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Homework Help: I need some help-Trig Identities

  1. May 2, 2008 #1
    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! :bugeye:

    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:
     
    Last edited: May 2, 2008
  2. jcsd
  3. May 2, 2008 #2
    First, do you know what [tex](a+b)^2=...???????[/tex] you have not expanded well. Give it another shot.

    [tex] (cosA+cosB)^2=...?....[/tex]

    [tex](sinA+sinB)^2=....?[/tex]

    After you expand this, remember also that:

    [tex] cos^2A+sin^2A=1, and, sin^2B+cos^2B=1[/tex]

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

    [tex]2+ 2cosBcosA+2sinAsinB[/tex] now the answer should be obvious, right?
     
    Last edited: May 2, 2008
  4. May 2, 2008 #3
    Urk, I thought that because in my third step on the left I had (cos^2A) and (-cos ^2A), that they would cancel...?
     
  5. May 2, 2008 #4
    [tex](sinA+sinB)^2=sin^2A+2sinAsinB+sin^2B....?[/tex]

    also

    [tex] (cosA+cosB)^2=cos^2A+2cosAcosB+cos^2B[/tex]


    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

    [tex]cos(A-B)=cosAcosB+sinAsinB[/tex]
     
  6. May 2, 2008 #5
    I think I can I didnt know that before, I will try to work it out now!
     
  7. May 2, 2008 #6
    I wrote in there everything you need to do that problem, you only need to put all the hints together.
     
  8. May 2, 2008 #7
    I got it, Thank you!
     
  9. May 2, 2008 #8
    You're welcome. THis is what PF is for!
     
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