Is prove-2sin(k-l)sin(k+l)=cos(2k)-cos(2l) ?

The answer to whether prove-2sin(k-l)sin(k+l)=cos(2k)-cos(2l) is True

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prove-2sin(k-l)sin(k+l)=cos(2k)-cos(2l)

prove

- 2 sin (k - l) sin (k + l) = cos (2 k) - cos (2 l)

True

Solution steps

- 2 sin (k - l) sin (k + l) = cos (2 k) - cos (2 l)

Manipulating right side

cos (2 k) - cos (2 l)

Rewrite using trig identities

cos (2 k) - cos (2 l)

Use the Sum to Product identity:

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

= - 2 sin (\frac{2 k + 2 l}{2}) sin (\frac{2 k - 2 l}{2})

Simplify

- 2 sin (\frac{2 k + 2 l}{2}) sin (\frac{2 k - 2 l}{2}) : - 2 sin (k + l) sin (k - l)

- 2 sin (\frac{2 k + 2 l}{2}) sin (\frac{2 k - 2 l}{2})

\frac{2 k + 2 l}{2} = k + l

\frac{2 k + 2 l}{2}

Factor out common term

2

= \frac{2 ( k + l )}{2}

Divide the numbers:

\frac{2}{2} = 1

= k + l

= - 2 sin (k + l) sin (\frac{2 k - 2 l}{2})

\frac{2 k - 2 l}{2} = k - l

\frac{2 k - 2 l}{2}

Factor out common term

2

= \frac{2 ( k - l )}{2}

Divide the numbers:

\frac{2}{2} = 1

= k - l

= - 2 sin (k + l) sin (k - l)

= - 2 sin (k + l) sin (k - l)

= - 2 sin (k + l) sin (k - l)

= - 2 sin (k - l) sin (k + l)

We showed that the two sides could take the same form

\Rightarrow True

2 tan (2 x) sin^{2} (x) = tan (2 x) - sin (2 x)

tan^{2} (x) + sec^{2} (x) = cot^{2} (x)

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

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

cos (θ - 6 0^{\circ}) + cos (θ + 6 0^{\circ}) = cos (θ)

Is prove-2sin(k-l)sin(k+l)=cos(2k)-cos(2l) ?
The answer to whether prove-2sin(k-l)sin(k+l)=cos(2k)-cos(2l) is True