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ajprincessBest ResponseYou've already chosen the best response.3
\(\cos2\theta=2\cos^2\theta1\) \(\sin2\theta=2\sin\theta\cos\theta\) using these see if u can do. If u find any difficulties and need some more help plz ask:)
 one year ago

AnimalAinBest ResponseYou've already chosen the best response.1
\[ (1+\cosθ+\cos2θ)/(\sinθ+\sin2θ)=\cotθ \implies\]\[ \cosθ+2\cos^2θ =\frac{\cos \theta}{\sin \theta}(\sin \theta+2 \sin \theta \cos \theta)=\cos \theta + 2 \cos^2 \theta\]
 one year ago

sarah110128Best ResponseYou've already chosen the best response.0
prove: (1+cosθ+cos2θ)/(sinθ+sin2θ)=cotθ (1+cosθ+cos2θ)/(sinθ+2sinθcosθ) (cosθ+cos θ )/(sinθ+sinθ) cos2θ/sin2θ (2〖cos〗^2 θ1)/2cosθsinθ (cosθ1)/sinθ not sure where i went wrong... :/
 one year ago

ajprincessBest ResponseYou've already chosen the best response.3
Taking the expression in the left hand side \[\frac{1+\cos\theta+\cos2\theta}{\sin\theta+\sin2\theta}\] \[=\frac{1+\cos\theta+2\cos^2\theta1}{\sin\theta+2\sin\theta\cos\theta}\] 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
 one year ago

ajprincessBest ResponseYou've already chosen the best response.3
Is that clear @sarah11028
 one year ago

sarah110128Best ResponseYou've already chosen the best response.0
Sometimes I wonder why I'm even doing maths....hahaha thank you for you help! :)
 one year ago

ajprincessBest ResponseYou've already chosen the best response.3
welcome:) it is really easy if practice a bit:)
 one year ago

sarah110128Best ResponseYou've already chosen the best response.0
haahha lets hope it get easy! :) And hopefully it'll be my last year of maths
 one year ago

sarah110128Best ResponseYou've already chosen the best response.0
Cos3θ=4cos^3θ3cosθ (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(1cos^2θ) cosθ = 4cos^3θ  3cosθ thanks
 one year ago

ajprincessBest ResponseYou've already chosen the best response.3
\[\sin2\theta=2\sin\theta\cos\theta\] \[\sin2\theta*\sin\theta=2\sin\theta\cos\theta*\sin\theta\] \[=2\sin\theta*\sin\theta*cos\theta\] \[=2\sin^2\theta\cos\theta\] Is that clear?
 one year ago

ajprincessBest ResponseYou've already chosen the best response.3
\[\sin2\theta=\sin(\theta+\theta)\] \[=\sin\theta\cos\theta+\cos\theta\sin\theta\] \[=2\sin\theta\cos\theta\]
 one year ago

ajprincessBest ResponseYou've already chosen the best response.3
getting it @sarah110128?
 one year ago

sarah110128Best ResponseYou've already chosen the best response.0
ohh thanks, you've been a great help! :)
 one year ago
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