Are there any trigonometric identity that can make $f(x) = \sin(x)+\cos(x)$ easier to plot? I have no idea how it becomes a sin graph shape in the end.
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$\begingroup$Special Case
It's not that hard. You should just use the summation formula for sines:
$$\sin (x + y) = \sin (x)\cos (y) + \cos (x)\sin (y)$$
This is how it works
$$\eqalign{ \sin (x) + \cos (x) &= \sqrt 2 \left( {{1 \over {\sqrt 2 }}\cos (x) + {1 \over {\sqrt 2 }}\sin (x)} \right) \cr &= \sqrt 2 \left( {\sin ({\pi \over 4})\cos (x) + \cos ({\pi \over 4})\sin (x)} \right) \cr &= \sqrt 2 \sin (x + {\pi \over 4}) \cr} $$
That's all you need to do for this case. If you are interested to tackle down the general case then read the sequel.
General Case
Consider a linear combination of $\sin \alpha x$ and $\cos \alpha x$ as follows
$$y = A \cos \alpha x + B \sin \alpha x$$
where $A$ and $B$ are some real constants. Then we rewrite $y$ in this way
$$y = \sqrt{A^2+B^2} \left( \frac{A}{\sqrt{A^2+B^2}} \cos \alpha x + \frac{B}{\sqrt{A^2+B^2}} \sin \alpha x \right)$$
Now, the magic comes in! We can find a unique angle $\phi$ in the interval $[0,2\pi)$ such that
$$\begin{array}{} \sin \phi = \dfrac{A}{\sqrt{A^2+B^2}} \\ \cos \phi = \dfrac{B}{\sqrt{A^2+B^2}} \end{array}$$
and hence
$$y = \sqrt{A^2+B^2} \left( \sin \phi \cos \alpha x + \cos \phi \sin \alpha x \right) \\$$
Finally, using the summation formula for sines we get
$$y = \sqrt{A^2+B^2} \sin(\alpha x+\phi)$$
$\endgroup$ 0 $\begingroup$Note Using the formula for $\sin(A)+\sin(B)$ you get
$$\sin(x)+\cos(x)=\sin(x)+\sin(\frac{\pi}{2}-x)=2\sin( \frac{x+\frac{\pi}{2}-x}{2}) \cos(\frac{x-\frac{\pi}{2}+x}{2})$$which is easy to calculate and plot.
Note also that $$\cos(x- \frac{\pi}{4})=\cos(\frac{\pi}{4}-x)=\sin( \frac{\pi}{2}-\frac{\pi}{4}+x)$$which is consistent with the other answer.
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