $g (x) = \sin 2 x,$ then $f (g (\frac{π}{12})) =$