# What is sin(x)+cos(x) in terms of sine?

Apr 15, 2015

Please see two possibilities below and another in a separate answer.

#### Explanation:

Using Pythagorean Identity

${\sin}^{2} x + {\cos}^{2} x = 1$, so ${\cos}^{2} x = 1 - {\sin}^{2} x$

$\cos x = \pm \sqrt{1 - {\sin}^{2} x}$

$\sin x + \cos x = \sin x \pm \sqrt{1 - {\sin}^{2} x}$

Using complement / cofunction identity

$\cos x = \sin \left(\frac{\pi}{2} - x\right)$

$\sin x + \cos x = \sin x + \sin \left(\frac{\pi}{2} - x\right)$

Aug 19, 2017

I've learned another way to do this. (Thanks Steve M.)

#### Explanation:

Suppose that $\sin x + \cos x = R \sin \left(x + \alpha\right)$

Then

$\sin x + \cos x = R \sin x \cos \alpha + R \cos x \sin \alpha$

$= \left(R \cos \alpha\right) \sin x + \left(R \sin \alpha\right) \cos x$

The coefficients of $\sin x$ and of $\cos x$ must be equal so

$R \cos \alpha = 1$
$R \sin \alpha = 1$

Squaring and adding, we get

${R}^{2} {\cos}^{2} \alpha + {R}^{2} {\sin}^{2} \alpha = 2$ so ${R}^{2} \left({\cos}^{2} \alpha + {\sin}^{2} \alpha\right) = 2$

$R = \sqrt{2}$

And now

$\cos \alpha = \frac{1}{\sqrt{2}}$
$\sin \alpha = \frac{1}{\sqrt{2}}$

so $\alpha = {\cos}^{-} 1 \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$

$\sin x + \cos x = \sqrt{2} \sin \left(x + \frac{\pi}{4}\right)$