Rewriting an expression in the form of $A \sin(x + C)$

Question

The problem asks to rewrite $$\sin(x) - \cos(x)$$ in the form of $A\sin(x + C)$, using the reduction formula.

The answer is supposed to be $\sqrt{2}\sin(x - \pi/4)$, or $\sqrt{2}\sin(x - 45)$ using degrees.

But from what the book is doing, I don't know what C is supposed to be or how to get it.

$\sin x - \cos x = \sqrt{2} \left({1 \over \sqrt 2} \sin x - {1 \over \sqrt 2} \cos x \right) $. Now use the formula for $\sin(x + y)$. — Simon S, Apr 05 '15 at 00:26
I wrote an answer elsewhere which explains how to make such reductions in general; it happens to address the particular case of $\cos(x)-\sin(x)$ which is just the negative of what you have. — Milo Brandt, Apr 05 '15 at 01:13
Here's a picture-proof of the general identity (with addition instead of subtraction) that could provide insights into why the formula is the way it is. — Blue, Apr 05 '15 at 01:16

score 3 · Accepted Answer · answered Apr 05 '15 at 00:37

This is an interesting question, because you wouldn't necessarily thing that the sum of two sine waves would give another sine wave and not some other wavy thing. The key to solving this problem is using two identities. The first is the complementary angle one: $\cos x = \sin (\pi/2-x)$ using radians. The second (which you may be calling the reduction formula) is the sum-to-product formula $\sin a - \sin b = 2 \cos ((a+b)/2) \sin ((a-b)/2)$. Applying these,

$$ \sin x - \cos x = \sin x - \sin (\pi/2 -x) = 2\cos(\pi/4) \sin (x-\pi/4) = \sqrt{2} \sin (x-\pi/4) $$

which is close to the answer you gave.

The "$C$" is essentially the phase shift of the new sign function that occurs because you are making a new function. Alternatively, you could take Simon's hint (which may or may not be more obvious) by expanding $\sqrt{2} \sin (x+\pi/4)$ and working backwards.

score 1 · Answer 2 · answered Apr 05 '15 at 12:39

Let $\varphi$ be an angle such that \begin{align*} \cos\varphi & = \frac{b}{\sqrt{a^2 + b^2}}\\ \sin\varphi & = \frac{a}{\sqrt{a^2 + b^2}} \end{align*} Then $b\sin\theta - a\cos\theta = \sqrt{a^2 + b^2}\sin(\theta - \varphi)$ since \begin{align*} \sqrt{a^2 + b^2}\sin(\theta - \varphi) & = \sqrt{a^2 + b^2}(\sin\theta\cos\varphi - \cos\theta\sin\varphi\\ & = \sqrt{a^2 + b^2}\left(\sin\theta \cdot \frac{b}{\sqrt{a^2 + b^2}} - \cos\theta \cdot \frac{a}{\sqrt{a^2 + b^2}}\right)\\ & = b\sin\theta - a\cos\theta \end{align*} In the expression $\sin x - \cos x$, $a = 1$ and $b = 1$. Thus, $\sqrt{a^2 + b^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$. Hence, \begin{align*} \cos\varphi & = \frac{1}{\sqrt{2}}\\ \sin\varphi & = \frac{1}{\sqrt{2}} \end{align*} Therefore, $$\varphi = \arccos\left(\frac{1}{\sqrt{2}}\right) = \arcsin\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$$ Substituting $1$ for $a$, $1$ for $b$, $x$ for $\theta$, and $\pi/4$ for $\varphi$ in the formula $$b\sin\theta - a\cos\theta = \sqrt{a^2 + b^2}\sin(\theta - \varphi)$$ yields $$\sin x - \cos x = \sqrt{2}\sin\left(x - \frac{\pi}{4}\right)$$

Rewriting an expression in the form of $A \sin(x + C)$

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