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Math Help - Simplifying a trig identity

  1. #1
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    Simplifying a trig identity

    Simplify the expression:

    cos(θ + π/6) - sin (θ + π/6)

    Write your answer as an exact value.

    I'm taking grade 12 math through an independent learning course and I'm really stuck on this question, I have no teacher to ask for help, could someone give me some pointers? I tried expanding the brackets but could only simplify to:

    √3/2 cos θ - 1/2 sin θ

    Which doesn't seem right to me...
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  2. #2
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    Re: Simplifying a trig identity

    When you expanded the brackets, what did you come up with?

    I you must've misinterpreted the meaning of the brackets. For instance, cos(theta + pi/6) means the cosine of the quantity "theta + pi/6', not "cosine theta plus cosine pi/6."

    If you need a table of trig identities, try

    Table of Trigonometric Identities

    I did come up with a different answer than you did.

    I just checked out my result on a CAS, and I'm pretty sure I have the right formulation. Now, I'm assuming that by simplifying you mean a form where you have terms containing only sin and cos theta with coefficients.

    If you have a calculator or CAS, one way to check your result is to substitute a value for theta, in both the original and simplified form, and see if they're equal.
    Last edited by zhandele; November 28th 2012 at 06:08 PM. Reason: Checked out the result
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Another approach would be to essentially use a linear combination identity as follows:

    \sqrt{2}\left(\sin\left(\frac{\pi}{4} \right)\cos\left(\theta+\frac{\pi}{6} \right)-\cos\left(\frac{\pi}{4} \right)\sin\left(\theta+\frac{\pi}{6} \right) \right)

    Now, within the brackets, use the angle-difference identity for sine to write:

    \sqrt{2}\sin\left(\frac{\pi}{4}-\left(\theta+\frac{\pi}{6} \right) \right)=\sqrt{2}\sin\left(\frac{\pi}{12}-\theta \right)

    Now, if we prefer, we may use the identity \sin(\pi-x)=\sin(x) to write this as:

    \sqrt{2}\sin\left(\pi-\left(\frac{\pi}{12}-\theta \right \right)=\sqrt{2}\sin\left(\theta+\frac{11\pi}{12} \right)
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    Re: Simplifying a trig identity

    I expanded and got:

    cos θ cos π/6 - sin θ sin π/6

    I know from special triangles that cos π/6 = √3/2 and sin π/6 = 1/2, which is where I got the solution that I posted at first
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    Re: Simplifying a trig identity

    Quote Originally Posted by MarkFL2 View Post
    Another approach would be to essentially use a linear combination identity as follows:

    \sqrt{2}\left(\sin\left(\frac{\pi}{4} \right)\cos\left(\theta+\frac{\pi}{6} \right)-\cos\left(\frac{\pi}{4} \right)\sin\left(\theta+\frac{\pi}{6} \right) \right)

    Now, within the brackets, use the angle-difference identity for sine to write:

    \sqrt{2}\sin\left(\frac{\pi}{4}-\left(\theta+\frac{\pi}{6} \right) \right)=\sqrt{2}\sin\left(\frac{\pi}{12}-\theta \right)

    Now, if we prefer, we may use the identity \sin(\pi-x)=\sin(x) to write this as:

    \sqrt{2}\sin\left(\pi-\left(\frac{\pi}{12}-\theta \right \right)=\sqrt{2}\sin\left(\theta+\frac{11\pi}{12} \right)
    I believe this is at a level beyond the lesson that I'm on, I have yet to hear the term "linear combination identity"
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Simplifying a trig identity

    Quote Originally Posted by Lethargic View Post
    I expanded and got:

    cos θ cos π/6 - sin θ sin π/6

    I know from special triangles that cos π/6 = √3/2 and sin π/6 = 1/2, which is where I got the solution that I posted at first
    That is the expansion for the first term only. You need to also apply the angle-sum identity for sine on the second term.
    Last edited by MarkFL; November 28th 2012 at 06:18 PM. Reason: To add quote for clarity
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    Re: Simplifying a trig identity

    OK, it looks like you applied the sum of cosines identity. After that, you still have to subtract sin(th + pi/6). That requires using the other identity, for the sine of a sum.
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    Re: Simplifying a trig identity

    Not needed!

    -Dan
    Last edited by topsquark; November 28th 2012 at 06:18 PM. Reason: Not needed
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    Re: Simplifying a trig identity

    Oh wow that was such a simple mistake, thank you so much everyone, I've been tearing my hair out.
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  10. #10
    MHF Contributor MarkFL's Avatar
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    Re: Simplifying a trig identity

    Quote Originally Posted by Lethargic View Post
    I believe this is at a level beyond the lesson that I'm on, I have yet to hear the term "linear combination identity"
    You may not have heard the term yet, but you can see that I multiplied the first by 1=\sqrt{2}\sin\left(\frac{\pi}{4} \right) and the second term by 1=\sqrt{2}\cos\left(\frac{\pi}{4} \right) and then the rest should make sense if you know standard the identities used.
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