maths question

hi, please could i have help with part bii? im trying to use the identity given 8cos^4(a) but wouldn't there be 2 lots of them =1? Thanks!

f() is sin+cos, so there are four trig terms on the left of the identity in 9i)?

sorry i meant ii) b. using f(4theta)+4f(2theta), i grouped together the terms in ii and then got 2*8cos^4(3a)=1 but the ms says 8cos^4(3a)=1

Sure, if you subbed f() on the left of the identity in 9i) you (should) get the left of 9iib)? Obv with 3alpha instead of just theta. If not, post what youve done.

Really not sure how you go from line 1 to line 2, but youre given
f(t) = sin(t+30) + cos(t+60)
and question part 9i) shows
f(4t) + 4f(2t) = 8cos^4(t)-3
So sub the first equation into the second you get
sin(4t+30)+cos(4t+60)+4sin(2t+30)+4cos(2t+60) = 8cos^4(t)-3
now sub t=3a to get... and equate to 1.

Ohh ok I see, one more thing the ms says 49.1 but I’m not sure where they got it from. I’ve attached my working, 101.9 is invalid but idk how they got two roots in that range