trig mark scheme

Can someone explain how you approach this question? I tried using double single by splitting 3A into 2A and A but it didn't go well. specifically I don't understand how you get from marking point 1 to 2 in the mark scheme attached... thank you!

(edited 1 year ago)

Screen Shot 2023-04-05 at 11.48.26.png58.7KB

Screen Shot 2023-04-05 at 11.51.02.png7.3KB

Reply 1

mqb2766

Just the harmonic identity using a triple angle (which is irrelevant) so
R sin(3A - alpha)
where R and alpha are given by the usual pythagoras and tan.

Reply 2

lavely

thank you so much!! When I worked out the hypotenuse for R, I got 2, so shouldn't it be 2sin(3A – 30°) = 1/4?

Original post by mqb2766

Just the harmonic identity using a triple angle (which is irrelevant) so
R sin(3A - alpha)
where R and alpha are given by the usual pythagoras and tan.