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Math Help - Inverse function same as original?

  1. #1
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    Inverse function same as original?

    Hi guys,

    Working on these last questions on my practice exam and I'm pulling my hair out...ha:

    Solve the following equations in [0, 2pi]:

    (a) sin^2 x + sin x = cos^2 x

    I ended up with one solution, x = 3pi/2, wondering if it's ok

    (b) sin 2x cos x + cos 2x sin x = 1

    This one got me pretty angry but I figure there is simple solution.

    I also have an aside: In my solutions manual there was a step that sort of threw me, it was about solving a trigonometric identity and they did something I hadn't seen before, they turned sin^2 x - 3 cos^2 x into sin^2 x + cos^2 x - 4 cos^2 x in order to use the pythagorean identity. So you're allowed to rewrite -3 cos^2 x as cos^2 x - 4cos^2 x? If so it's news to me...

    Thanks for any help, it's much appreciated.
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  2. #2
    MHF Contributor
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    Hi

    Quote Originally Posted by DannyMath View Post
    Hi guys,

    Working on these last questions on my practice exam and I'm pulling my hair out...ha:

    Solve the following equations in [0, 2pi]:

    (a) sin^2 x + sin x = cos^2 x

    I ended up with one solution, x = 3pi/2, wondering if it's ok
    Not exactly
    sinē x + sin x = cosē x
    sinē x + sin x = 1 - sinē x
    2 sinē x + sin x - 1 = 0
    which is a quadratic in sin x
    sin x = -1 or sin x = 1/2
    x = 3pi/2 or x = pi/6 or x = 5pi/6

    Quote Originally Posted by DannyMath View Post
    (b) sin 2x cos x + cos 2x sin x = 1

    This one got me pretty angry but I figure there is simple solution.
    Use sin(a+b) = sin a . cos b + sin b . cos a
    with a = 2x and b = x

    Quote Originally Posted by DannyMath View Post
    I also have an aside: In my solutions manual there was a step that sort of threw me, it was about solving a trigonometric identity and they did something I hadn't seen before, they turned sin^2 x - 3 cos^2 x into sin^2 x + cos^2 x - 4 cos^2 x in order to use the pythagorean identity. So you're allowed to rewrite -3 cos^2 x as cos^2 x - 4cos^2 x? If so it's news to me...
    You can factor out cosē x
    cosē x - 4 cosē x = cosē x (1-4) = -3 cosē x

    It is exactly the same as -3y = y - 4y with y = cosē x
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