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Math Help - Proving and finding values.

  1. #1
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    Proving and finding values.

    I am having some problems with these questions,can anybody help?

    2) It is given that sinA = 5/13 where 90<A<180(degrees) , and that cosB = -33/65 , where 180<B<270. Find the exact value of cos2(A+B)

    How do i expand cos2(A+B) ? apparently when i use my calculator cos2(A+B) = cos(2A + 2B) but again,the expansion gets too complicated and i derived a six digit fraction which is a far cry from the answer (7/25)

    3)A right pyramid has a square horizontal base ABCD of side 2x and a vertex O. Given that Angle AOB = 2\theta , show that the inclination of OA to the horizontal is arcsin(\sqrt cos2\theta) and find the inclination of the plane OAB to the horizontal.

    I can't think of any way to start on this problem where I am able to ultilize all of the information given.

    1)(solved) If  A + B + C = \dfrac{\pi}{2} \,radians , show that  tanA tanB + tanB tanC + tanC tanA = 1

    For this I was thinking to make the pi/2 into 1 we have to introduce sine to A+B+C but it just gets more and more complicated when i try to expand everything and I feel that there should be a simpler way which I have overlooked.

    Any help at all is appreciated
    Last edited by arccos; November 27th 2010 at 07:07 PM.
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  2. #2
    MHF Contributor harish21's Avatar
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    1)  A+B+C=\dfrac{\pi}{2}

     A = \dfrac{\pi}{2} - (B+C)

     \tan (A) = \tan \bigg(\dfrac{\pi}{2} - (B+C)\bigg)

    Since tan(\frac{\pi}{2} - x) = \cot (x)

    \tan (A) = \cot (B+C)

    convert cot into tan and finish..
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  3. #3
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    Hello, arccos!


    \text{2) It is given that }\sin A = \frac{5}{13}\,\text{ where }90^o < A < 180^o

    \text{and that }\cos B = -\frac{33}{65}\,\text{ where }180^o <B<270^o

    \text{Find the exact value of: }\:\cos2(A+B)

    You need these identities: . \cos2\theta \;=\;2\cos^2\!\theta-1

    . . . . . . . . . . . . . . . . \cos(A + B) \;=\;\cos A\cos B - \sin A\sin B


    We are given: . \sin A \,=\,\frac{5}{13} in Quadrant 2.
    . . Hence: . \cos A \:=\:-\frac{12}{13}

    We are given: . \cos B \,=\,-\frac{33}{65} in Quadrant 3.
    . . Hence: . \sin B \,=\,-\frac{56}{65}


    \cos2(A+B) \;=\;2\cos^2(A+B) - 1

    . . . . . . . . . . =\; 2\bigg[\cos A\cos B - \sin A\sin B\bigg]^2 - 1

    . . . . . . . . . . =\; 2\bigg[\left(\text{-}\frac{12}{13}\right)\left(\text{-}\frac{33}{65}\right) - \left(\frac{5}{13}\right)\left(\text{-}\frac{56}{65}\right) \bigg] - 1

    . . . . . . . . . . =\;2\bigg[\frac{396}{845} + \frac{280}{845}\bigg]^2 - 1 \;\;=\;\;2\bigg[\frac{676}{845}\bigg]^2 - 1

    . . . . . . . . . . =\;\frac{913,952}{714,025} - 1 \;\;=\;\;\frac{199,927}{714,025} \;\;=\;\;\dfrac{7}{25}

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  4. #4
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    thank you! i totally overlooked the -1
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