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Math Help - Trig help

  1. #1
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    Exclamation Trig help

    Hey guys,
    I am confused with these two questions (8 and 9). I have no idea where to begin off.
    These questions are where one trigonometric function is given and find another.

    It would really be help if you could post how to work out this questions.

    Thanks
    Trig help-wp_20140715_004.jpg
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  2. #2
    MHF Contributor
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    Re: Trig help

    you can note that

    $\sin(\theta+90^\circ)=\sin(\theta)\cos(90^\circ)+ \cos(\theta)\sin(90^\circ)=\cos(\theta)$

    and that

    $\cos(\theta+90^\circ)= \cos(\theta)\cos(90^\circ)-\sin(\theta)\sin(90^\circ)=-\sin(\theta)$

    so

    $\tan(\theta + 90^\circ)=\dfrac{\cos(\theta)}{-\sin(\theta)}$

    $\theta$ is obtuse so

    $\sin(\theta) > 0$ and $\cos(\theta)<0$

    so

    $\cos(\theta)=-\sqrt{1-sin^2(\theta)}=-\sqrt{1-k^2}$

    and thus

    $\tan(\theta+90^\circ)=\dfrac{-\sqrt{1-k^2}}{-k}=\dfrac{\sqrt{1-k^2}}{k}$

    for (9)

    $\sec^2(\theta)=1+\tan^2(\theta)$

    $\sec^2(\theta)-1=a^2-\dfrac 1 2 + \dfrac 1 {16a^2}=\left(a-\dfrac 1 {4a}\right)^2$

    so

    $\tan(\theta)=a - \dfrac 1 {4a}$

    $\sec(\theta)+\tan(\theta)=\left(a + \dfrac 1 {4a}\right) + \left(a - \dfrac 1 {4a}\right)=2a$
    Thanks from rainbowstar
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  3. #3
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    Re: Trig help

    Quote Originally Posted by romsek View Post
    you can note that

    $\sin(\theta+90^\circ)=\sin(\theta)\cos(90^\circ)+ \cos(\theta)\sin(90^\circ)=\cos(\theta)$

    and that

    $\cos(\theta+90^\circ)= \cos(\theta)\cos(90^\circ)-\sin(\theta)\sin(90^\circ)=-\sin(\theta)$

    so

    $\tan(\theta + 90^\circ)=\dfrac{\cos(\theta)}{-\sin(\theta)}$

    $\theta$ is obtuse so

    $\sin(\theta) > 0$ and $\cos(\theta)<0$

    so

    $\cos(\theta)=-\sqrt{1-sin^2(\theta)}=-\sqrt{1-k^2}$

    and thus

    $\tan(\theta+90^\circ)=\dfrac{-\sqrt{1-k^2}}{-k}=\dfrac{\sqrt{1-k^2}}{k}$

    for (9)

    $\sec^2(\theta)=1+\tan^2(\theta)$

    $\sec^2(\theta)-1=a^2-\dfrac 1 2 + \dfrac 1 {16a^2}=\left(a-\dfrac 1 {4a}\right)^2$

    so

    $\tan(\theta)=a - \dfrac 1 {4a}$

    $\sec(\theta)+\tan(\theta)=\left(a + \dfrac 1 {4a}\right) + \left(a - \dfrac 1 {4a}\right)=2a$
    Thankyou very much. That helped a lot!
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