Is cos(6x)+cos(2x)=2cos(4x)cos(2x) ?

The answer to whether cos(6x)+cos(2x)=2cos(4x)cos(2x) is True

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prove cos(6x)+cos(2x)=2cos(4x)cos(2x)

prove

cos (6 x) + cos (2 x) = 2 cos (4 x) cos (2 x)

True

Solution steps

cos (6 x) + cos (2 x) = 2 cos (4 x) cos (2 x)

Manipulating left side

cos (6 x) + cos (2 x)

Rewrite using trig identities

cos (6 x) + cos (2 x)

Use the Sum to Product identity:

cos (s) + cos (t) = 2 cos (\frac{s + t}{2}) cos (\frac{s - t}{2})

= 2 cos (\frac{6 x + 2 x}{2}) cos (\frac{6 x - 2 x}{2})

Simplify

2 cos (\frac{6 x + 2 x}{2}) cos (\frac{6 x - 2 x}{2}) : 2 cos (4 x) cos (2 x)

2 cos (\frac{6 x + 2 x}{2}) cos (\frac{6 x - 2 x}{2})

\frac{6 x + 2 x}{2} = 4 x

\frac{6 x + 2 x}{2}

Add similar elements:

6 x + 2 x = 8 x

= \frac{8 x}{2}

Divide the numbers:

\frac{8}{2} = 4

= 4 x

= 2 cos (4 x) cos (\frac{6 x - 2 x}{2})

\frac{6 x - 2 x}{2} = 2 x

\frac{6 x - 2 x}{2}

Add similar elements:

6 x - 2 x = 4 x

= \frac{4 x}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 x

= 2 cos (4 x) cos (2 x)

= 2 cos (4 x) cos (2 x)

= 2 cos (4 x) cos (2 x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e cos (x) + cot (x) + sin (x) = csc (x)

p ro v e cos (- θ) csc (- θ) tan (- θ) = 1

p ro v e sec (x) - 2 tan (x) = 0

p ro v e cos (3 θ) = cos (θ) (cos^{2} (θ) - 3 sin^{2} (θ))

p ro v e cos (θ) = sec (θ) - (sin (θ)) (tan (θ))

Is cos(6x)+cos(2x)=2cos(4x)cos(2x) ?
The answer to whether cos(6x)+cos(2x)=2cos(4x)cos(2x) is True