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

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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)#