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hey i figured out the answer to this question,http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_31.pdf
question number 6 (ii) ? how to do it?
√10 cos (x-71.57) = cos 2θ + 3sin 2θ and from that we can say that √10 cos (x-71.57)
let x = 2θ and let S = x-71.57 just to make things simpler
now we can say √10 cos S = 2
S = cos^-1 (2/√10)
S = 50.8 and to get another value we will have -50.8
now that S is 2θ-71.57 we can say that,
2θ-71.57 = 50.8 and 2θ-71.57 = -50.8
θ = 61.2 and θ = 10.4
i hope u got it!