Prove the following:1. cos(3π/2 + x) cos(2π + x) [cot(3π/2 - x) + cot(2π + x)] = 1 2. sin(n+1)x.sin(n+2)x + cos(n+1)x.cos(n+2)x = cos x 3....

1. We'll transform cot(3π/2 - x) + cot(2π + x) into a
product:

cot(3π/2 - x) + cot(2π + x)=sin(2π + x+3π/2 -
x)/-sinx*cosx

sin(2π + x+3π/2 - x)=sin(2π +
3π/2)=sin2πcos3π/2+sin3π/2cos2π=0*0-1*1=-1

cot(3π/2 - x) +
cot(2π + x)=-1/-sinx*cosx=1/sinx*cosx

On the other
hand,

cos(3π/2 + x) = cos3π/2cos x - sin3π/2sin
x=sinx

cos(2π + x) = cos2π cos x - sin2π sin x = cos
x

The given expression will
become:

sinx*cos x*(1/sinx*cosx) =
1

It is clear that, after simplifying, the
result will be:

1=1, so the identity is
true!

2. Here, we'll use the
formula:

cos A cos B + sin A sin B = cos
(A-B)

In our case, A = (n+1)x and B =
(n+2)x

cos(n+2)xcos(n+1)x+sin(n+2)xsin(n+1)x=cos[(n+2)x-(n+1)x]

cos[(n+2)x-(n+1)x]=cosx(n+2-n-1)=cos
x

So, cos(n+2)xcos(n+1)x+sin(n+2)xsin(n+1)x =
cos x

3. For solving the
difference between these 2 trigonometric functions, we'll use the following
formula:

cos a-cos b =
-2sin[(a+b)/2]sin[(a-b)/2]

In our case, a=(3π/4 + x) and
b=(3π/4 - x)

cos(3π/4 + x) - cos(3π/4 -
x)=

=2sin[(3π/4 + x+3π/4 - x)/2]sin[(3π/4 - x-3π/4 -
x)/2]

After simplifying, we'll
have:

cos(3π/4 + x) - cos(3π/4 -
x)=2sin(6π/8)sin(-x)

sin(6π/8) =
sin(3π/4)=sin(π/4+π/2)=sinπ/4cosπ/2+sinπ/2cosπ/4

sin(6π/8)
= sqrt 2/2 *0+1*sqrt2/2

sin(6π/8) =
sqrt2/2

So,
2sin(6π/8)sin(-x)=-2*(sqrt2/2)*sinx=-sqrt2*sinx, q.e.d.