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Homework Help: Trig identities (is my method correct?)

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove the following identities

    31c) sin([itex]\frac{\pi}{2}[/itex]+x)=cosx

    2. Relevant equations


    3. The attempt at a solution

    The idea here is to prove the identity by making LS=RS

    so here is what i have done, but im not sure if it is the right way, since the book shows it went about it in a different way.

    My method:



    now we use the pythagreon identity


    If we move the cos to the right side, we are left with sinx=-cosx

    and we use another identity


    therefore LS=RS and we have proved the identity?

    Book method:


    LS=sin([itex]\frac{\pi}{2}[/itex]+x) RS=cosx




    LS=RS therefore sin([itex]\frac{\pi}{2}[/itex]+x)=cosx

    Is my method correct? Also at the point where they do "=sin[itex]\frac{\pi}{2}[/itex]cosx+cosx+cos[itex]\frac{\pi}{2}[/itex]sinx" Are they still just dealing with LS only?


    Also if someone could please explain how the book got to the solution it did, what thought process does one have to use to get to the solution the way book did? Thanks
    Last edited: May 29, 2013
  2. jcsd
  3. May 29, 2013 #2


    Staff: Mentor

    The algebra in your method is not quite right:

    sin(pi/2 + x) =/= sin(pi/2) + sin(x) _________ (i.e. =/= 1 + sin(x) )

    with sin(pi/2) = 1

    the sine of the sum of two angles is:

    sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
  4. Jun 19, 2013 #3
    The mistake is in this line :
    From where you got this equation?
  5. Jun 19, 2013 #4


    User Avatar
    Homework Helper

    The book's method is not what you should be using, you should be using the CAST rule and the rule about cofunctions.
  6. Jun 19, 2013 #5


    Staff: Mentor

    The most straightforward approach to this problem is to use the identity for the sine of a sum of angles: sin(A + B) = sinA*cosB + cosA*sinB. You don't have it listed in your relevant identities/equations, but should.

    The mistake you made is thinking that sin(A + B) "distributes" over a sum to result in sinA + sinB. That is definitely NOT true.
  7. Jun 30, 2013 #6
    I was wondering the same thing.
    Perhaps, you meant to write:
    ## 1 - sin^2(x) = cos^2(x) ##

    Otherwise I am not sure.

    ## sin(A+B) = sin(A)cos(B) + sin(B)cos(A) ##
  8. Jul 1, 2013 #7


    Staff: Mentor

    I believe this is what the OP was doing:
    ##sin(\pi/2 + x) = cos(x)##
    ##sin(\pi/2) + sin(x) = cos(x)##
    1 + sin(x) = cos(x)

    IOW, the OP was to distribute in a situation for which it wasn't applicable.
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