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Math Help - Trig nd AP

  1. #1
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    Trig nd AP

    sec (a-x) , sec a , sec (a+x) are in AP.... Proove, cos a=Root2 (cos (x/2))
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  2. #2
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    Quote Originally Posted by shaurya View Post
    sec (a-x) , sec a , sec (a+x) are in AP.... Proove, cos a=Root2 (cos (x/2))
    According to the properties of AP
    2sec(a) = sec(a+x) + sec(a-x) = 1/cos(a+x) + 1/cos(a-x)
    2/cos(a) = [cos(a-x) + cos(a+x)]/[cos(a-x)*cos(a+x)]
    Can you simplify the right hand side by using product to sum and sum to product formula?
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  3. #3
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    i cant simplify thats why i am askin
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  4. #4
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    2/cos(a) = 1/cos(a+x) + 1/cos(a-x)
    = [cos(a-x) +cos(a+x)]/cos(a+x)cos(a-x)
    = 2cos(a)cos(x)/[(1/2)(cos(2a) + cos(2x)
    cos(2a) + cos(2x) = 2cos^2(a)*cos(x)
    2cos^2(a) - 1 +2cos^x - 1 = 2cos^2(a)*cos(x)
    cos^2(a) +cos^2(x) -1 = cos^2*cos(x)
    cos^2(x) -1 = cos^2a*cos(x) - cos^2(a)
    [cos(x) + 1][cos(x) - 1] = cos^2(a)[cos(x) - 1]
    cos(x) + 1 = cos^2(a)
    2cos^2(x/2) = cos^2(a)
    cos(a) = sqrt(2)*cos(x/2)
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