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Math Help - Finding angles

  1. #1
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    Finding angles

    Find all the angles between 0^\circ and 360^\circ which satisfy the equation sec(2x+30^\circ)=sec(2x-60^\circ)
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    Quote Originally Posted by Punch View Post
    Find all the angles between 0^\circ and 360^\circ which satisfy the equation sec(2x+30^\circ)=sec(2x-60^\circ)
    \sec{(2x + 30^\circ)} = \sec{(2x - 60^\circ)}

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


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


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

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

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

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

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

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

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

    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

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

    Using this identity, what does \tan{15^\circ} equal? (It will help you solve your equation...)
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    Quote Originally Posted by Punch View Post
    Find all the angles between 0^\circ and 360^\circ which satisfy the equation sec(2x+30^\circ)=sec(2x-60^\circ)
    HI

    \cos (2x-60)=\cos (2x+30)

    2x-60=2n(180) \pm (2x+30)

    2x-60=2n(180) -2x-30

    2n(180)=4x-30

    180n=2x-15 , where n=0,1,2,3
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  4. #4
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    Quote Originally Posted by Prove It View Post
    \sec{(2x + 30^\circ)} = \sec{(2x - 60^\circ)}

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


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


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

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

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

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

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

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

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

    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

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

    Using this identity, what does \tan{15^\circ} equal? (It will help you solve your equation...)
    I can't continue
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    Quote Originally Posted by Punch View Post
    I can't continue
    I told you how to continue.

    Use the final identity I gave you to evaluate \tan{15^\circ}.

    \tan{15^\circ} = \tan{\frac{30^\circ}{2}}.

    It will help you.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    I told you how to continue.

    Use the final identity I gave you to evaluate \tan{15^\circ}.

    \tan{15^\circ} = \tan{\frac{30^\circ}{2}}.

    It will help you.
    may i ask where did u get the 15 from?
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  7. #7
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    Experience.

    Try it.
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  8. #8
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    Quote Originally Posted by Prove It View Post
    Experience.

    Try it.
    I know you are an expert lol but i need to know the way in solving the problems and not by luck... i can't have u sitting beside me in the exam right
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  9. #9
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    or do you tangent inverse to find what 2x is equals to? or using special angle
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    I would advise

    1. Stop arguing with the person who is trying to help you, instead do as you've been instructed so that you learn something and hopefully will be able to remember this for an exam.

    2. Use the half-angle identities for sine, cosine and tangent to find the exact surd answers when your angle is 15^\circ. You can also find exact values when your angle is 22.5^{\circ}, or any successive half angle.
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    Quote Originally Posted by Punch View Post
    or do you tangent inverse to find what 2x is equals to? or using special angle
    If you thought more about post #2 your life would be less stressful.
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    Sorry if my words offended you but i truly don't understand and am not arguing with u, i know u r trying to help me and i am truly thankful to u however i really do not understand where the 15 came from...
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  13. #13
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    If you work out what tan15 is equal to, it will make sense where the 15 comes from and why it is being used.

    alternatively
    2 - root3 = tan2x
    2 - root3 = (2tanx)/(1 - tan^2x)

    simplify to find tanx
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  14. #14
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    Oh so it's similar to using a special angle... 2 - \sqrt{3} = \tan{15}
    then use double angle formula to solve for x

    Don't mind me asking but why is this type of question coming up when I don't learn anything about the special angles of tan15 and such? I have only heard of tan 30, tan 60, tan 90 and so on..
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  15. #15
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    Quote Originally Posted by Punch View Post
    Oh so it's similar to using a special angle... 2 - \sqrt{3} = \tan{15}
    then use double angle formula to solve for x

    Don't mind me asking but why is this type of question coming up when I don't learn anything about the special angles of tan15 and such? I have only heard of tan 30, tan 60, tan 90 and so on..
    hi again punch ,

    nowdays questions don't confine to only whatever you have learnt so be prepared . It's good that you learnt the extras . However , in this case , tan 15 is related to whatever you have learnt , the special angle and double angle formula .

    \tan 30=\frac{1}{\sqrt{3}}

    \tan 2(15)=\frac{1}{\sqrt{3}}

    \frac{2\tan 15}{1-\tan^2 15}=\frac{1}{\sqrt{3}}

    1-\tan^2 15=2\sqrt{3}\tan 15

    \tan^2 15+2\sqrt{3}\tan 15 -1=0

    Using the quadratic formula ,

     <br />
\tan 15=\frac{-2\sqrt{3}\pm \sqrt{12-4(-1)}}{2}<br />

    =\pm 2-\sqrt{3}

    -2-\sqrt{3} is not possible because tan 15 is in the first quadrant , everything here will be positive .

    therefore , \tan 15=2-\sqrt{3}
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