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Thread: Find the value

  1. #1
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    Find the value

    If sin A + sin B = 1 and cos A + cos B = (3)1/2
    A and B < 90o
    Find a value for (A+B) ?
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  2. #2
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    Re: Find the value

    Hey Dilan.

    Do you know how to relate sin and cos (A+B) to sin(A),sin(B),cos(A),cos(B)?

    [Hint - look up a couple of trig identities for more information].
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  3. #3
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    Re: Find the value

    Quote Originally Posted by Dilan View Post
    If sin A + sin B = 1 and cos A + cos B = (3)1/2
    A and B < 90o
    Find a value for (A+B) ?
    Try using a couple of the sum to product identities below ...

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  4. #4
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    Re: Find the value

    Let's make a silly assumption and assume $a=b$ (this would be a special case, but who knows, it might pan out).

    Then we have:

    $2\sin A = 1$ and $2\cos A = \sqrt{3}$

    So, $\sin A = \dfrac{1}{2}$ and $\cos A = \dfrac{\sqrt{3}}{2}$

    This is a well known angle. Plug it in for $A$, and since we set $B$ equal to the same, it is trivial to find $A+B$.
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  5. #5
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    Re: Find the value

    $\dfrac{\sin{A}+\sin{B}}{\cos{A}+\cos{B}} = \dfrac{2\sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2} \right)}{2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)} = \tan\left(\dfrac{A+B}{2}\right) = \dfrac{1}{\sqrt{3}}$

    note that $\tan(30^\circ) = \dfrac{1}{\sqrt{3}}$
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