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Math Help - 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: \sin\left(-\frac{5\pi}{12}\right)

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

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

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

    . . . . . . . . = \;-\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]

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

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


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


    ** . \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

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

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

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

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

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

    the second question

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

    first thing you should know

    sin(-\theta)=-sin(\theta)

    you can write the angle

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

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

    5. \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)

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

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

    Now you know that  tan(\beta)=\frac{12}{5}  , so this gives  \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. \cos{0}=1=a and \sin^{-1}(\frac{\pi}{6})=.5507=b. You do the rest.
    Just a friendly reminder: in this case, leave \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

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

    use these formulas

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

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

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


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

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

    \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: . \frac{\sin[2(x-y)]}{2\cos(x)\cos(y)\cos(x-y)}
    In the numerator, use: . \sin2\theta \:=\:2\sin\theta\cos\theta

    We have: . \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)}

    . . = \;\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)}

    . . =\;\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

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

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

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

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

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

    the second question

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

    first thing you should know

    sin(-\theta)=-sin(\theta)

    you can write the angle

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

    then use the formula \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)
    \begin{aligned}<br />
2\sin^2 x &= 1 \\<br />
\sin^2 x &= \frac{1}{2} \\<br />
\sin x &= \pm \frac{\sqrt{2}}{2}<br />
\end{aligned}

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

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


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