Show in the form $A\sin(x+c)=\sin x - \cos x$?

Question

Not sure what identity I should be using here: My gut tells me to use the Sin sum formula: $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$, but can't figure out how to.

Answer 1

Yes, you want to use that form. You have: $$A\sin(x+y)=A\sin x \cos y + A\sin y \cos x=\sin x - \cos x$$

So that means that $$A\cos y=1, A\sin y=-1$$

squaring the equations and adding them, we see that $$A^2\cos^2 y+A^2\sin^2 y=2$$ $$A^2=2$$ $$A=\pm\sqrt{2}$$

Let's take $A=\sqrt{2}$. Putting this into our first two equations, this implies that:

$$\sqrt{2}\cos y=1$$ $$\cos y=\frac 1 {\sqrt 2}$$

And similarly $$\sin y=\frac{-1}{\sqrt 2}$$

$y=-\pi/4$ solves both of these. So our answer is

$$\sin x - \cos x = A\sin(x+c)=\sqrt{2}\sin(x-\frac{\pi}4)$$

Note that there are other solutions that will work, but this is probably the simplest.

Answer 2

We can always express $a\sin x+b\cos x$ in the form $R\sin(x+\theta)$ where $R \ge 0$

If $a\sin x+b\cos x=R\sin(x+\theta)$

Or if $a\sin x+b\cos x=R\sin x\cos\theta + R\cos x\sin\theta$

Comparing the coefficients of $\sin x$ and $\cos x$,

$a=R\cos\theta$ and $b=R\sin\theta$.

Squaring & adding we get, $R^2=a^2+b^2=>R=\sqrt{a^2+b^2}$ as $R\ge 0$

Diving we get $\frac{R\sin\theta}{R\cos\theta}=\frac{b}{a}$,

or $\tan\theta=\frac{b}{a}\Rightarrow\theta=\tan^{-1}(\frac{b}{a})$.

Here, $a=1$, $b=-1$, so $R=\sqrt2$

and $\theta=\tan^{-1}(\frac{-1}{1})=\tan^{-1}(-1)$

Now this demands a bit care as $\tan^{-1}(-1)=n\pi-\frac{\pi}{4}$
=> $\theta$ can lie in the 2nd or in the 4th quadrant.

Observe that here $\cos\theta=\frac{1}{\sqrt2}$ and $\sin\theta=-\frac{1}{\sqrt2}$

So, $\theta$ lies in the 4th quadrant.

So, $\theta=2m\pi-\frac{\pi}{4}$ where m is any integer.