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Thread: 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
    $\displaystyle y = Rcos(\theta - \alpha)$


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

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

    $\displaystyle make 2arccos(x) = \theta$

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

    $\displaystyle cos\theta = 2x$

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

    so that means

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

    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; Apr 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 $\displaystyle y=cos\theta + sin\theta$ in the form
    $\displaystyle y = \sqrt(2)[cos(\theta)\sqrt(20 - sin(\theta)\sqrt(2) $
    $\displaystyle + \sqrt(2)[cos(\frac{\theta-\pi}{4})$

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

    $\displaystyle \Rightarrow R = \sqrt2$

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

    $\displaystyle \Rightarrow \alpha = \pi/4$

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

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

    $\displaystyle =2x^2-1$, 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 . . .


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

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

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

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

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

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

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

    $\displaystyle =2x^2-1$

    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 $\displaystyle \cos2A$ would equal $\displaystyle 2\cos A$. 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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