Hard Trigonometric Identities?

Question

Prove the two identities below.

I found these two to be very hard.

(1) sin(x)(sin3(x) + sin5(x)) = cos(x)(cos(cos3(x) - cos5(x))

(2) (cos(a) + cos(b))/(cos(a) - cos(b)) = - (cot(a + b)/2)(cot(a - b)/2)

Answer 1

I guess the 'cos' was repeated inadvertently.
Note: cos(A+B) = cosA cosB - sinA sinB ; cos(-A) = cosA

⑴ sin x (sin 3x + sin 5x)
= sinx · sin3x + sinx · sin5x
= [sinx sin3x - cosx cos3x] + cosx cos3x + [sinx sin5x + cosx cos5x] - cosx cos5x
= [-cos(x+3x)] + cosx cos3x + [cos(x-5x)] - cosx cos5x
= -cos4x + cosx cos3x + cos4x - cosx cos5x
= cosx cos3x - cosx cos5x
= cosx (cos3x - cos5x)

⑵ Note: cosθ + cosω = 2cos[½(θ+ω)] cos[½(θ-ω)]
while ... cosθ - cosω = -2sin[½(θ+ω)] sin[½(θ-ω)]
(these formulas are conversion of sum to product)

Getting their quotient will immediately yield the result.




Otherwise a = [ (a+b)/2 + (a-b)/2 ] ....... while ... b = [ (a+b)/2 - (a-b)/2]
So cos(a) = cos(a+b)/2 cos(a-b)/2 - sin(a+b)/2 sin(a-b)/2
cos(b) = cos(a+b)/2 cos(a-b)/2 + sin(a+b)/2 sin(a-b)/2

Thus, cos(a) + cos(b) = 2 cos(a+b)/2 cos(a-b)/2,
which is the above conversion formula. It is easy to see the result for the difference.

I hope this is clear for you.

d:

Answer 2

is this your homework, or what? they are no longer no longer elementary in any respect... don't be lazy, in simple terms think of roughly each and each situation and resolve it. Like cos + sin*tan = sec cos + sin*tan = a million/cos cos^2 + cos*sin*(sin/cos) = a million cos^2 + sin^2 = a million

Answer 3

remember the formulas
sinp+sinq=?
cosp-cosq=?