stuck on trig identities questions :(?

Alright, stick with me here...

cos(A + B) = cosAcosB - sinAsinB
cos(A - B) = cosAcosB + sinAsinB

Subtract the two identities

cos(A + B) - cos(A - B) = -2sinAsinB

Let (A + B) = P and Let (A-B) = Q

cosP - cos Q = -2sinAsinB

Now, with these 2 equations for P and Q, let's solve for A and B in terms of P and Q

A+B=P
A=P - B

Now on the equation of Q, substitute A with (P-B)

A - B = Q
(P-B) - B = Q
P - 2B = Q
-2B = Q - P
B = (-Q + P)/2

Then, solve for A too in terms of P and Q.

A + B = P
B = P - A

since A - B = Q
A - (P-A) = Q
A - P + A = Q
2A = P + Q
A= (P+Q)/2

so now that we have A and B in terms of P and Q

A = (P+Q)/2 B = (P-Q)/2

substitute them on the equation above:

cosP - cosQ = -2sinAsinB

cosP - cos Q = -2 sin[(P+Q)/2] sin[(P-Q)/2]

whoo... that was long.

Alright, let's go to b

cos 5x + sin3x - cosx = 0

since: 2sinAsinB = -[cos(A+B) - cos(A-B)]

we can assume that cos5x = cos(3x + 2x) and cos x = cos (3x - 2x) thus A= 3x B =2x

with that we can simplify cos5x - cos x = 2sin3xsin2x

put that back in the equation:

cos 5x - cosx + sin3x = 0
2sin3xsin2x + sin3x = 0

factor this

sin3x (2sin2x + 1) = 0

thus...

sin3x = 0
3x = arcsin 0
arcsin 0 = 0
thus x= 0

also..

2sin2x + 1 = 0
2sin2x = -1
sin2x = -1/2
2x = arcsin (-1/2)
arcsin (1/2) is 30 degrees, but it's negative so it's going to be in the 3rd or 4th quadrant and it's going to be 210 degrees and 330 degrees.

2x = 210, 330
x= 210/2 , 330/2
x = 105, 165

to sum it up

x = 0, 105, 165 (in degrees)

So long, this is what you call a mathematical essay. Anyways, hope it helped.

-2 sin P+Q/2 sin P-Q/2 that makes no sense.... write it properly and so we can actually have a go..