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the question

How to simplify $\sin(x-y)\cos(y)+\cos(x-y)\sin(y)$

my steps

I tried to use trig identities on the $\sin(x-y)$ and $\cos(x-y)$ and tried to distribute the others in but it didn't work. Any ideas?

Michael Hardy
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John Rawls
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    Use the identity the other way around: sin(a+ b)= sin(a)cos(b)+ cos(a)sin(a+ b) with a= x- y, b= y. – user247327 Jul 13 '16 at 23:57
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    Your question is very easy. Another way, use a plotter with slider control for the curve $\sin(x-a)\cos(a)+\cos(x-a)\sin(a)$ and see that the curve shown is not altered with the variation of $a$. What do you suggest? – Piquito Jul 14 '16 at 00:24

4 Answers4

8

Let's do the (harder) method attempted by the OP, "but it didn't work". $$ \sin(x-y)\cos(y)+\cos(x-y)\sin(y) \\ = \big[\sin(x)\cos(y)-\cos(x)\sin(y)\big]\cos(y)+\big[\cos(x)\cos(y)+\sin(x)\sin(y)\big]\sin(y) \\ = \sin(x)\cos(y)\cos(y)-\cos(x)\sin(y)\cos(y)+\cos(x)\cos(y)\sin(y)+\sin(x)\sin(y)\sin(y) \\= \sin(x)\cos^2(y)+\sin(x)\sin^2(y) \\= \sin(x)\big[\cos^2(y)+\sin^2(y)\big] \\= \sin(x)\big[ 1 \big] \\=\sin(x) $$

GEdgar
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HINT:

Recall that $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$. Now, let $a=x-y$ and $b=y$.

Mark Viola
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5

Knowing the solution (using methods posted by others) suggests an alternate solution. Apply $\frac{\partial}{\partial y}$ to your expression and you get:

$$-\cos(x-y)\cos(y)-\sin(x-y)\sin(y)+\sin(x-y)\sin(y)+\cos(x-y)\cos(y)=0$$

So the expression is actually constant in $y$. It takes on the same value for all values of $y$ as for when $y=0$:

$$\begin{align} &\sin(x - 0)\cos(0)+\cos(x-0)\sin(0)\\ &=\sin(x)\end{align}$$

2'5 9'2
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by using well known identity we get

$$\\ \sin { x=\sin { \left( x-y+y \right) = } } \sin (x-y)\cos (y)+\cos (x-y)\sin (y)$$

haqnatural
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