A calculus problem by Christopher Boo

Calculus Level 3

d d x cos ⁡ x = − d d x sin ⁡ x d d x ( cos ⁡ x ) 2 = − d d x ( sin ⁡ x ) 2 d d x ( cos ⁡ x ) 3 = − d d x ( sin ⁡ x ) 3 \begin{array} {l c l } \dfrac d{dx} \cos x &= &- \dfrac d{dx} \sin x \\ \dfrac d{dx} (\cos x)^2 &=&- \dfrac d{dx}(\sin x)^2 \\ \dfrac d{dx} (\cos x)^3 &=& - \dfrac d{dx}(\sin x)^3 \end{array}

Which of the equations above is an identity?

d d x ( cos ⁡ x ) 3 = − d d x ( sin ⁡ x ) 3 \frac d{dx} (\cos x)^3 = - \frac d{dx}(\sin x)^3 d d x cos ⁡ x = − d d x sin ⁡ x \frac d{dx} \cos x = - \frac d{dx} \sin x d d x ( cos ⁡ x ) 2 = − d d x ( sin ⁡ x ) 2 \frac d{dx} (\cos x)^2 = - \frac d{dx}(\sin x)^2

Christopher Boo by Christopher Boo , 24, Singapore

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1 solution

A nice approach would be to integrate the sides first: (Ignoring the arbitrary constants of integration)

cos ⁡ x = − sin ⁡ x ⟹ cos ⁡ x + sin ⁡ x = 1 \cos x = -\sin x \implies \cos x + \sin x = 1 , which is not true.

cos ⁡ 2 x = − sin ⁡ 2 x ⟹ cos ⁡ 2 x + sin ⁡ 2 x = 1 \cos^2 x = -\sin^2 x \implies \cos^2 x + \sin^2 x = 1 , which is true.

cos ⁡ 3 x = − sin ⁡ 3 x ⟹ cos ⁡ 3 x + sin ⁡ 3 x = 1 \cos^3 x = -\sin^3 x \implies \cos^3 x + \sin^3 x = 1 , which is not true.

Hence, only the second one is correct.

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