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Math Help - Trig help

  1. #1
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    Nov 2008
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    Trig help

    hey can someone show me how to solve these questions.....thanks

    (1) Prove 1 + sin theta - cos theta = 2 sin (1/2 theta) [ cos (1/2 theta) + sin (1/2 theta) ]

    (2) hence or otherwise, solve this equation

    1 + sin theta - cos theta = 0.


    thanks
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  2. #2
    Super Member

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    Hello, coremaths!

    (1) Prove: . 1 + \sin\theta - \cos\theta \;=\;  2\sin\tfrac{\theta}{2}\left(\cos\tfrac{\theta}{2} + \sin\tfrac{\theta}{2}\right)

    On the right: . \underbrace{2\sin\tfrac{\theta}{2}\cos\tfrac{\thet  a}{2}} + \underbrace{2\sin^2\!\tfrac{\theta}{2}}

    . . . . . . . . = \quad\;\sin\theta \quad+\quad 1 - \cos\theta




    2) Hence or otherwise, solve this equation: . 1 + \sin\theta - \cos\theta \;=\; 0

    We have: . 2\sin\tfrac{\theta}{2}\left(\cos\tfrac{\theta}{a} - \sin\tfrac{\theta}{2}\right) \;=\;0


    Then: . \sin\tfrac{\theta}{2} = 0 \quad\Rightarrow\quad \tfrac{\theta}{2} = \pi n \quad\Rightarrow\quad \boxed{\theta \:=\:2\pi n}

    And: . \cos\tfrac{\theta}{2} - \sin\tfrac{\theta}{2} \:=\:0\quad\Rightarrow\quad \sin\tfrac{\theta}{2} \:=\:\cos\tfrac{\theta}{2} \quad\Rightarrow\quad \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} \:=\:1

    . . \tan\tfrac{\theta}{2} \:=\:1\quad\Rightarrow\quad \tfrac{\theta}{2} \:=\:\tfrac{\pi}{4} + \pi n \quad\Rightarrow\quad\boxed{\theta \:=\:\tfrac{\pi}{2} + 2\pi n}

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  3. #3
    Newbie
    Joined
    Nov 2008
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    hi there

    thanks for the reply, would you mind just explaining question 1, in a bit more depth as i don't understand how u got the sin theta and 1 - cos theta.

    thanks
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