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Math Help - trigonometry (1)

  1. #1
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    trigonometry (1)

    Show that sin^2x + \frac{1}{2}cos x = \frac{17}{16}-(\cos x-\frac{1}{4})^2

    I can do this part .

    Hence , state the range

    \frac{1}{2\sin^2 x+\cos x-1} for 0\leq x\leq 2\pi

    Need help with this part . Thanks .
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  2. #2
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    Show that sin^2x + \frac{1}{2}cos x = \frac{17}{16}-(\cos x-\frac{1}{4})^2

    I can do this part .

    Hence , state the range

    \frac{1}{2\sin^2 x+\cos x-1} for 0\leq x\leq 2\pi

    Need help with this part . Thanks .
    \frac{1}{2\sin^2 x+\cos x-1}= \frac{\frac12}{\sin^2x+\tfrac12\cos x - \tfrac12}

    =\frac{\frac12}{\frac{17}{16}-(\cos x - \frac14)^2-\tfrac12}

    =\frac{\frac12}{\frac{9}{16}-(\cos x - \frac14)^2}

    Now use the fact that 0\le(\cos x -\tfrac14)^2\le\tfrac{25}{16} to find the range of values of this expression. (Can you see why?)

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello thereddevils \frac{1}{2\sin^2 x+\cos x-1}= \frac{\frac12}{\sin^2x+\tfrac12\cos x - \tfrac12}

    =\frac{\frac12}{\frac{17}{16}-(\cos x - \frac14)^2-\tfrac12}

    =\frac{\frac12}{\frac{9}{16}-(\cos x - \frac14)^2}

    Now use the fact that 0\le(\cos x -\tfrac14)^2\le\tfrac{25}{16} to find the range of values of this expression. (Can you see why?)

    Grandad
    Thanks Grandad , but i don see how you got that range ? Can you help me a little further
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  4. #4
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    Thanks Grandad , but i don see how you got that range ? Can you help me a little further
    The cosine of any angle lies between -1 and +1. So the maximum value of (\cos x - \tfrac14) is 1-\tfrac14=\tfrac34 and its minimum value is -1-\tfrac14=-\tfrac54.

    But when we square it, (\cos x - \tfrac14)^2 is always positive, and now lies between 0 and (-\tfrac54)^2 = \tfrac{25}{16}.

    OK?

    Grandad
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