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    Find the values of R and a, where R > 0 and 0 < a < 90 for which 5/(8cosx - 6sinx) =Rsec(x+a)

    so far I know that it can be changed to R/Cos(x+a) but I'm not too sure where to go from there
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    (Original post by not_lucas1)
    Find the values of R and a, where R > 0 and 0 < a < 90 for which 5/(8cosx - 6sinx) =Rsec(x+a)

    so far I know that it can be changed to R/Cos(x+a) but I'm not too sure where to go from there
    If you take the inverse of both sides, you want to find \frac{8}{5}\cos x - \frac{6}{5}\sin x = \frac{1}{R} \cos (x+ \alpha) which is a standard problem.

    In particular: \frac{1}{R^2} = \frac{8^2}{5^2} + \frac{6^2}{5^2}, can you find \alpha?
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    (Original post by Zacken)
    If you take the inverse of both sides, you want to find \frac{8}{5}\cos x - \frac{6}{5}\sin x = \frac{1}{R} \cos (x+ \alpha) which is a standard problem.

    In particular: \frac{1}{R^2} = \frac{8^2}{5^2} + \frac{6^2}{5^2}, can you find \alpha?
    I ended up with 24/25 Cos(x+53.13) which i changed to 25/24 Sec(x+53.13)
    not sure if its right
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    (Original post by not_lucas1)
    I ended up with 24/25 Cos(x+53.13) which i changed to 25/24 Sec(x+53.13)
    not sure if its right
    Don't think it is.
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    well 1/cosx = secx so if youve got 5/Zcos(x +a) then your answer will be 5/Z sec (x + a), where 5/Z is equal to R.
 
 
 
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