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anonymous

  • one year ago

Verify: sin (x+y)cos(x-y)=sin(x) cos(x)+cos(y)sin(y)

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  1. welshfella
    • one year ago
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    use the compound angle formula to expand the left side then simplify. I think you will be able to make use of the identity sin^2 x + cos^2 x = 1

  2. welshfella
    • one year ago
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    sin(x + y) = sin x cos y + sin y cos x cos (x - y) = cos x cos y + sin x sin y

  3. anonymous
    • one year ago
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    Well once it's broken down where do I go from there?

  4. welshfella
    • one year ago
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    you should get what is on the right side so proving the identity

  5. welshfella
    • one year ago
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    its a bit long winded but that is the way to do it

  6. anonymous
    • one year ago
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    Can u show me all the way to the answer? My brain is ready to fall outta my ears from trying to figure this out.

  7. zepdrix
    • one year ago
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    Ehhh I can maybe show you a few steps -_- let's see here....

  8. zepdrix
    • one year ago
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    \[\large\rm \color{orangered}{\sin(x + y) = \sin x \cos y + \sin y \cos x}\]\[\large\rm \color{royalblue}{\cos (x - y) = \cos x \cos y + \sin x \sin y}\]We'll apply these identities to our problem:\[\large\rm \color{orangered}{\sin(x+y)}\color{royalblue}{\cos(x-y)}\]Which will give us:\[\large\rm \color{orangered}{\left[\sin x \cos y + \sin y \cos x\right]}\color{royalblue}{\left[\cos x \cos y + \sin x \sin y\right]}\]Ok with that first step? :)

  9. anonymous
    • one year ago
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    Yes I understand that!

  10. zepdrix
    • one year ago
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    So hmm.. I guess we have to FOIL from here.

  11. zepdrix
    • one year ago
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    \[\rm =\color{orangered}{\sin x \cos y}\color{royalblue}{\cos x \cos y}+\color{orangered}{\sin x \cos y}\color{royalblue}{\sin x \sin y}\\+\color{orangered}{\sin y \cos x}\color{royalblue}{\cos x \cos y}+\color{orangered}{\sin y \cos x}\color{royalblue}{\sin x \sin y}\]Which becomes this I suppose.

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