Question

prove: (1+cosθ+cos2θ)/(sinθ+sin2θ)=cotθ

Answer 1

\(\cos2\theta=2\cos^2\theta-1\)
\(\sin2\theta=2\sin\theta\cos\theta\)
using these see if u can do. If u find any difficulties and need some more help plz ask:)

Answer 2

\[ (1+\cosθ+\cos2θ)/(\sinθ+\sin2θ)=\cotθ \implies\]\[ \cosθ+2\cos^2θ =\frac{\cos \theta}{\sin \theta}(\sin \theta+2 \sin \theta \cos \theta)=\cos \theta + 2 \cos^2 \theta\]

Answer 3

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... :/

Answer 4

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

Answer 5

Is that clear @sarah11028

Answer 6

Sometimes I wonder why I'm even doing maths....hahaha thank you for you help! :)

Answer 7

welcome:) it is really easy if practice a bit:)

Answer 8

haahha lets hope it get easy! :) And hopefully it'll be my last year of maths

Answer 9

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

Answer 10

@ajprincess

Answer 11

\[\sin2\theta=2\sin\theta\cos\theta\]
\[\sin2\theta*\sin\theta=2\sin\theta\cos\theta*\sin\theta\]
\[=2\sin\theta*\sin\theta*cos\theta\]
\[=2\sin^2\theta\cos\theta\]
Is that clear?

Answer 12

\[\sin2\theta=\sin(\theta+\theta)\]
\[=\sin\theta\cos\theta+\cos\theta\sin\theta\]
\[=2\sin\theta\cos\theta\]

Answer 13

getting it @sarah110128?

Answer 14

ohh thanks, you've been a great help! :)

Answer 15

welcome:)