# What is the angle x if the identity sin2x-(cos 2x)/2 -1/2=cosx-2cosx*tanx?

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### 3 Answers

We have to solve for x given sin2x-(cos 2x)/2 -1/2=cosx-2cosx*tanx

sin 2x - (cos 2x)/2 -1/2 = cos x - 2cos x * tan x

=> 2 sin x cos x - (1 - 2 (sin x)^2)/2 - 1/2 = cos x - 2cos x * sin x/ cos x

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

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

=> cos x ( 2 sin x - cos x) = cos x - 2 sin x

=> - cos x = 1

=> cos x = -1

=> x = arc cos -1

=> x = pi

also 2 sin x - cos x = 0

=> 2 sin x = cos x

=> 2 tan x = 1

=> x = arc tan 1/2

**Therefore x = arc tan (1/2) and x = pi**

To determine the angle x, we'll have to solve the equation.

We'll write the first term of the equation as the function sine of a double angle.

We'll apply the formula for the double angle:

sin 2a = sin (a+a)=sina*cosa + sina*cosa=2sina*cosa

We'll replace 2a by 2x and we'll get:

sin 2a = 2sin x*cos x

We'll re-write the equation:

2sin x*cos x - (cos 2x)/2 -1/2 = cosx-2cosx*tanx

We'll factorize by -1/2 the last 2 terms from the left side:

2sin x*cos x - (1-cos 2x)/2 = cosx-2cosx*tanx

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

2sin x*cos x - (cos x)^2 = cosx-2cosx*sinx/cos x

We'll simplify and we'll get:

2sin x*cos x - (cos x)^2 = cosx - 2sinx

We'll factorize by cos x to the left side:

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

We'll move all the terms to the left side:

cos x(2sin x - cos x) + (2sin x - cos x) = 0

We'll factorize by (2sin x - cos x):

(2sin x - cos x)(cos x + 1) = 0

We'll set each factor as zero:

cos x + 1 = 0

We'll add -1 both sides:

cos x = -1

x = arccos (-1)

x = pi

2sin x - cos x = 0

We'll divide by cos x:

2tan x - 1 = 0

tan x = 1/2

x = arctan (1/2)

**The angle x has the following values: { pi ; arctan (1/2) }.**

SORRY!

The identity is

sin2x-(cos 2x)/2 -1/2=cosx-2cosx*tanx