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Math Help - Trig identites question

  1. #1
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    Trig identites question

    Hi
    I need help on the following question:

    1)Express the function y=cos\theta + sin\theta in the form
    y = Rcos(\theta - \alpha)


    2)Express the following as algebraic expressions in terms of x: cos(2arccos(x))

    This is what i have done, but i don't understand how this is incorrect.

    make 2arccos(x) = \theta

    2cos(arccos(x)) = cos\theta

    cos\theta = 2x

    cos\theta = \frac{adj}{hyp} = \frac{2x}{1}

    so that means

    cos\theta = \frac{adj}{hyp} = 2x

    P.S
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  2. #2
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    1)Express the function y=cos\theta + sin\theta in the form
    y = sqrt(2)[cos(theta)/sqrt(20 - sin(theta)sqrt(2)
    = sqrt(2)[cos(theta-pi/4)

    2) 2(arccos(x) = theta
    arccos(x) = theta/2
    (x) = cos(theta/2)
    cos(theta) = 2cos^2(theta/2) - 1 = 2x^2 -1
    Last edited by sa-ri-ga-ma; April 9th 2010 at 07:24 PM.
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  3. #3
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    Quote Originally Posted by sa-ri-ga-ma View Post
    1)Express the function y=cos\theta + sin\theta in the form
    y = \sqrt(2)[cos(\theta)\sqrt(20 - sin(\theta)\sqrt(2)
    + \sqrt(2)[cos(\frac{\theta-\pi}{4})

    2)  2(arccos(x) = \theta
    arccos(x) = \frac{\theta}{2}
    (x) = cos(\frac{\theta}{2})
    cos(\theta) = 2cos^2(\frac{\theta}{2}) - 1 = 2x^2 -1
    For the first question i don't quite understand what you have done. Where did the square roots come from?

    For the second question how did you get
    cos(\theta) = 2cos^2(\frac{\theta}{2}) - 1
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  4. #4
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    Hello Paymemoney
    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help on the following question:

    1)Express the function y=cos\theta + sin\theta in the form
    y = Rcos(\theta - \alpha)
    Let
    \cos\theta+\sin\theta=R\cos(\theta-\alpha)
    Expand the RHS using the formula \cos(A-B)=\cos A \cos B + \sin A \sin B:
    \Rightarrow\cos\theta+\sin\theta=R\cos\theta\cos\a  lpha + R\sin\theta\sin\alpha
    Now compare the coefficients of \cos\theta and \sin\theta on both sides of the equation:
    \left\{\begin{array}{l l}1 = R\cos\alpha&\text{... (1)}\\1 = R\sin\alpha&\text{... (2)}\end{array}\right.
    Square (1) and (2) and add, noting that \cos^2\alpha+\sin^2\alpha = 1:
    1^2+1^2=R^2

    \Rightarrow R = \sqrt2

    Divide (2) by (1):
    \frac{R\sin\alpha}{R\cos\alpha}=\tan\alpha = 1

    \Rightarrow \alpha = \pi/4

    Substituting these values of R and \alpha back into the original equation gives:
    \cos\theta+\sin\theta= \sqrt2\cos(\theta-\pi/4)
    2)Express the following as algebraic expressions in terms of x: cos(2arccos(x))
    Let
    2\arccos(x) = \theta
    Then:
    \arccos(x) = \tfrac12\theta

    \Rightarrow x = \cos(\tfrac12\theta) ...(3)
    Also:
    \cos(2\arccos(x)) = \cos(\theta)
    =2\cos^2(\tfrac12\theta)-1, using the formula \cos2A = 2\cos^2A-1

    =2x^2-1, from (3)
    Grandad
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  5. #5
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    so from the formula: cos2A are you making arccos(x) = A?

    I though that cos(2arccos(x)) = 2x, why wouldn't this work?
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  6. #6
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    Hello, Paymemoney!

    The first one requires an imaginative step . . .


    \text{1) Express the function: }\:y \:=\:\cos\theta + \sin\theta\,\text{ in the form: }\:y \:=\: R\cos(\theta - \alpha)

    We are given: . y \;=\;\cos\theta + \sin\theta

    Divide by \sqrt{2}\!:\;\;\frac{1}{\sqrt{2}}\,y \;=\;\frac{1}{\sqrt{2}}\,\cos\theta + \frac{1}{\sqrt{2}}\,\sin\theta .[1]

    . . We know that: . \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}},\;\; \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}

    Substitute into [1]: . \frac{1}{\sqrt{2}}\,y \;=\;\cos\frac{\pi}{4}\cos\theta + \sin\frac{\pi}{4}\sin\theta

    And this becomes: . \frac{1}{\sqrt{2}}\,y \;=\; \cos\left(\theta - \frac{\pi}{4}\right)

    Multiply by \sqrt{2}:\;\;y \;=\;\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right)

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  7. #7
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    Hello Paymemoney
    Quote Originally Posted by Paymemoney View Post
    so from the formula: cos2A are you making arccos(x) = A?
    OK. Let's see if we can make it easier. I'll set it out slightly differently.

    Let
    \arccos(x)=A
    Now \arccos... means 'the angle whose cosine is ...'. So
    \arccos(x) = A
    means
    the angle whose cosine is x is A
    In other words
    \cos A = x, which I'll call equation (1)
    Now the question asks us for
    \cos(2\arccos(x))
    In other words, for
    \cos2A
    So we use the well-known double angle formula
    \cos2A=2\cos^2A-1
    and then use equation (1) to replace \cos A by x. So we get:
    \cos(2\arccos(x)) = \cos2A
    =2\cos^2A -1

    =2x^2-1

    OK now?
    Quote Originally Posted by Paymemoney View Post
    I though that cos(2arccos(x)) = 2x, why wouldn't this work?
    I'm not sure why you should think this. This would mean that if you double an angle you double its cosine. So \cos2A would equal 2\cos A. Which it doesn't, does it?

    Grandad
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  8. #8
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    yep thanks i understand now.
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