How to show that $f(x)=\sin(x)+\cos(x)$ is sinusoidal(alternative)

Question

There was this question in our trig homework; it was for plotting a graph but I found it far more interesting than that. When drawing the graph of $\sin(x)+\cos(x)$ (by hand, which I find rather pointless), I found that it looked like some sort of sine or cosine graph. So I set out with all my trig identities to prove this.

$f(x)=\sin(x)+\cos(x)$

$\begin{align} f^2(x)=\sin^2 (x) + \cos^2 (x)+2\sin(x)\cos(x)&=1+\sin(2x) \\ &= 1+\cos\left(\dfrac{\pi}{2} -2x\right) \\&= 1+\cos\left(2x-\dfrac{\pi}{2}\right) \\ &= 1+\cos\left(2\left(x-\dfrac{\pi}{4}\right) \right) \\ &= 1+2\cos^2 \left(x-\dfrac{\pi}{4}\right)-1 \\ &= 2\cos^2 \left(x-\dfrac{\pi}{4}\right) \end{align}$

So that means that $f(x)=\sqrt{2}\cos \left(x-\dfrac{\pi}{4}\right)$

Is there another shorter way to arrive at this result? Also, is there a geometric interpretation of this that can be explained to someone who doesn't know most of the identities I've used?

PS: If this question has already been asked please leave a link for it. I honestly tried to search for a similar question before asking this.

Answer 1

Geometric interpretation:

Rotate the right triangle. The projected length of the $\sqrt 2$ hypothenuses will vary following a sinusoid.

At the same time, this projected length is $\cos\theta+\sin\theta$.

Answer 2

Using the cosine of angle difference formula,

$\sqrt2\cos\left(x-\dfrac\pi4\right)=\sqrt2\cos(x)\cos\left(\dfrac\pi4\right)+\sqrt2\sin(x)\sin\left(\dfrac\pi4\right)=\cos x+\sin x.$

Answer 3

Yes. Indeed all linear combinations of the form $a\cos x+b\sin x$ are sinusoids, as shown below:

Write this as $$\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}\left(a\cos x+b\sin x\right)=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\right)=R(\sin\phi\cos x+\cos\phi\sin x)=R\sin(\phi+x),$$ where $R=\sqrt{a^2+b^2}$ and $\tan\phi=a/b.$

In your case, you have $a=b=1,$ so that you have $$\sqrt 2\sin(π/4+x).$$