November 21st, 2008, 12:55 AM
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| Senior Member | | Join Date: Jul 2008
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 Trig Problem
 (a) Explain why AP = 2cosθ. (b) Find a similar expression for BP in terms of θ. (c) Explain why the coordinates of pt P is (cos2θ, sin2θ). (d) Prove that triangle APQ is similar to triangle ABP. (e) Hence deduce that (i) sin2θ = 2sin θcosθ (ii) cos2θ = 2cos^2 θ - 1 (iii) tan2θ = (2tanθ)/(1-tan^ 2 θ) Could someone please help me on this problem?? I've worked out parts a and b but added them in just in case it is required for the other parts. |