Is cos(4x)-cos(3x)cos(x)=sin(2x)-4cos(2x)sin(2x) ?

The answer to whether cos(4x)-cos(3x)cos(x)=sin(2x)-4cos(2x)sin(2x) is False

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prove cos(4x)-cos(3x)cos(x)=sin(2x)-4cos(2x)sin(2x)

prove

cos (4 x) - cos (3 x) cos (x) = sin (2 x) - 4 cos (2 x) sin (2 x)

False

Solution steps

cos (4 x) - cos (3 x) cos (x) = sin (2 x) - 4 cos (2 x) sin (2 x)

Show that the two sides are not equal

For

cos (4 x) - cos (3 x) cos (x) = sin (2 x) - 4 cos (2 x) sin (2 x)

plug in

x = 1

cos (4 \cdot 1) - cos (3 \cdot 1) cos (1) = - 0.11874 \dots

cos (4 \cdot 1) - cos (3 \cdot 1) cos (1)

Refine to a decimal form

= - 0.11874 \dots

sin (2 \cdot 1) - 4 cos (2 \cdot 1) sin (2 \cdot 1) = 2.42290 \dots

sin (2 \cdot 1) - 4 cos (2 \cdot 1) sin (2 \cdot 1)

Refine to a decimal form

= 2.42290 \dots

The two sides are not equal

\Rightarrow False

2 - 2 sin (x) - cos^{2} (x) = (sin (x) - 1)^{2}

cos (4 x) = 1 - 8 sin^{2} (x) cos (x)

\frac{( 1 )}{( 2 )} (cot (θ) - tan (θ)) = cot (2 θ)

1 - cos (x) = \frac{1 - cos ^{2} ( x )}{2}

\frac{1}{sin ^{2} ( x )} = 1 - tan^{2} (x)

Is cos(4x)-cos(3x)cos(x)=sin(2x)-4cos(2x)sin(2x) ?
The answer to whether cos(4x)-cos(3x)cos(x)=sin(2x)-4cos(2x)sin(2x) is False