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Thread: trigonometry (1)

  1. #1
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    trigonometry (1)

    Show that $\displaystyle 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

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

    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 $\displaystyle 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

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

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

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

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

    Now use the fact that $\displaystyle 0\le(\cos x -\tfrac14)^2\le\tfrac{25}{16}$ 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$\displaystyle \frac{1}{2\sin^2 x+\cos x-1}= \frac{\frac12}{\sin^2x+\tfrac12\cos x - \tfrac12}$

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

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

    Now use the fact that $\displaystyle 0\le(\cos x -\tfrac14)^2\le\tfrac{25}{16}$ 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 $\displaystyle -1$ and $\displaystyle +1$. So the maximum value of $\displaystyle (\cos x - \tfrac14)$ is $\displaystyle 1-\tfrac14=\tfrac34$ and its minimum value is $\displaystyle -1-\tfrac14=-\tfrac54$.

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

    OK?

    Grandad
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