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Thread: Having trouble with these 5 trig exercises

  1. #1
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    Smile Having trouble with these 5 trig exercises

    Hello everybody. I was solving my set of exercises, but got stuck with these 5 exercises... Could anyone help me with it? Even if you can help with only one of them, I would be really glad.. Thank you all so much.


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  2. #2
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    Hello, wdma!

    Here's #2 . . .


    2) Find the exact value of: $\displaystyle \sin\left(-\frac{5\pi}{12}\right)$

    $\displaystyle \sin\left(-\frac{5\pi}{12}\right) \;=\; -\sin\left(\frac{5\pi}{12}\right)$

    . . . . . . . .$\displaystyle = \;-\sin\left(\frac{\pi}{6} + \frac{\pi}{4}\right)$ .**

    . . . . . . . .$\displaystyle = \;-\bigg[\sin\frac{\pi}{6}\cos\frac{\pi}{4} + \sin\frac{\pi}{4}\cos\frac{\pi}{6}\bigg] $

    . . . . . . . .$\displaystyle = \;-\bigg[\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\r ight) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3 }}{2}\right)\bigg] $

    . . . . . . . .$\displaystyle =\;-\bigg[\frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}}\bigg]$

    . . . . . . . .$\displaystyle = \;-\,\boxed{\frac{1+\sqrt{3}}{2\sqrt{2}}}$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    ** . $\displaystyle \frac{5\pi}{12} \;=\;\frac{2\pi}{12} + \frac{3\pi}{12} \;=\;\frac{\pi}{6} + \frac{\pi}{4}$ . . . see?

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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by wdma View Post
    Hello everybody. I was solving my set of exercises, but got stuck with these 5 exercises... Could anyone help me with it? Even if you can help with only one of them, I would be really glad.. Thank you all so much.



    I prefer to give you hints and if you can't solve it ask for help

    for the first question

    $\displaystyle \cos^{-1}a=0 \Rightarrow \cos(\cos^{-1}a)=cos(0) \Rightarrow a=1 $ since $\displaystyle \cos(\cos^{-1}a)=a $ the same thing for

    $\displaystyle sin^{-1}(\frac{\pi}{6})=b $

    $\displaystyle \frac{\pi}{6}=\sin(b) $
    now you have a=1 and you have sin(b) now to find the value of

    $\displaystyle \sin(a\pi+b)=sin(\pi+b)$ since a=1

    use $\displaystyle \sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$

    the second question

    $\displaystyle sin(\frac{-5\pi}{12})=??$

    first thing you should know

    $\displaystyle sin(-\theta)=-sin(\theta)$

    you can write the angle

    $\displaystyle \frac{5\pi}{12}=\frac{3\pi}{12}+\frac{2\pi}{12}$

    then use the formula $\displaystyle \sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by wdma View Post
    Hello everybody. I was solving my set of exercises, but got stuck with these 5 exercises... Could anyone help me with it? Even if you can help with only one of them, I would be really glad.. Thank you all so much.



    1. $\displaystyle \cos{0}=1=a$ and $\displaystyle \sin^{-1}(\frac{\pi}{6})=.5507=b$. You do the rest.

    5. $\displaystyle \sin{x}=\pm\frac{\sqrt{2}}{2}$. You do the rest.
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  5. #5
    Senior Member Twig's Avatar
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    Hi

    Here is 4)

    $\displaystyle tan(\beta)=\frac{12}{5} $

    $\displaystyle tan(2\beta)=tan(\beta+\beta)=\frac{tan(\beta)+tan( \beta)}{1-tan(\beta)\cdot tan(\beta)} $

    Now you know that $\displaystyle tan(\beta)=\frac{12}{5} $ , so this gives $\displaystyle \frac{\frac{24}{5}}{1-\left(\frac{12}{5}\right)^{2}} = -\frac{120}{119}$
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    1. $\displaystyle \cos{0}=1=a$ and $\displaystyle \sin^{-1}(\frac{\pi}{6})=.5507=b$. You do the rest.
    Just a friendly reminder: in this case, leave $\displaystyle \sin^{-1}\left(\frac{\pi}{6}\right)$ as an exact value, because something will work out nicely in the simplification...
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  7. #7
    MHF Contributor Amer's Avatar
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    question three

    $\displaystyle \frac{sin(2(x-y))}{2\cos(x)\cos(y)\cos(x-y)}$

    use these formulas

    $\displaystyle \sin(x-y)=\sin(x)\cos(y)-\sin(y)\cos(x)---------(1)$

    $\displaystyle \sin2x=2sin(x)cos(x)------------------(2)$

    $\displaystyle \cos(x-y)=\cos(x)\cos(y)+\sin(x)sin(y)--------(3)$


    $\displaystyle \frac{2\sin(x-y)\cos(x-y)}{2\cos(x)\cos(y)\cos(x-y)}$

    $\displaystyle \frac{2\sin(x-y)}{2\cos(x)cos(y)}$

    $\displaystyle \frac{sin(x)cos(y)-sin(y)cos(x)}{cos(x)cos(y)}$
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  8. #8
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    Hello again, wdma!

    Here's my version of #3 . . .


    3) Simplify as much as possible: .$\displaystyle \frac{\sin[2(x-y)]}{2\cos(x)\cos(y)\cos(x-y)} $
    In the numerator, use: .$\displaystyle \sin2\theta \:=\:2\sin\theta\cos\theta$

    We have: .$\displaystyle \frac{{\color{red}\rlap{/}}2\sin(x-y){\color{red}\rlap{/////////}}\cos(x-y)}{{\color{red}\rlap{/}}2\cos(x)\cos(y){\color{red}\rlap{/////////}}\cos(x-y)} \;=\;\frac{\sin(x-y)}{\cos(x)\cos(y)}$

    . . $\displaystyle = \;\frac{\sin(x)\cos(y) - \cos(x)\sin(y)}{\cos(x)\cos(y)} \;=\;\frac{\sin(x){\color{red}\rlap{/////}}\cos(y)}{\cos(x){\color{red}\rlap{/////}}\cos(y)} - \frac{{\color{red}\rlap{/////}}\cos(x)\sin(y)}{{\color{red}\rlap{/////}}\cos(x)\cos(y)} $

    . . $\displaystyle =\;\frac{\sin(x)}{\cos(x)} - \frac{\sin(y)}{\cos(y)} \;=\;\tan(x) - \tan(y) $

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  9. #9
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    Wink

    Quote Originally Posted by Amer View Post
    I prefer to give you hints and if you can't solve it ask for help

    for the first question

    $\displaystyle \cos^{-1}a=0 \Rightarrow \cos(\cos^{-1}a)=cos(0) \Rightarrow a=1 $ since $\displaystyle \cos(\cos^{-1}a)=a $ the same thing for

    $\displaystyle sin^{-1}(\frac{\pi}{6})=b $

    $\displaystyle \frac{\pi}{6}=\sin(b) $
    now you have a=1 and you have sin(b) now to find the value of

    $\displaystyle \sin(a\pi+b)=sin(\pi+b)$ since a=1

    use $\displaystyle \sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$

    the second question

    $\displaystyle sin(\frac{-5\pi}{12})=??$

    first thing you should know

    $\displaystyle sin(-\theta)=-sin(\theta)$

    you can write the angle

    $\displaystyle \frac{5\pi}{12}=\frac{3\pi}{12}+\frac{2\pi}{12}$

    then use the formula $\displaystyle \sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$

    I'm still struggling with #1 and #5. Could anyone help me further with them?
    Also, I would like to thank all of you so much, for being so kind and helpful!
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  10. #10
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    Quote Originally Posted by wdma View Post
    I'm still struggling with #1 and #5. Could anyone help me further with them?
    Also, I would like to thank all of you so much, for being so kind and helpful!
    5)
    $\displaystyle \begin{aligned}
    2\sin^2 x &= 1 \\
    \sin^2 x &= \frac{1}{2} \\
    \sin x &= \pm \frac{\sqrt{2}}{2}
    \end{aligned}$

    You should have recognized the ratios $\displaystyle \pm\frac{\sqrt{2}}{2}$ (the alternate form is $\displaystyle \frac{1}{\sqrt{2}}$).

    Ask yourself, the sine of what angles is $\displaystyle \frac{\sqrt{2}}{2}$? (There are two answers). Then ask yourself, the sine of what angles is $\displaystyle -\frac{\sqrt{2}}{2}$? (There are two more answers). If you have to, draw some triangles in the cartesian plane.


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