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

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

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

prove

cos (4 x) cos (2 x) = cos^{2} (3 x) - cos^{2} (x)

False

Solution steps

cos (4 x) cos (2 x) = cos^{2} (3 x) - cos^{2} (x)

Show that the two sides are not equal

For

cos (4 x) cos (2 x) = cos^{2} (3 x) - cos^{2} (x)

plug in

x = 0

cos (4 \cdot 0) cos (2 \cdot 0) = 1

cos (4 \cdot 0) cos (2 \cdot 0)

Simplify:

4 \cdot 0 = 0

4 \cdot 0

Apply rule

0 \cdot a = 0

= 0

Simplify:

2 \cdot 0 = 0

2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

cos (0) cos (0) = cos^{2} (0)

cos (0) cos (0)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (0) cos (0) = cos^{1 + 1} (0)

= cos^{1 + 1} (0)

Add the numbers:

1 + 1 = 2

= cos^{2} (0)

= cos^{2} (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= 1^{2}

Simplify

= 1

cos^{2} (3 \cdot 0) - cos^{2} (0) = 0

cos^{2} (3 \cdot 0) - cos^{2} (0)

Refine to a decimal form

= 0

The two sides are not equal

\Rightarrow False

csc (x) - \frac{sin ( x )}{1 - cos ( x )} = cot (x)

cos (θ - \frac{3 π}{2}) = - sin (θ)

cos (4 x) - sin (4 x) = 1 - 2 sin (2 x)

cos (x) tan^{3} (x) = sin (x) tan^{2} (x)

sin (x) = \frac{2}{2}

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