Hi guys,

1.

(a)

Answer: Negative,

(b) Taking

I actually haven't gotten any answers for this ^^, as I have little idea of how to do the question.

2. Hence, or otherwise, solve 2cos

Thanks in advance!

P.S. π is actually

1.

(a)

*****I do not need help with this part*****Consider the statement cos[(*π*/2) +*A*] = +/- sin*A*. For which sign is this statement true for all*A*? For which*A*is the statement true for both signs?Answer: Negative,

*A*= 0,*n**π*(*n*is an integer)(b) Taking

*A*= 5*θ*in (a), write down the value of*B*such that -sin5*θ*= cos*B*. Hence find the least value of*θ*between 0 and 2*π*such that sin5*θ*+ cos8*θ*= 0.I actually haven't gotten any answers for this ^^, as I have little idea of how to do the question.

2. Hence, or otherwise, solve 2cos

*θ*+ 2cos(*θ*+ (*π*/3)) = 3, for 0 <*θ*< 2*π*. (This is the second part of the question. The first part was to express the expression on the LHS in the form*R*cos(*θ*+*α*), where*R*>0 and*α*is between 0 and 90 degrees (not including them). The answer was 2√3cos(*θ*+ (*π*/6)).) I keep getting 5*π*/3, and I don't know where I'm going wrong (I'm sure I'm wrong).Thanks in advance!

P.S. π is actually

**pi**, just before anyone asks, to prevent people wasting time asking this.
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