How do you simplify #cos(u+v)cosv+sin(u+v)sinv#?

Question

11628 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-05T00:44:03

The expression can be simplified to #cosu#.

We need to start by expanding the #cos(A + B)# and the #sin(A + B)# using the sum and difference identities, as shown in the following image.

![http://www.regentsprep.org/regents/math/algtrig/att14/http://formulalesson.htm](https://useruploads.socratic.org/5YltLONvT86Xt1YrQQBu_formul30.gif)

Expanding, we have:

#=>(cosucosv - sinusinv)(cosv) + (sinucosv + cosusinv)(sinv)#

#=>cosucos^2v - sinusinvcosv + sinucosvsinv + cosusin^2v#

#=>cosucos^2v + cosusin^2v#

#=>cosu(cos^2v + sin^2v)#

Applying the pythagorean identity #sin^2x + cos^2x = 1#:

#=>cosu#

Hopefully this helps!