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    How did I get +24cos(4theta) instead of -24cos(4theta)? Name:  ImageUploadedByStudent Room1464009192.884314.jpg
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    Posted from TSR Mobile
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    Could you post the question?
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    (Original post by TSRforum)
    How did I get +24cos(4theta) instead of -24cos(4theta)? Name:  ImageUploadedByStudent Room1464009192.884314.jpg
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    Posted from TSR Mobile
    Several things are wrong.

    1. It should be \int_0^{\pi/6} + 48\cdots \, \mathrm{d} \theta

    2. \sin \theta \sin 3\theta = \frac{1}{2}(\cos 2x - \cos 4x). i.e: \cos(-2x) = \cos 2x \neq -\cos 2x as you seem to think.
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    (Original post by Zacken)
    Several things are wrong.

    1. It should be \int_0^{\pi/6} + 48\cdots \, \mathrm{d} \theta

    2. \sin \theta \sin 3\theta = \frac{1}{2}(\cos 2x - \cos 4x). i.e: \cos(-2x) = \cos 2x \neq -\cos 2x as you seem to think.
    So when it says (A-B) in the trig identity I should look at that as the difference between A and B?


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    (Original post by TSRforum)
    So when it says (A-B) in the trig identity I should look at that as the difference between A and B?


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    That's generally what subtraction means, yes.
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    (Original post by Zacken)
    That's generally what subtraction means, yes.
    Not sure, difference of x and y might mean |x-y|.

    If by their last post OP means A-B should be taken as |A-B| then yes but only because \cos(x) \equiv \cos(-x).
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    (Original post by IrrationalRoot)
    Not sure, difference of x and y might mean |x-y|.

    If by their last post OP means A-B should be taken as |A-B| then yes but only because \cos(x) \equiv \cos(-x).
    Good point, I wasn't sure what the OP was asking.
 
 
 
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