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Thread: Is this true?

  1. #1
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    Is this true?

    Hi, I am revising on Trig at the moment.

    Would you say;

    R sin (x+36.9) = R cos (x-53.1)

    I have just transformed these equations from 3 cos x - 4 sin x using the alpha method.
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  2. #2
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    Quote Originally Posted by Googl View Post
    Hi, I am revising on Trig at the moment.

    Would you say;

    R sin (x+36.9) = R cos (x-53.1)

    I have just transformed these equations from 3 cos x - 4 sin x using the alpha method.
    Huh?

    What was the original problem?

    $\displaystyle 3\cos(x)-4\sin(x)=0\text{?}$

    If so, solve

    $\displaystyle \displaystyle\tan(x)=\frac{3}{4}$
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    Huh?

    What was the original problem?

    $\displaystyle 3\cos(x)-4\sin(x)=0\text{?}$

    If so, solve

    $\displaystyle \displaystyle\tan(x)=\frac{3}{4}$
    Would you be able to draw a cos and sin graph with that? Try to convert it to a cos/sin equation first and see what equation you end up with.
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  4. #4
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    Quote Originally Posted by Googl View Post
    Would you be able to draw a cos and sin graph with that? Try to convert it to a cos/sin equation first and see what equation you end up with.
    What was the exact question you were given?
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  5. #5
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    Try to write $\displaystyle y = 3 cos x - 4 sin x $ as;

    $\displaystyle y = R cos(x-[alpha])$

    and

    $\displaystyle y = R sin(x+[alpha])$

    [alpha] is the alpha symbol

    That
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  6. #6
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    Hello, Googl!

    $\displaystyle \text{Try to write }y \:=\:3\cos x - 4\sin x$

    $\displaystyle \text{as: }\;\begin{Bmatrix}y \:=\: R\cos(x-\alpha) \\ \text{and} \\ y \:=\: R\sin(x+\alpha) \end{Bmatrix}$

    We have: .$\displaystyle y \;=\;3\cos x - 4\sin x $


    Multiply by $\displaystyle \frac{5}{5}\!:$

    . . $\displaystyle y \;=\;\frac{5}{5}(3\cos x - 4\sin x) \quad\Rightarrow\quad y \;=\; 5\left(\frac{3}{5}\cos x - \frac{4}{5}\sin x\right)$ .[1]


    Let $\displaystyle \,\alpha$ be an acute angle is a 3-4-5 right triangle.

    Code:
    
                *
               /|
              / |
           5 /  |4
            /   |
           /α   |
          *-----*
             3

    And we have: .$\displaystyle \begin{Bmatrix}\cos\alpha \:=\:\frac{3}{5} \\ \\[-3mm] \sin\alpha \:=\:\frac{4}{5} \end{Bmatrix}$


    Substitute into [1]:

    . . $\displaystyle y \;=\;5\left(\cos\alpha \cos x - \sin\alpha\sin x)$


    Therefore: .$\displaystyle y \;=\;5\cos(x + \alpha)$

    . . where $\displaystyle \alpha \:=\:\sin^{\text{-}1}\left(\frac{4}{5}\right) \:\approx\:53.1^o$

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  7. #7
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    I forgot to give you the R values sorry. R is 5 in my original post and it's similar to what you have here.

    Would you say; $\displaystyle 5 sin (x+36.9) = 5 cos (x-53.1)$

    Try to change it into the form;

    R sin (x + [alpha])

    alpha is the alpha symbol.
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  8. #8
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    Quote Originally Posted by Googl View Post
    Hi, I am revising on Trig at the moment.

    Would you say;

    R sin (x+36.9) = R cos (x-53.1)
    36.9+ 53.1= 90.0 so if x is in degrees, yes, that is true. If we let y= x+ 36.9, then x- 53.1= y- 36.9- 53.1= y- 90.

    But cos(y- 90)= cos(90- y) (since cosine is an even function) and it is always true that sin(y)= cos(90- y).

    I have just transformed these equations from 3 cos x - 4 sin x using the alpha method.
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  9. #9
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    Thanks.
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