Can You Solve This Trigonometric Identity?

[stuck]

Prove the following identities:

(cos A = cos B)^2 + (sin A + sin B)^2 = 2[1+cos(A-B)]


I'm really a mess at this stuff. I missed a few important days and fell behind, so I don't reeeally know what to do when things start getting squared and whatnot., but I tried!

Left Side
(cos A + Cos B)^2 + (sin A + sin B)^2
(cos(^2)A+cos(^2)B = Sin(^2)A+sin(^2)B
(cos (^2)A+cos(^2)B+ (1+-cos(^2)A + (1+-cos(^2)B)
2

Right Side
2+2cos(A-B)
2(cosAcosB+sinAsinB) +2

But that's as far as I can get. I can't find a way to make both sides equal, and I'm not even sure if my left side is correct...

Please help me if you can! :shy:

 

[sutupidmath]

First, do you know what $$(a+b)^2=...??$$ you have not expanded well. Give it another shot.

$$(cosA+cosB)^2=...?...$$

$$(sinA+sinB)^2=...?$$

After you expand this, remember also that:

$$cos^2A+sin^2A=1, and, sin^2B+cos^2B=1$$

And somewhere in between you will end up with sth like this

$$2+ 2cosBcosA+2sinAsinB$$ now the answer should be obvious, right?

 

[stuck]

Urk, I thought that because in my third step on the left I had (cos^2A) and (-cos ^2A), that they would cancel...?

 

[sutupidmath]

$$(sinA+sinB)^2=sin^2A+2sinAsinB+sin^2B...?$$

also

$$(cosA+cosB)^2=cos^2A+2cosAcosB+cos^2B$$


Can you go from here now?

Just use the hints that i gave you on my post #2

also, another hint, altough you should have figured this out by yourself

$$cos(A-B)=cosAcosB+sinAsinB$$

 

[stuck]

I think I can I didnt know that before, I will try to work it out now!

 

[sutupidmath]

I think I can I didnt know that before, I will try to work it out now!

I wrote in there everything you need to do that problem, you only need to put all the hints together.

 

[stuck]

I got it, Thank you!

 

[sutupidmath]

I got it, Thank you!

You're welcome. THis is what PF is for!