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Math Help - Can someone check my trig work please?

  1. #1
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    Can someone check my trig work please?

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  2. #2
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    Well, to start

    -\frac{25\pi}{4} = -\frac{24\pi}{4} - \frac{\pi}{4}

     = -6\pi - \frac{\pi}{4}.


    So which basic angle do you think -\frac{25\pi}{4} is the same as?
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  3. #3
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    It is preferred you post one question per thread.

    Consider -2\pi gives the same postition as  0 so

    \frac{-25\pi}{4} =\frac{(-8-8-8-1)\pi}{4}=\frac{-8\pi}{4}+\frac{-8\pi}{4}+\frac{-8\pi}{4}+\frac{-1\pi}{4} =-2\pi-2\pi-2\pi-\frac{\pi}{4} = -\frac{\pi}{4}
    Last edited by pickslides; June 30th 2010 at 09:48 PM. Reason: formatting
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    Quote Originally Posted by pickslides View Post
    It is preferred you post one question per thread.

    Consider -2\pi gives the same postition as  0 so

    \frac{-25\pi}{4} =\frac{(-8-8-8-1)\pi}{4}=\frac{(-8-8-8-1)\pi}{4}=\frac{-8)\pi}{4}+\frac{-8)\pi}{4}+\frac{(-8)\pi}{4}+\frac{-1\pi}{4} -2/pi-2/pi-2/pi-\frac{\pi}{4} = -\frac{-1\pi}{4}
    Be careful with your equal signs...

    -\frac{25\pi}{4} \neq -\frac{\pi}{4}.

    But you can say that the POSITION of the angle -\frac{25\pi}{4} is the same POSITION as -\frac{\pi}{4}.
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  5. #5
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    Quote Originally Posted by Prove It View Post

    But you can say that the POSITION of the angle -\frac{25\pi}{4} is the same POSITION as -\frac{\pi}{4}.
    But I did mention the position ;0)
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    Quote Originally Posted by pickslides View Post
    But I did mention the position ;0)
    Yes, but you're still using equals signs where they shouldn't be used...

    Anyway, Question 2 is correct.

    In Question 3, you have \sin{\theta} correct. For the rest you could have used the identities:

    \sin^2{\theta} + \cos^2{\theta} = 1

    \frac{\sin{\theta}}{\cos{\theta}} = \tan{\theta}

    \tan^2{\theta} + 1 = \sec^2{\theta}

    \cot{\theta} = \frac{1}{\tan{\theta}}.


    If you were going to use the triangle, you have used Pythagoras' Theorem incorrectly.

    It should be A^2 = H^2 - O^2. You added them.
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    thanks! correcting.

    Sorry about more than one Q per post.
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  8. #8
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    Quote Originally Posted by Endymion View Post
    thanks! correcting.

    Sorry about more than one Q per post.
    Your Question 4 and 5 are correct.


    Q. 6. (a) is correct but I would have done

    \csc{\frac{\pi}{3}} = \frac{1}{\sin{\frac{\pi}{3}}} = \frac{1}{\sin{60^{\circ}}}.

    (b) is correct.


    Q. 7. Remember that \sec^2{\frac{\pi}{6}} = \frac{1}{\cos^2{\frac{\pi}{6}}} = \frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{\frac{3}{4}} = \frac{4}{3}.

    So \sec^2{\frac{\pi}{3}} -  4 = \frac{4}{3} - 4 = \frac{4}{3} - \frac{12}{3} = -\frac{8}{3}.
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