# 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. cos(3π/4 + x) - cos(3π/4 - x) = - √2 sin x

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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.

1. cos(3π/2 + x) cos(2π + x) [cot(3π/2 - x) + cot(2π + x)]

We use the 1st quadrant values and rewrite the LHS

LHS: (sinx)(cosx){tanx+cotx) = sin^2x +cos^2x = 1 = RHS.

2. sin(n+1)x.sin(n+2)x + cos(n+1)x.cos(n+2)x = cos x

LHS:(1/2) [Cos (x)-cos(2n+3)+ (1/2)[cos[(2n+3)+ cosx]

(1/2)(cosx+cosx)= cosx = RHS.

3. cos(3π/4 + x) - cos(3π/4 - x) = - √2 sin x

LHS:

cos(3n/4+x)-cos(3n/4-x) = 2sin (-2x)cos(3n/2+0)

= 2sin(-2x/2)*sin ((3n/2)/2)

= (-2sinx)sin (3n/4)

=(-2sinx)(1/sqrt2)

=-[2^(1/2) ]sinx.

Note: We used Cos(A+B)-cos(A-B) = 2sinA sin B in Q2 and Q3.