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    I can't seem to get the answer for this question - or understand the mark scheme.

    (a) Use the identities for cos (A + B) and cos (A − B) to prove that
    2 cos A cos B ≡ cos (A + B) + cos (A − B).
    I did this easily

    (b) Hence, or otherwise, find in terms of π the solutions of the equation
    2 cos (x + \frac{\pi}{2}) = sec (x + \frac{\pi}{6}),
    for x in the interval 0 ≤ x ≤ \pi.

    I opened up the RHS, turned sec into 1/cos and multiplied through by cos and then simplified down to:

    -2sinx(cosxcos\frac{\pi}{6}-sinxsin\frac{\pi}{6})=1

    now, I'm stuck. I'm sure I went wrong somewhere, but I have no clue where.
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    Think you might have made a mistake using the identity. Consider this

     2 cos (x + \frac{\pi}{2}) cos (x + \frac{\pi}{6}) = cos (2x + \frac{2\pi}{3}) + cos(\frac{\pi}{3}) where i've used the identity for 2 cos A cos B you gave, defining

     A = x + \frac{\pi}{2}


     B = x + \frac{\pi}{6}
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    The question says hence, therefore this question relates to the identity on the previous question, you are expected to apply that identity to this question.

    Hint: let A = x + pie/2, and B = x + pie/6
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    (Original post by spread_logic_not_hate)
    Think you might have made a mistake using the identity. Consider this

     2 cos (x + \frac{\pi}{2}) cos (x + \frac{\pi}{6}) = cos (2x + \frac{2\pi}{3}) + cos(\frac{\pi}{3}) where i've used the identity for 2 cos A cos B you gave, defining

     A = x + \frac{\pi}{2}


     B = x + \frac{\pi}{6}
    I actually ignored the identity given, because I had no clue where to use it! :p: But now, it makes perfect sense. thank you!
 
 
 
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