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Thread: Sum and differences of angles?

  1. #1
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    Sum and differences of angles?

    I'm having a little trouble, here's the question:

    If sin $\displaystyle x= 7/25$ and cos $\displaystyle y= 3/5$, and $\displaystyle x$ and $\displaystyle y$are positive acute angles, find

    a.$\displaystyle sin (x+y)$
    b.$\displaystyle cos (x-y)$
    c.$\displaystyle tan (x+y)$
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  2. #2
    MHF Contributor
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    First question - do you know the sum and difference formulas?
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  3. #3
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    Yeah.
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  4. #4
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    Hello, eh501!

    Okay, you know the three formulas:

    . . $\displaystyle \begin{array}{ccc}\sin(x+y) &=&\sin x\cos y + \cos x\sin y \\ \\[-3mm] \cos(x+y) &=&\cos x\cos y - \sin x\sin y \\ \\[-3mm] \tan(x+y) &=&\dfrac{\tan x + \tan y}{1 - \tan x\tan y} \end{array}$


    If $\displaystyle \sin x= \tfrac{7}{25}\text{ and }\cos y= \tfrac{3}{5}$, and $\displaystyle x$ and $\displaystyle y$ are positive acute angles, find

    . . $\displaystyle a)\;\;\sin (x+y)\qquad b)\;\;\cos (x-y) \qquad c)\;\;\tan (x+y)$
    To use those formulas, we need more trig values.


    We can use: .$\displaystyle \sin^2\!\theta + \cos^2\!\theta \:=\:1$


    Since $\displaystyle \sin x = \tfrac{7}{25}\!:\;\;(\tfrac{7}{25})^2 + \cos^2\!x \:=\:1 \quad\Rightarrow\quad\cos^2\!x \:=\:1-\tfrac{49}{625}$

    . . . . $\displaystyle \cos^2\!x\:=\:\tfrac{576}{625} \quad\Rightarrow\quad \cos x \:=\:\tfrac{24}{25}$

    Then: .$\displaystyle \tan x \;=\;\frac{\sin x}{\cos x} \;=\;\frac{\frac{7}{25}}{\frac{24}{25}} \;=\;\tfrac{7}{24}$

    . . Hence, we have: .$\displaystyle \begin{Bmatrix} \sin x &=& \frac{7}{25} \\ \\[-4mm] \cos x &=& \frac{24}{25} \\ \\[-4mm] \tan x &=& \frac{7}{24} \end{Bmatrix} $


    Use the same procedure on $\displaystyle \cos y = \tfrac{3}{5}$ to find $\displaystyle \sin y\text{ and }\tan y$

    . . then you can use those formulas . . .

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