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Thread: Odd Trig Identity

  1. #1
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    Odd Trig Identity

    How do I manipulate $\displaystyle sin(x)+cos(x)$ to show that it equals $\displaystyle \sqrt2cos(\frac{\pi}{4}-x)$

    Thanks!
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  2. #2
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    Hello, RubyRed!

    How do I manipulate $\displaystyle \sin x +\cos x $ to show that it equals $\displaystyle \sqrt{2}\,\cos\left(\frac{\pi}{4}-x\right)$
    We have: .$\displaystyle \sin x + \cos x$


    Multiply by $\displaystyle \frac{\sqrt{2}}{\sqrt{2}}\!:\quad \frac{\sqrt{2}}{\sqrt{2}}(\sin x + \cos x) \;=\;\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x\right)\;\;{\color{blue}[1]} $


    Note that: .$\displaystyle \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}\;\;\text{ and }\;\;\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} $


    Substitute into [1]: . $\displaystyle \sqrt{2}\left(\sin\frac{\pi}{4}\sin x + \cos\frac{\pi}{4}\cos x\right) $

    . . . . . . . . . . . . . $\displaystyle = \;\sqrt{2}\left(\cos\frac{\pi}{4}\cos x + \sin\frac{\pi}{4}\sin x\right) $

    . . . . . . . . . . . . . $\displaystyle = \;\sqrt{2}\,\cos\left(\frac{\pi}{4} - x\right) \;\;{\color{red}**}$

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    ** . The last step uses the identity:

    . . . . $\displaystyle \cos A\cos B + \sin A\sin B \;=\;\cos(A - B)$

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