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Trigonometric identities? watch

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    Im supposed to be using the trigonometic identities:
    \tan Q = \frac{\sin Q}{\cos Q}
    \sin^2Q + \cos^2Q = 1

    to solve:
    1.(1+\sin x)^2 +(1-\sin x)^2 + 2\cos^2x
    2.\sin^4Q+\sin^2Q\cos^2Q

    Once again... not a clue where to start(and yes ive read the book, the examples are crap)
    1. = 4 and 2. = 1
    For number 1, i know that 1-\sin^2x = \cos^2x
    so (1+\sin x)^2 +\cos^2x+2\cos^2x
    so(1+\sin^2x)+3\cos^2x
    But i cant re-arrange the formula to make 1+sinx the subject formula

    For number 2 i think i might need to use the 'difference between two squares'

    Any help much appreciated, just hints etc
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    What you have done seems wrong, you need to remember (a+b)^2=a^2+2ab+b^2 so expanding e.g. (1+sin(x))^2 gives 1+2sin(x)+sin^2(x).

    Expand everything correctly and use c^2+s^2=1 to rearrange it and then factorise into something nice


    For question 2 sub cos^2(x)=1-sin^2(x)
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    Thanks!
    so, for 1:
    (1+2\sin x+\sin^2x)+(1-2\sin x+\sin^2x)+2\cos^2x
    = 2 + 2\sin^2x + 2\cos^2x = 4

    Then for 2:
    \sin^4Q+sin^2Q(1-\sin^2Q)
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    yup, you've got 1 correct

    I'm not totally sure what your question wants in 2 - multiplying together what you have will leave sin^2(Q) - is that okay?
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    Yea sorry, diddnt finish it off on here
    Thats alot +rep
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    An alternative solution to 2 would be to factorise:

    \sin^4 x + \sin^2 x \cos^2 x = \sin^2 x (\sin^2 x +\cos^2 x) = \sin^2 x
 
 
 
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Updated: August 31, 2007

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