The Student Room Group

Trig functions and equation

i need help with the question in this link...
im not sure how to work it out..

https://imgur.com/a/rIB0A
Original post by Chez 01
i need help with the question in this link...
im not sure how to work it out..

https://imgur.com/a/rIB0A


(a) All you need to do is firstly express the equation entirely in terms of sinθ\sin \theta, then just use the given substitution x=sinθx= \sin \theta to obtain what they have.

(b) (i) That's just a quadratic to solve, I'm sure you can do this.

(ii) Use the fact that we denoted x=sinθx = \sin \theta, and along with the solutions to (i) to then solve for θ\theta in the given domain.
Reply 2
Original post by Chez 01
i need help with the question in this link...
im not sure how to work it out..

https://imgur.com/a/rIB0A

3cos2θ=3(1sin2θ)3\cos^2 \theta = 3(1-\sin^2 \theta)

and when you expand that you get

33sin2θ3 - 3\sin^2 \theta

You just wrote 3sin2θ3\sin^2\theta in your working.
Reply 3
Original post by RDKGames
(a) All you need to do is firstly express the equation entirely in terms of sinθ\sin \theta, then just use the given substitution x=sinθx= \sin \theta to obtain what they have.

(b) (i) That's just a quadratic to solve, I'm sure you can do this.

(ii) Use the fact that we denoted x=sinθx = \sin \theta, and along with the solutions to (i) to then solve for θ\theta in the given domain.


thankyou for replying....so i got 90 and 19.5 and 160.5 degrees for the last question...is that right?
Reply 4
Original post by Notnek
3cos2θ=3(1sin2θ)3\cos^2 \theta = 3(1-\sin^2 \theta)

and when you expand that you get

33sin2θ3 - 3\sin^2 \theta

You just wrote 3sin2θ3\sin^2\theta in your working.


oh okay...thankyou

Quick Reply

Latest