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Math Help - Sum or Difference Formulas

  1. #1
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    Sum or Difference Formulas

    This is not for homework, nor for a teacher to see, just a test correction so that I know what to do, or what I am doing wrong.

    Use the sum or Difference formulas to find the exact value of each expression.

    1. sin (165 degrees)

    2. cos (5π/12)

    I tried to use my calculator, a TI-89 ended up with the same answer both, and it was wrong unfortunately, and I was possibly wondering what I could do to get the correct answer.

    my TI-89 gave me this:

    ((√3-1)*√(2))/(4)
    Radical 3 - 1 * radical 2 all over 4.

    And my best guess as the answer for both is [√6 + √2] / 4. Just not sure how to get that.

    Edit: this might be my only trig question on here, as the 1st semester is almost gone, and I just wanted to figure these problems out.
    Last edited by Nickod777; January 4th 2012 at 10:33 PM. Reason: something important to say
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Sum or Difference Formulas

    1.) \sin\left(165^{\circ}\right)

    Use the reflection \sin\left(180^{\circ}-x\right)=\sin(x)

    \sin\left((180-165)^{\circ}\right)=\sin\left\((45-30)^{\circ}\)

    Now apply the angle difference identity for sine.

    Or alternately:

    \sin\left(165^{\circ} \right)=\sin\left(15^{\circ}\right)=\cos\left(75^{  \circ}\right)=\cos\left((45+30)^{\circ}\right)

    Now apply the angle sum identity for cosine.

    2.) \cos\left(\frac{5\pi}{12}\right)

    Apply the identity: \cos(x)=\sin\left(\frac{\pi}{2}-x\right) to get:

    \cos\left(\frac{5\pi}{12} \right)=\sin\left(\frac{\pi}{2}-\frac{5\pi}{12}\right)

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

    \cos\left(\frac{5\pi}{12} \right)=\sin\left(\frac{\pi}{4}-\frac{\pi}{6}\right)

    Now apply the angle difference identity for sine.

    Or, alternately:

    \cos\left(\frac{5\pi}{12}\right)=\cos\left(\frac{(  3+2)\pi}{12}\right)=\cos\left(\frac{\pi}{4}+\frac{  \pi}{6}\right)

    Now apply the angle sum identity for cosine.
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  3. #3
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    Re: Sum or Difference Formulas

    Hello, Nickod777!

    Use the Sum or Difference formulas to find the exact value of each expression.

    . . 1.\;\sin (165^o) \qquad 2.\;\cos\left(\tfrac{5\pi}{12}\right)

    Did it occur to you to use the Sum or Difference Formulas?
    Do you know the Sum or Difference Formulas?


    1.\;\sin(165^o) \;=\;\sin(45^o + 120^o)

    . . . . . . . . . =\; \sin(45^o)\cos(120^o) + \cos(45^o)\sin(120^o)

    . . . . . . . . . =\;\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3  }}{2}\right)

    . . . . . . . . . =\; -\frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}} \;=\;\frac{\sqrt{3}-1}{2\sqrt{2}} \quad\begin{array}{c}^{\text{Usually this is an acceptable answer,}} \\ ^{\text{but often they rationalize the denominator.}}\end{array}

    . . . . . . . . . =\;\frac{\sqrt{2}}{\sqrt{2}}\cdot \frac{\sqrt{3}-1}{2\sqrt{2}} \;=\; \frac{\sqrt{2}\left(\sqrt{3}-1\right)}{4} \;=\;\frac{\sqrt{6} - \sqrt{2}}{4}



    2.\;\cos\left(\frac{5\pi}{12}\right) \;=\;\cos\left(\frac{\pi}{6} + \frac{\pi}{4}\right)

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

    . . . . . . . . . =\;\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\  \sqrt{2}}\right) - \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}} \right)

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

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  4. #4
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    Re: Sum or Difference Formulas

    Quote Originally Posted by Soroban View Post
    Hello, Nickod777!


    Did it occur to you to use the Sum or Difference Formulas?
    Do you know the Sum or Difference Formulas?


    1.\;\sin(165^o) \;=\;\sin(45^o + 120^o)

    . . . . . . . . . =\; \sin(45^o)\cos(120^o) + \cos(45^o)\sin(120^o)

    . . . . . . . . . =\;\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{2}\right) + \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3  }}{2}\right)

    . . . . . . . . . =\; -\frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}} \;=\;\frac{\sqrt{3}-1}{2\sqrt{2}} \quad\begin{array}{c}^{\text{Usually this is an acceptable answer,}} \\ ^{\text{but often they rationalize the denominator.}}\end{array}

    . . . . . . . . . =\;\frac{\sqrt{2}}{\sqrt{2}}\cdot \frac{\sqrt{3}-1}{2\sqrt{2}} \;=\; \frac{\sqrt{2}\left(\sqrt{3}-1\right)}{4} \;=\;\frac{\sqrt{6} - \sqrt{2}}{4}



    2.\;\cos\left(\frac{5\pi}{12}\right) \;=\;\cos\left(\frac{\pi}{6} + \frac{\pi}{4}\right)

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

    . . . . . . . . . =\;\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\  \sqrt{2}}\right) - \left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}} \right)

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

    Unfortunately I do not know the formulas.

    Thanks for the help though.
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  5. #5
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    Re: Sum or Difference Formulas

    Quote Originally Posted by Nickod777 View Post
    Unfortunately I do not know the formulas.

    Thanks for the help though.
    Here are the formulas MarkFL2 & Soroban used.

    Trigonometric Formulas: Sum and Difference
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  6. #6
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    Re: Sum or Difference Formulas

    Quote Originally Posted by Nickod777 View Post
    Unfortunately I do not know the formulas.
    I find it hard to believe that you were asked to use formulae if you haven't been taught them...
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  7. #7
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    Re: Sum or Difference Formulas

    Quote Originally Posted by Prove It View Post
    I find it hard to believe that you were asked to use formulae if you haven't been taught them...

    in my class, we are given the formulas, not learning how to use them, instead we have to learn them ourselves.

    Anyway, let us just end this gentlemen.
    Last edited by Nickod777; January 5th 2012 at 08:22 PM.
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