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Math Help - Word problems on trigonometric ratios.

  1. #1
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    Word problems on trigonometric ratios.

    Hi MHF, I need some help with 2 problems that I have been stuck with please

    1) If \displaystyle sin(\dfrac{\pi}{4} - \theta) = \dfrac {5}{13} and
    0 <= theta <= pi/4 , find the value of \displaystyle \dfrac {cos2\theta}{cos(\dfrac{\pi}{4} + \theta)} .

    2) In triangle ABC,the angle B is twice the size as compared to that of angle A.
    Show that
    a) \displaystyle b = 2a cos A
    b) the area of triangle ABC = \displaystyle a^2 cosA sin3A
    (a is the side opposite angle A and b....angle B .....)

    For question 1 , I could not draw \displaystyle sin(\dfrac{\pi}{4} - \theta) = \dfrac {5}{13} in terms of a triangle in the quadrant and hence I tried to manipulate it and got to cos\theta - sin\theta = \dfrac{5(\sqrt2)}{13} (cheated a little by changing radians to degrees here) and got stuck. I am totally lost on this from here on and my approach is wrong too i guess,since we need to find the value of  cos2\theta .

    For question 2, i tried to use the law of cosines (to find length of side) to see if i could derive anything but i am stuck with  2b^2 = c^2 + a^2 cos2A

    Any help would be greatly appreciated.
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  2. #2
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    1) If sin(π/4 - θ) = 5/13, then cos(π/4 - θ) = 12/13.

    Again sin(π/4 - θ) = sin[π/2 - (π/4 + θ)] = cos(π/4 + θ)

    coa2θ= sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ)

    Substitue hese values and find the result.

    2) B = 2A

    So sinB = sin2A = 2sinAcosA.

    Now proceed.
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  3. #3
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    Quote Originally Posted by sa-ri-ga-ma View Post
    1) If sin(π/4 - θ) = 5/13, then cos(π/4 - θ) = 12/13.

    Again sin(π/4 - θ) = sin[π/2 - (π/4 + θ)] = cos(π/4 + θ)

    coa2θ= sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ)

    Substitue hese values and find the result.

    2) B = 2A

    So sinB = sin2A = 2sinAcosA.
    Now proceed.

    why is sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ) and not cos(2θ)?
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  4. #4
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    Quote Originally Posted by arccos View Post
    why is sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ) and not cos(2θ)?
    I didn't understand youe question. In fact sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ) = cos(2θ)
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  5. #5
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    Quote Originally Posted by sa-ri-ga-ma View Post
    I didn't understand youe question. In fact sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ) = cos(2θ)
    Oh yes sorry. My bad. I overlooked sin2A = 2sinAcosA. I will attempt the questions now and let you know in a bit.
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  6. #6
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    Quote Originally Posted by sa-ri-ga-ma View Post
    1) If sin(π/4 - θ) = 5/13, then cos(π/4 - θ) = 12/13.

    Again sin(π/4 - θ) = sin[π/2 - (π/4 + θ)] = cos(π/4 + θ)

    coa2θ= sin(π/2 - 2θ) = 2sin(π/4 - θ)cos(π/4 - θ)

    Substitue hese values and find the result.

    2) B = 2A

    So sinB = sin2A = 2sinAcosA.

    Now proceed.
    I am still unsure of what to do after we made sinB = sin2A = 2sinAcosA.
    Could you please give me more hints?

    EDIT: I have solved part 1(b = 2a cosA). Now i'm stuck on part 2.
    Last edited by arccos; November 12th 2010 at 07:39 PM.
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  7. #7
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    sinB = 2sinAcosA

    So b = 2a*cosA

    A + B + C = 180

    C = 180 - (A +B) = 180 - 3A

    Area od the triangle = 1/2a*b*sinC = 1/2*a*2a*cosA*sin[180 - (A + B)] = a^2*cosA*sin3A.
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  8. #8
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    Quote Originally Posted by sa-ri-ga-ma View Post
    sinB = 2sinAcosA

    So b = 2a*cosA

    A + B + C = 180

    C = 180 - (A +B) = 180 - 3A

    Area od the triangle = 1/2a*b*sinC = 1/2*a*2a*cosA*sin[180 - (A + B)] = a^2*cosA*sin3A.
    Thank you so much! I was trying to figure out angle C. Totally overlooked that.
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