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Thread: a*b and a+b

  1. #1
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    a*b and a+b

    given that sinx * cosx = 60/169

    find sinx + cosx
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle (\sin x+\cos x)^2=\sin^2x+\cos^2x+2\sin x\cos x=1+2\sin x\cos x=1+\frac{120}{169}=\frac{289}{169}$
    $\displaystyle \Rightarrow \sin x+\cos x=\pm\frac{17}{13}$
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  3. #3
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    Hello, DINOCALC09!

    There is a back-door approach to this one, too . . .


    Given that: .$\displaystyle \sin x\cdot\cos x \:=\:\frac{60}{169}$

    find: .$\displaystyle \sin x + \cos x$

    Let: $\displaystyle y \;=\;\sin x + \cos x$

    Square: .$\displaystyle y^2\;=\;\sin^2\!x + 2\sin x\cos x + \cos^2x$

    $\displaystyle \text{And we have: }\;y^2 \;=\;\underbrace{\sin^2\!x+\cos^2\!x}_{\text{This is 1}}\;+\;2\underbrace{\sin x\cos x}_{\text{This is }\frac{60}{169}}$

    Hence: .$\displaystyle y^2\;=\;1 + \frac{120}{169} \;=\;\frac{289}{169}\quad\Rightarrow\quad y \;=\;\sqrt{\frac{289}{169}} \;=\;\frac{17}{13}$

    . . Therefore: .$\displaystyle \boxed{\sin x + \cos x \;=\;\frac{17}{13}}$

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