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sarah110128
prove: (1+cosθ+cos2θ)/(sinθ+sin2θ)=cotθ
\(\cos2\theta=2\cos^2\theta-1\) \(\sin2\theta=2\sin\theta\cos\theta\) using these see if u can do. If u find any difficulties and need some more help plz ask:)
\[ (1+\cosθ+\cos2θ)/(\sinθ+\sin2θ)=\cotθ \implies\]\[ \cosθ+2\cos^2θ =\frac{\cos \theta}{\sin \theta}(\sin \theta+2 \sin \theta \cos \theta)=\cos \theta + 2 \cos^2 \theta\]
prove: (1+cosθ+cos2θ)/(sinθ+sin2θ)=cotθ (1+cosθ+cos2θ)/(sinθ+2sinθcosθ) (cosθ+cos θ )/(sinθ+sinθ) cos2θ/sin2θ (2〖cos〗^2 θ-1)/2cosθsinθ (cosθ-1)/sinθ not sure where i went wrong... :/
Taking the expression in the left hand side \[\frac{1+\cos\theta+\cos2\theta}{\sin\theta+\sin2\theta}\] \[=\frac{1+\cos\theta+2\cos^2\theta-1}{\sin\theta+2\sin\theta\cos\theta}\] factor out \(\cos\theta\) from numerator and \(\sin\theta\) from denominator \[=\frac{\cos\theta+2\cos^2\theta}{\sin\theta+2\sin\theta\cos\theta}\] \[=\frac{\cos\theta(1+2\cos\theta)}{\sin\theta(1+2\cos\theta)}\] cancelling the common factor \((1+2\cos\theta)\) \[=\frac{\cos\theta}{\sin\theta}\] \[=\cot\theta\] nw we have shown that lhs=rhs hence proved
Is that clear @sarah11028
Sometimes I wonder why I'm even doing maths....hahaha thank you for you help! :)
welcome:) it is really easy if practice a bit:)
haahha lets hope it get easy! :) And hopefully it'll be my last year of maths
Cos3θ=4cos^3θ-3cosθ (I am trying to prove this identity) cos3θ = cos(θ+2θ) = cosθ cos2θ - sinθ sin2θ = cosθ (2cos^2θ - 1) - 2sin^2θ cosθ (basically here can you please explain how to go from sinθ sin2θ ->2sin^2θ cosθ ) = 2cos^3θ - cosθ - 2(1-cos^2θ) cosθ = 4cos^3θ - 3cosθ thanks
\[\sin2\theta=2\sin\theta\cos\theta\] \[\sin2\theta*\sin\theta=2\sin\theta\cos\theta*\sin\theta\] \[=2\sin\theta*\sin\theta*cos\theta\] \[=2\sin^2\theta\cos\theta\] Is that clear?
\[\sin2\theta=\sin(\theta+\theta)\] \[=\sin\theta\cos\theta+\cos\theta\sin\theta\] \[=2\sin\theta\cos\theta\]
getting it @sarah110128?
ohh thanks, you've been a great help! :)