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Thread: Finding angles

  1. #16
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    And from that we see

    $\displaystyle \tan{15^\circ} = \tan{2x}$

    What do you think $\displaystyle x$ must be?
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  2. #17
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    Quote Originally Posted by Prove It View Post
    And from that we see

    $\displaystyle \tan{15^\circ} = \tan{2x}$

    What do you think $\displaystyle x$ must be?
    In fact, since the tangent function is positive in the first and third quadrants, we actually have

    $\displaystyle \tan{(\{15^\circ, 195^\circ\} + 360^\circ n)} = 2 - \sqrt{3}$, where $\displaystyle n$ is an integer representing the number of times around the unit circle.


    So solve $\displaystyle \tan{(\{15^\circ, 195^\circ\} + 360^\circ n)} = \tan{2x}$.
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  3. #18
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    Quote Originally Posted by Prove It View Post
    $\displaystyle \sec{(2x + 30^\circ)} = \sec{(2x - 60^\circ)}$

    $\displaystyle \cos{(2x + 30^\circ)} = \cos{(2x - 60^\circ)}$.


    Use the identity $\displaystyle \cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$


    So $\displaystyle \cos{2x}\cos{30^\circ} - \sin{2x}\sin{30^\circ} = \cos{2x}\cos{60^\circ} + \sin{2x}\sin{60^\circ}$

    $\displaystyle \frac{\sqrt{3}}{2}\cos{2x} - \frac{1}{2}\sin{2x} = \frac{1}{2}\cos{2x} + \frac{\sqrt{3}}{2}\sin{2x}$

    $\displaystyle \frac{\sqrt{3} - 1}{2}\cos{2x} = \frac{\sqrt{3} + 1}{2}\sin{2x}$

    $\displaystyle \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \tan{2x}$

    $\displaystyle \frac{(\sqrt{3} - 1)^2}{3 - 1} = \tan{2x}$

    $\displaystyle \frac{3 - 2\sqrt{3} + 1}{2} = \tan{2x}$

    $\displaystyle \frac{4 - 2\sqrt{3}}{2} = \tan{2x}$

    $\displaystyle 2 - \sqrt{3} = \tan{2x}$



    This might seem a bit off track, but if you're good with your half angle identities, you will know that

    $\displaystyle \tan{\frac{\theta}{2}} = \frac{1 - \cos{\theta}}{\sin{\theta}}$.

    Using this identity, what does $\displaystyle \tan{15^\circ}$ equal? (It will help you solve your equation...)
    Taking it from here, tan2x=tan^-1\frac{1 - \cos{\theta}}{\sin{\theta}}
    = 15

    Since tan2x=positive, it is in quadrant 1 and 3

    So x=15, 195, 360+15, 360+195
    = 15, 195, 375, 555
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  4. #19
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    Quote Originally Posted by Punch View Post
    Taking it from here, tan2x=tan^-1\frac{1 - \cos{\theta}}{\sin{\theta}}
    = 15

    Since tan2x=positive, it is in quadrant 1 and 3

    So x=15, 195, 360+15, 360+195
    = 15, 195, 375, 555
    No.

    You have $\displaystyle \{15^\circ, 195^\circ\} + 360^\circ n = 2x$

    So $\displaystyle x = \{7.5^\circ, 97.5^\circ\} + 180^\circ n$.


    Also, don't try and evaluate all possibilities. Since you do not have a restricted domain, you would need an infinity of them.

    Just write the answer in terms of $\displaystyle n$.
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  5. #20
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    Oh yes, I forgot to devide it by 2 to get x values... And no, there is a domain and that is between 0 and 360
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  6. #21
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    Quote Originally Posted by Punch View Post
    Find all the angles between $\displaystyle 0^\circ$ and $\displaystyle 360^\circ$ which satisfy the equation $\displaystyle sec(2x+30^\circ)=sec(2x-60^\circ)$
    I thought of something and that is,

    does the domain of this question applies for $\displaystyle 0^\circ<x<360^\circ$ or $\displaystyle 0^\circ<2x<360^\circ $?
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