Use of trig formulas

Question

a) 1+cos(8a)=?, given that 1+cos(2a)=2cos²(a)

b) sin²(4a)=?, given that sin(2a)=2sin(a)*cos(a)

Answer 1

let's say the given is just cos(a) value, and from sin(a)=+/-√(1-cos²(a)) you have +/- the sin value.

finding cos(8a) can be complicated since cos^-1(k)=(+/-)a+2πn,

if cos(a)=k,,,, cos(8a) could be cos(8*-a) or cos( 8*a)

1 + cos(8a) = 1 + cos[2(4a)] = 2cos²(4a)

=2*(cos²(2a)-sin²(2a))²

=2*(cos⁴(2a)-2sin²(2a)cos²(2a)+sin⁴(2a))

=2*((cos²(a)-sin²(a))⁴-(4*sin(a)cos(a))²*(cos²(a)-sin²(a))²+(2*sin(a)cos(a))⁴)

since all sin(a) are squared or to the fourth power the +/- ambiguity for sin(a)=+/-√(1-cos²(a))

can be resolved by choosing either.

sin²(4a)=(2*sin(2a)cos(2a))²

=(2*(2*sin(a)cos(a))*(cos²(a)-sin²(a)))²

as can be seen, the ambiguity for cos(a)=+/-√(1-sin²(a))

is not of concern since cos(a) is alsways raised to even poower.

agian, if given sin(a)=k then a=sin^-1(k)+2πn or a=π-sin^-1(k)+2πn

example sin(π/18)=.173, given sin(a)=.173, a=π/18 or a=17π18

so sin^-1(k) cannot just be multiplied by 4.

Answer 2

1 + cos(8a) = 1 + cos[2(4a)] = 2cos²(4a)