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

Question

286220 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-15T18:47:39

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

Using Pythagorean Identity

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

$cosx = +- sqrt (1-sin^2x)$

$sinx + cosx = sinx +- sqrt (1-sin^2x)$

Using complement / cofunction identity

$cosx = sin(pi/2-x)$

$sinx + cosx = sinx + sin(pi/2-x)$

Answer 2 · 2017-08-19T14:01:13

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

Suppose that $sinx+cosx=Rsin(x+alpha)$

Then

$sinx+cosx=Rsinxcosalpha+Rcosxsinalpha$

$=(Rcosalpha)sinx+(Rsinalpha)cosx$

The coefficients of $sinx$ and of $cosx$ must be equal so

$Rcosalpha = 1$
$Rsinalpha=1$

Squaring and adding, we get

$R^2cos^2alpha+R^2sin^2alpha = 2$ so $R^2(cos^2alpha+sin^2alpha) = 2$

$R = sqrt2$

And now

$cosalpha = 1/sqrt2$
$sinalpha = 1/sqrt2$

so $alpha = cos^-1(1/sqrt2) = pi/4$

$sinx+cosx = sqrt2sin(x+pi/4)$