How do you prove {sin(x+y)sin(x)cos(y)}=1cot(x)tan(y)?

2 Answers
Apr 7, 2017

Q.E.D. See below.

Explanation:

I'm going to assume you meant sin(x+y)sinxcosy=1+cotxtany and just forgot parentheses on the left side denominator and the plus on the right side, otherwise the identity is false.

Let's start working the left hand side by using the sine addition formula sin(α+β)=sinαcosβ+cosαsinβ
sin(x+y)sinxcosy=sinxcosy+cosxsinysinxcosy

Then split the numerator over the denominator
sinxcosy+cosxsinysinxcosy=sinxcosysinxcosy+cosxsinysinxcosy

Simplify and factor
sinxcosysinxcosy+cosxsinysinxcosy=1+(cosxsinx)(sinycosy)

And knowing that tanθ=sinθcosθ and cotθ=cosθsinθ,
1+(cosxsinx)(sinycosy)=1+cotxtany

Therefore the left side is equal to the right side. Q.E.D.

Apr 7, 2017

Assuming the questioner meant sin(x+y)sinxcosy=1+cotxtany, then the proof is as follows:

Explanation:

sin(x+y)sinxcosy=sinxcosy+cosxsinysinxcosy=

sinxcosysinxcosy+cosxsinx×sinycosy=

1+cotxtany QED/ΟΕΔ