Question

By expressing cos2θ and sin2θ in terms of tanθ,solve the following eqn in the range 0≤θ≤360 .

a) cos2θ−2sin2θ=2

Answer 1

@mathmale

Answer 2

Sorry I didn't get to your question tonight (Thursday). Possibly tomorrow.

Answer 3

sure,no prob.

Answer 4

use the identities

cos 2θ = (1 - tan^2 θ ) / ( 1 + tan^2 θ )
and
sin 2θ = 2 tan θ / ( 1 + tan^2 θ)

Answer 5

Hmm,can u show me how u get that?

I thought
\(cos2\theta=cos^2\theta-sin^2\theta\)
\(=2cos^2\theta-1\)
\(=1-2sin^2\theta\)

\(sin2\theta=2sin\theta~cos\theta\)

Answer 6

(1 - tan^2 θ ) / ( 1 + tan^2 θ ) - 2* 2 tan θ / ( 1 + tan^2 θ) = 2

multiply through by ( 1 + tan^2 θ):-
1 - tan^2 θ - 4 tan θ = 2(1 + tan^2 θ)

3 tan^2 θ + 4 tan θ + 1 = 0

solve this for tan θ
- then you can find the required values of θ.

Answer 7

with \(t = \tan \theta\), you can grind these out:

\(\sin 2 \theta = \dfrac{2t}{t^2 + 1} \qquad \triangle\)

\(\cos 2 \theta = \dfrac{1-t^2}{t^2 + 1}\)

Eg: \(\triangle\) follows from:

\(\sin 2 \theta = 2 \sin \theta \cos \theta\)

\( = 2 \tan \theta \cos^2 \theta\)

\( = 2 \dfrac{\tan \theta }{\sec^2 \theta}\)

\( = 2 \dfrac{\tan \theta }{\tan^2 \theta + 1} = \dfrac{2t}{t^2 + 1} \)

Then have a play

Answer 8

I think I know how to do...

Answer 9

Thank you @celticcat @sillybilly123