Translation turning sine into cosine

Problem

Two functions are given: $\;y=\sin\!\left(3x-\dfrac{\pi}{4}\right)\;$ and $\;y=\cos\!\left(3x+\dfrac{\pi}{2}\right)$. a) Prove that the second is obtained by translating the first by a certain amount along the $x$-axis, and find that amount. b) Explain the equality in terms of associated arcs. c) Apply the translation you found to the sine function and check that you obtain the cosine function.

Solution

a) Using $\cos\theta=\sin\!\left(\theta+\tfrac{\pi}{2}\right)$ we get $\cos\!\left(3x+\tfrac{\pi}{2}\right)=\sin\!\left(3x+\pi\right)$. Translating the first by $h$ along $x$ gives $\sin\!\left(3(x-h)-\tfrac{\pi}{4}\right)=\sin\!\left(3x-3h-\tfrac{\pi}{4}\right)$. Imposing $-3h-\tfrac{\pi}{4}=\pi$ yields $h=-\frac{5\pi}{12}\approx-1{,}309.$ b) Sine and cosine differ by an arc of $\tfrac{\pi}{2}$: $\cos\theta=\sin\!\left(\theta+\tfrac{\pi}{2}\right)$; here $\cos\!\left(3x+\tfrac{\pi}{2}\right)=\sin\!\left(3x+\pi\right)=-\sin 3x$. c) $\sin\!\left(3\!\left(x+\tfrac{5\pi}{12}\right)-\tfrac{\pi}{4}\right)=\sin\!\left(3x+\pi\right)=\cos\!\left(3x+\tfrac{\pi}{2}\right)$, as required (translation of $\tfrac{5\pi}{12}$ to the left). $\boxed{\,h=-\dfrac{5\pi}{12}\approx-1{,}309\,}$