In my study of functions I've briefly learnt about transformation of functions wherein we can reflect, stretch , translate or compress a base function (informally) by adding out multiplying a constant to the variable or function itself. Suddenly the graph of trigonometric functions clicked my mind ( ... which I haven't studied in detail ) and I wondered whether function $\sin x$ is a transformation of funcion $\cos x$.
Algebraically $ \ sin^2 x + cos^2 x = 1 $ which shows a relation between the two functions and may be they could be transformed to each other. But here as I don't have a thorough understanding of transformation I don't know whether squaring a function also yields a transformation.
Though I am a bit unsure about it algebraically, geometrically I am firm about my question. This is so because if the cosine function is shifted towards right, it can certainly be transformed to sine function what I feel intuitively. I may be wrong but these two aspects ( geometric and algebraic) are pestering me to ask this question .
Please help and thanks.
As people have commented, yes cosine is a translation of sine (and vice versa) and to be precise $\sin(x + \pi/2) = \cos(x)$ and $\sin(x) = \cos(\pi/2 - x)$. This can be seen if you graph the two function and you will see that they do indeed coincide with one another. If you want a more algebraic proof of this, you can use the sum/difference formulas, which state: \begin{eqnarray*} \sin(\alpha \pm \beta) & = & \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\ \cos(\alpha \pm \beta) & = & \cos \alpha \cos \beta \mp \sin \alpha \sin \beta \end{eqnarray*} Using these formulas, we have \begin{eqnarray*} \sin(x + \pi/2) & = & \sin x \cos \pi/2 + \cos x \sin \pi/2 \\ & = & \sin x \cdot 0 + \cos x \cdot 1 \\ & = & \cos x \end{eqnarray*} and \begin{eqnarray*} \cos(\pi/2 - x) & = & \cos \pi/2 \cos x + \sin \pi/2 \sin x\\ & = & 0 \cdot \cos x + 1 \cdot \sin x\\ & = & \sin x \end{eqnarray*}