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Math Help - trigonometry problems

  1. #1
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    trigonometry problems

    1.[SOLVED]
    Given that \frac{cos(A-B)}{cos(A+B)} = \frac{7}{5}, prove that cos A cos B = 6 sin A sin B and deduce a relationship between tan A and tan B. Given further that A+B= 45^\circ, calculate the value of tan A + tan B.


    2.
    If A, B and C are the angles of a triangle, prove that sin A - sin(B+C)=0.
    Hence, prove that sin A - sin(B-C)= 2 cos B sin C .
    Last edited by wintersoltice; August 16th 2008 at 07:25 AM.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by wintersoltice View Post
    1.
    Given that \frac{cos(A-B)}{cos(A+B)} = \frac{7}{5}, prove that cos A cos B = 6 sin A sin B and deduce a relationship between tan A and tan B. Given further that A+B= 45^\circ, calculate the value of tan A + tan B.
    \cos (A-B) = \cos (A)\cos (B) + \sin (A) \sin (B)

     \cos (A+B) = \cos (A)\cos (B) - \sin (A) \sin (B)<br />

    so 5 \cos (A-B) = 7\cos (A+B) ...

    after that, \frac{6\sin (A) \sin (B)}{\cos (A) \cos(B)} = 1...

    and \tan (A + B) = \frac{\tan (A) + \tan (B)}{1 - \tan(A)\tan(B)}

    Quote Originally Posted by wintersoltice View Post
    2.
    If A, B and C are the angles of a triangle, prove that sin A - sin(B+C)=0.
    Hence, prove that sin A - sin(B-C)= 2 cos B sin C .
    A+B+C = \pi so, A = \pi - (B+C)...
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  3. #3
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    Quote Originally Posted by wintersoltice View Post
    1.
    Given that \frac{cos(A-B)}{cos(A+B)} = \frac{7}{5}, prove that cos A cos B = 6 sin A sin B

    Mr F says: Substitute from the appropriate double angle formula in the numerator and denominator, cross multiply to get rid of fractions, and simplify.

    and deduce a relationship between tan A and tan B.

    Mr F says: Divide both sides of cos A cos B = 6 sin A sin B by cos A cos B .....

    Given further that A+B= 45^\circ, calculate the value of tan A + tan B.

    Mr F says: {\color{red}\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}}. And from the previous result you know what tan A tan B is equal to.


    2.
    If A, B and C are the angles of a triangle, prove that sin A - sin(B+C)=0.

    Mr F says: Here's a start: {\color{red}A + B + C = 180 \Rightarrow A = 180 - B - C } {\color{red}\Rightarrow \sin A = \sin (180 - B - C) = \sin ([180 - B] - C) \Rightarrow } ......

    Hence, prove that sin A - sin(B-C)= 2 cos B sin C .

    Mr F says: It's not difficult to use the compound angle formulae to show that sin (B + C) = sin (B - C) + 2 sin C cos B ......
    ..
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  4. #4
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    i still don't get it for question 2

    A + B + C = 180
    A=180 - B - C

    sin A -sin(B+C)=0
    sin A = sin (B+C)

    LHS
    = sin A
    =sin (180 - B- C)
    =sin(-B-C)
    = - sin (B+C)

    but that doesn't equal to RHS......
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  5. #5
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    its very simple apply componendo n dividendo.
    cos(a-b) = cosa.cosb + sina.sinb
    cos(a+b) = cosa.cosb - sina.sinb
    so by componendo n dividendo u will get the ans....
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  6. #6
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    its very simple apply componendo n dividendo.
    cos(a-b) = cosa.cosb + sina.sinb
    cos(a+b) = cosa.cosb - sina.sinb
    so by componendo n dividendo u will get the ans....

    by that relation you will get tanA.tanB=1/6.
    if A+B = 45 YOU CAN GET value of tanA+TanB=tAN45(1-TANA.TANB)
    WHICH IS EQUAL TO 5/6 IF YOU PUT VALUE OF TANA.TANB IN ABOVE EQUATION....
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  7. #7
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    Quote Originally Posted by vishalgarg View Post
    its very simple apply componendo n dividendo.
    cos(a-b) = cosa.cosb + sina.sinb
    cos(a+b) = cosa.cosb - sina.sinb
    so by componendo n dividendo u will get the ans....

    by that relation you will get tanA.tanB=1/6.
    if A+B = 45 YOU CAN GET value of tanA+TanB=tAN45(1-TANA.TANB)
    WHICH IS EQUAL TO 5/6 IF YOU PUT VALUE OF TANA.TANB IN ABOVE EQUATION....

    thanks for your help but i have solved question 1....
    question 2......still can't get it ....
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  8. #8
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    Quote Originally Posted by wintersoltice View Post
    thanks for your help but i have solved question 1....
    question 2......still can't get it ....
    Quote Originally Posted by wintersoltice View Post
    [snip]
    2.
    If A, B and C are the angles of a triangle, prove that sin A - sin(B+C)=0.

    Mr F says: Here's a start: {\color{red}A + B + C = 180 \Rightarrow A = 180 - B - C } {\color{red}\Rightarrow \sin A = \sin (180 - B - C) = \sin ([180 - B] - C) \Rightarrow }

    [snip]
    \sin A = \sin [180 - B] \cos C - \sin C \cos [180 - B]

    Now substitute \sin [180 - B] = \sin B and \cos [180 - B] = - \cos B ....
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  9. #9
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    Quote Originally Posted by kalagota View Post
    A+B+C = \pi so, A = \pi - (B+C)...
    an easier approach is..

    \sin A - \sin (B+C) = \sin (\pi - (B+C)) - \sin (B+C) = ...

    and by reduction formula, for any angle x, \sin (\pi - x) = \sin\pi \cos x - \sin x \cos\pi = - (-1)\sin x = \sin x

    hence, \sin (\pi - (B+C)) =\sin (B+C)...
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  10. #10
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    Quote Originally Posted by kalagota View Post
    an easier approach is..

    \sin A - \sin (B+C) = \sin (\pi - (B+C)) - \sin (B+C) = ...

    and by reduction formula, for any angle x, \sin (\pi - x) = \sin\pi \cos x - \sin x \cos\pi = - (-1)\sin x = \sin x

    hence, \sin (\pi - (B+C)) =\sin (B+C)...
    You say tomayto, I say tomarto ..... The technique is obvious enough that the OP can decide which way to go.
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