(a) (cos 3A) / 2 cos 2A - 1) = cos A and hence find cos 15 degree (b) [(sin^3 (x) + sin 3x) / sin x] + [(cos^3 (x) - cos 3x) /cos x] = 3 (c)...

b) [(sin^3 (x) + sin 3x) / sin x] + [(cos^3 (x) - cos 3x)
/cos x] = 3

We'll write the formula for sin
3x:

sin 3x = sin (2x+x) = sin 2x*cos x + sin x*cos
2x

sin 2x = 2sin x*cos x

cos
2x = 1 - 2(sin x)^2

We'll substitute sin 2x and cos 2x into
the formula of sin 3x:

sin 3x = 2 sinx*(cos x)^2 + sin
x*[1-2(sin x)^2]

We'll substitute (cos x)^2 = 1 - (sin x)^2
(fundamental formula of trigonometry)

sin 3x = 2sin x*[1 -
(sin x)^2] + sin x - 2(sin x)^3

We'll remove the
brackets:

sin 3x = 2sin x - 2(sin x)^3 + sin x - 2(sin
x)^3

We'll combine like
terms:

sin 3x = 3sin x - 4(sin
x)^3

We'll calculate the numerator of the first
ratio:

(sin x)^3 + sin 3x = (sin x)^3 + 3sin x - 4(sin
x)^3

We'll combine like
terms:

(sin x)^3 + sin 3x = 3sin x - 3(sin
x)^3

We'll factorize by 3 sin
x:

3sin x - 3(sin x)^3 = 3 sin x[1 - (sin
x)^2]

But 1 - (sin x)^2 = (cos
x)^2

3 sin x[1 - (sin x)^2] = 3 sin x (cos
x)^2

The first ratio will
become:

[(sin^3 (x) + sin 3x) / sin x] = 3 sin x (cos
x)^2/sin x

We'll
simplify:

 3 sin x (cos x)^2/sin x =  3 (cos
x)^2

Now we'll calculate the
second ratio:

[(cos x)^3 - cos 3x) /cos
x]

First, we'll calculate the formula for cos
3x:

cos 3x = cos (2x + x) = cos 2x*cos x - sin 2x*sin
x

cos 3x = 4 (cos x)^3 - 3 cos
x

We'll substitute cos 3x to the
numerator:

(cos x)^3 - cos 3x = (cos x)^3 - 4 (cos x)^3 + 3
cos x

(cos x)^3 - cos 3x = -3  (cos x)^3 + 3 cos
x

We'll factorize by -3 cos
x

-3  (cos x)^3 - 3 cos x = 3 cos x[1 - (cos
x)^2]

But 1 - (cos x)^2 = (sin
x)^2

3 cos x[1 - (cos x)^2] = 3 cos x (sin
x)^2

The second ratio will
become:

[(cos x)^3 - cos 3x) /cos x] = 3 cos x (sin
x)^2/cos x

We'll
simplify:

3 cos x (sin x)^2/cos x = 3 (sin
x)^2

The final expression will
be:

[(sin x)^3+ sin 3x) / sin x] + [(cos x)^3 - cos 3x)
/cos x] = 3 (cos x)^2 + 3 (sin x)^2

We'll factorize by
3:

3 (cos x)^2 + 3 (sin x)^2 = 3[(cos x)^2 + (sin
x)^2]

But (cos x)^2 + (sin x)^2 =
1

3[(cos x)^2 + (sin x)^2] =
3

[(sin x)^3+ sin 3x) / sin x] + [(cos x)^3 -
cos 3x) /cos x] =
3

 

a)
Now we'll prove the identity:

 (cos 3A) / 2
cos 2A - 1) = cos A

We'll take the formula for cos 3A from
the previous point b):

cos 3A = 4 (cos A)^3 - 3 cos
A

 cos 2A = 2 (cos A)^2 -
1

So, the difference 2 cos 2A - 1 = 2 (cos A)^2 - 1 -
1

2 (cos A)^2 - 1 - 1 = 2 (cos A)^2 -
2

We'll factorize by 2:

2 (cos
A)^2 - 2 = 2[(cos A)^2 - 1]

But (cos A)^2 - 1 = (sin
A)^2

2[(cos A)^2 - 1] = 2(sin A)^2