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Math Help - How do you derive the following trig identity?

  1. #1
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    How do you derive the following trig identity?



    Cheers. Any help is appreciated.
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  2. #2
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    Hello, Kaitosan!

    \sin x + \cos x \:=\:\sqrt{2}\,\sin(x + \tfrac{\pi}{4})
    Start with the right side . . .


    \sqrt{2}\,\sin(x + \tfrac{\pi}{4}) \;=\;\sqrt{2}\left(\sin x\cos\tfrac{\pi}{4} + \cos x\sin\tfrac{\pi}{4}\right)

    . . . . . . . . . = \;\sqrt{2}\,\bigg[\sin x\cdot\frac{1}{\sqrt{2}} + \cos x\cdot\frac{1}{\sqrt{2}}\bigg]

    . . . . . . . . . = \;\sin x + \cos x

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  3. #3
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    Actually... I'd rather you try with the left side (I have specific reasons). Can you do that....?

    This is as far as I can go -

    (sinx)^2 + (cosx)^2 = 1

    (sinx + cosx)^2 = 1 - sin2x

    sinx + cosx = sqrt(1 - sin2x)

    .................................................. ............
    Last edited by Kaitosan; January 31st 2009 at 12:00 AM.
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  4. #4
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    Hi

    This is not the good way to proceed.

    a \sin x + b \cos x = \sqrt{a^2+b^2} \left(\frac{a}{\sqrt{a^2+b^2}}  \sin x + \frac{b}{\sqrt{a^2+b^2}} \cos x \right)

    \frac{a}{\sqrt{a^2+b^2}} and \frac{b}{\sqrt{a^2+b^2}} can be seen as the cos and the sin of an angle \theta because the sum of their square is equal to 1

    a \sin x + b \cos x = \sqrt{a^2+b^2} \left(\sin x \cos \theta+ \cos x \sin \theta\right) = \sqrt{a^2+b^2} \:\sin (x + \theta)
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  5. #5
    MHF Contributor red_dog's Avatar
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    \sin x+\cos x=\sin x+\sin\left(\frac{\pi}{2}-x\right)=

    =2\sin\frac{x+\frac{\pi}{2}-x}{2}\cos\frac{x-\frac{\pi}{2}+x}{2}=

    =2\sin\frac{\pi}{4}\cos\left(x-\frac{\pi}{4}\right)=

    =2\frac{\sqrt{2}}{2}\cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}\sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=

    =\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)
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  6. #6
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    Quote Originally Posted by Kaitosan View Post
    Actually... I'd rather you try with the left side (I have specific reasons). Can you do that....?

    This is as far as I can go -

    (sinx)^2 + (cosx)^2 = 1

    (sinx + cosx)^2 = 1 - sin2x

    sinx + cosx = sqrt(1 - sin2x)

    .................................................. ............
    Can't you just take Soroban's answer and play it in reverse.

    For example. First take \sqrt{2} out as a factor

    \sin x + \cos x = \sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x\right)

    Then observe that \sin \frac{\pi}{4} = \cos \frac{\pi}{4}= \frac{1}{\sqrt{2}}, and therefore

    \qquad = \sqrt{2}\left(\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}\right)

    And then from the compound angle formula for sin

    \qquad = \sqrt{2}\sin \left(x + \frac{\pi}{4}\right)
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  7. #7
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    Quote Originally Posted by red_dog View Post
    \sin x+\cos x=\sin x+\sin\left(\frac{\pi}{2}-x\right)=

    =2\sin\frac{x+\frac{\pi}{2}-x}{2}\cos\frac{x-\frac{\pi}{2}+x}{2}=

    =2\sin\frac{\pi}{4}\cos\left(x-\frac{\pi}{4}\right)=

    =2\frac{\sqrt{2}}{2}\cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}\sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=

    =\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)
    Wow. Um... I'm completely lost on the second step. Hmm.
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  8. #8
    MHF Contributor red_dog's Avatar
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    In the second step I used the identity

    \sin x+\sin y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}
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  9. #9
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    Interesting! Could you please explain how that identity comes into being because I've never seen it? Right now I'm in calculus and I just want to make sure I know all the identities.
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