Is cos(10x)-cos(6x)=-2sin(8x)sin(2x) ?

The answer to whether cos(10x)-cos(6x)=-2sin(8x)sin(2x) is True

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prove cos(10x)-cos(6x)=-2sin(8x)sin(2x)

prove

cos (10 x) - cos (6 x) = - 2 sin (8 x) sin (2 x)

True

Solution steps

cos (10 x) - cos (6 x) = - 2 sin (8 x) sin (2 x)

Manipulating left side

cos (10 x) - cos (6 x)

Rewrite using trig identities

cos (10 x) - cos (6 x)

Use the Sum to Product identity:

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

= - 2 sin (\frac{10 x + 6 x}{2}) sin (\frac{10 x - 6 x}{2})

Simplify

- 2 sin (\frac{10 x + 6 x}{2}) sin (\frac{10 x - 6 x}{2}) : - 2 sin (8 x) sin (2 x)

- 2 sin (\frac{10 x + 6 x}{2}) sin (\frac{10 x - 6 x}{2})

\frac{10 x + 6 x}{2} = 8 x

\frac{10 x + 6 x}{2}

Add similar elements:

10 x + 6 x = 16 x

= \frac{16 x}{2}

Divide the numbers:

\frac{16}{2} = 8

= 8 x

= - 2 sin (8 x) sin (\frac{10 x - 6 x}{2})

\frac{10 x - 6 x}{2} = 2 x

\frac{10 x - 6 x}{2}

Add similar elements:

10 x - 6 x = 4 x

= \frac{4 x}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 x

= - 2 sin (8 x) sin (2 x)

= - 2 sin (8 x) sin (2 x)

= - 2 sin (8 x) sin (2 x)

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{1 + cos ( 2 x )}{cos ( x )} = 2 cos (x)

sec (y) - cos (y) = tan (y) sin (y)

csc (x) - sin (x) = cot (x) - cos (x)

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

Is cos(10x)-cos(6x)=-2sin(8x)sin(2x) ?
The answer to whether cos(10x)-cos(6x)=-2sin(8x)sin(2x) is True