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