# Is sin 2x sin x - cos x = sqrt 3/4 true for x = 7pi/6?

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For the sine function : sin (x + 2pi) = sin x and sin (x + pi)= -sin x and for the cosine function , cos ( x + pi) = -cos x.

sin 2x sin x - cos x, for x = 7pi/6

=> sin 14pi/6 * sin 7pi/6 - cos 7pi/6

=> sin ( 2pi/6 + 2pi) * sin (pi/6 + pi) - cos( pi + pi/6)

=> - sin 2pi/6 * sin pi/6 + cos pi/6

Now sin 2pi/6 = .5 and cos pi/6 = sqrt 3/2

=> -(sqrt 3/ 2)(1/2) + sqrt 3/2

=> -sqrt 3 / 4 + 2*sqrt 3 / 4

=> sqrt 3 / 4

For x = 7pi/6, therefore** sin 2x sin x - cos x = sqrt 3/ 4 **

We'll express si2x =2sinx*cosx

We'll re-write the equation:

2sinx*cosx*sinx - cosx = sqrt3/4

2cos x*(sin x)^2 - cos x = sqrt3/4

cos x(2(sin x)^2 - 1) = sqrt3/4 (1)

Now, we'll re-write the value 7pi/6 = (6pi/6 + pi/6) = (pi + pi/6)

The angle 7pi/6 is in the third quadrant, where the functions sine and cosine are both negative.

cos (pi + pi/6) = -cos pi/6 = -sqrt3/2

sin (pi + pi/6) = -sin pi/6 = -1/2

(sin x)^2 = 1/4

2(sin x)^2 = 2/4 = 1/2

We'll substitute all values in (1):

**(-sqrt3/2)(1/2 - 1) = (-sqrt3/2)(-1/2) = sqrt3/4 q.e.d.**

So, the given identity is holding for x = 7pi/6.

We put x = 7pi/6 in the given equation sin2x*sinx-cosx = (sqrt3)/4and see if it verifies the equation.

7pi/6 = (pi+pi/6)

sinx= (p+pi/6) = -sin(pi/6) = - 1/2

sin2x = sin(2pi+pi/3) = sin (pi/3) = sqrt3/2.

cos(pi+pi/6) = (sqrt3)/2

LHS = sin (2*7i/6)sin (7i/6)-cos7pi/6) = {(1/2)sqrt3}*(-1/2) - (1/2)(-sqrt3) = (sqrt3)/2 - sqrt3)/4 = **(sqrt3)/4 = RHS.**

So the given equality verifies for x = pi/6.