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Math Help - Basic Identities

  1. #1
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    Basic Identities

    Hello there!

    I have this problem on proving identities. I've been working on it for hours, but can't seem to get it right. The sad thing is that it's not even a double angle formula

    cos x + sin x = [cos x / (1 - tan x)] + [sin x / (1 - cot x)]

    Any help would be greatly appreciated. Thanks!
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  2. #2
    MHF Contributor red_dog's Avatar
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    Start from the right side and write \tan x and \cot x in terms of \sin x and \cos x.
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  3. #3
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    Quote Originally Posted by red_dog View Post
    Start from the right side and write \tan x and \cot x in terms of \sin x and \cos x.
    Hi red_dog,

    Yes I did that but I end up getting an extremely long sequence which ends up with a factor of three for sin and cos, that's when I get stuck.
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  4. #4
    MHF Contributor red_dog's Avatar
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    \displaystyle\frac{\cos x}{1-\tan x}+\frac{\sin x}{1-\cot x}=

    =\displaystyle\frac{\cos x}{1-\displaystyle\frac{\sin x}{\cos x}}+\frac{\sin x}{1-\displaystyle\frac{\cos x}{\sin x}}=

    =\displaystyle\frac{\cos x}{\displaystyle\frac{\cos x-\sin x}{\cos x}}+\frac{\sin x}{\displaystyle\frac{\sin x-\cos x}{\sin x}}=

    =\displaystyle\frac{\cos^2x}{\cos x-\sin x}+\frac{\sin^2x}{\sin x-\cos x}=

    =\displaystyle\frac{\cos^2x-\sin^2x}{\cos x-\sin x}=

    =\displaystyle\frac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x-\sin x}=\cos x+\sin x
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  5. #5
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    Hey thanks red_dog!

    I realised my mistake now. I cross multiplied the fourth line (which was why i was getting the cube) instead of multiplying -1 to the second term.
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