Question

sin(x+y)sin(x-y)=2sinxcosy

Answer 1

Frankyyyyy, eyyyyy.
What do we do? Prove this identity?

Answer 2

yes sir

Answer 3

I guess we could apply our `Angle Sum Formula` and `Angle Difference Formula` for sine, and expand out all the stuff we get.

Shouldn't be too bad, just a lot of multiplying.

Answer 4

i expanded it to sinxcosy-cosxsiny+sinxcosy-cosxsiny.

Answer 5

oh ok

Answer 6

So that's what you did.. ok ok interesting.

Answer 7

i just followed the sum and difference formulas

Answer 8

You messed up a sign somewhere, hmm.

Answer 9

\[\rm \color{red}{\sin(x+y)}\color{royalblue}{\sin(x-y)}\]So our formulas give us,\[\rm \color{red}{[\sin x \cos y+\sin y \cos x]}\color{royalblue}{[\sin x \cos y- \sin y \cos x]}\]

Answer 10

Oh oh oh, these are being added together right?\[\large\rm \sin(x+y)+\sin(x-y)\]Otherwise we're getting a big huge mess I think.
Was that just a typo?

Answer 11

yes

Answer 12

\[\rm \color{red}{\sin(x+y)}+\color{royalblue}{\sin(x-y)}\]So then our expression becomes,\[\rm \color{red}{\sin x \cos y+\sin y \cos x}+\color{royalblue}{\sin x \cos y- \sin y \cos x}\]

Answer 13

Ya I think you had the wrong sign on the second term,
should be positive, ya?

Answer 14

yeah

Answer 15

Understand what the next step is?

Answer 16

no thats where im stuck

Answer 17

We actually have some like-terms,
even though it might not seem like it.

Notice that we have a siny*cosx and a -siny*cosx, ya?

Answer 18

oh yeah they just cancel out

Answer 19

Good good good.
And you're left with sinx*cosy + sinx*cosy right? :)

Answer 20

right

Answer 21

again, combining like-terms, just two of those.
yay team!

Answer 22

wow. i thought too much of it. thanks.

Answer 23

can we do just 1 more?

Answer 24

tan(x+pi) - tan(pi-x) = 2tanx

Answer 25

Do they want us to use our Angle Sum/Difference formulas again?
The tangent ones are kind of annoying.

We can do something different in this case if you're comfortable with it.
Tangent is `periodic in pi`.

Which means that if you add pi to an angle, you get the same result as before,\[\large\rm \tan(x+\pi)=\tan x\]

Answer 26

my teacher wants me too, but I too think it's very annoying.

Answer 27

Ughhh D:

Answer 28

lets just do what you want to do though. For times sake

Answer 29

Lemme finish explaining it this way first,
it'll take like 1 minute,
and then we can move onto the longer approach if you like. :)

Answer 30

\[\large\rm \tan\left[\pi-x\right]=\tan\left[-(x-\pi)\right]\]Tangent is an `odd function`,
so it follows this property: \(\rm tan(-a)=-tan(a)\)
the negative can come outside.\[\large\rm \tan\left[-(x-\pi)\right]=-\tan(x-\pi)\]And again, since tangent is periodic in pi, x-pi is the same as x,\[\large\rm =-\tan(x)\]

Answer 31

\[\large\rm \color{orangered}{\tan(x+\pi)}-\color{royalblue}{\tan(\pi-x)}\quad=\quad \color{orangered}{\tan x}-\color{royalblue}{(-\tan x)}\]

Answer 32

=2tanx.

Ok so maybe that way is a little more complicated than I was letting on :)

Answer 33

i find this much better.

Answer 34

Ya as long as your teacher is ok with it :) hehe

Answer 35

im sure she doesnt mind seeing different ways that this problem can be solved. Thanks a ton!

Answer 36

np