Is tan(2x)-2tan(2x)sin^2(x)=sin(2x) ?

The answer to whether tan(2x)-2tan(2x)sin^2(x)=sin(2x) is True

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prove tan(2x)-2tan(2x)sin^2(x)=sin(2x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

Simplify

\frac{sin ( 2 x )}{cos ( 2 x )} - 2 sin^{2} (x) \frac{sin ( 2 x )}{cos ( 2 x )} : \frac{sin ( 2 x ) - 2 sin ^{2} ( x ) sin ( 2 x )}{cos ( 2 x )}

\frac{sin ( 2 x )}{cos ( 2 x )} - 2 sin^{2} (x) \frac{sin ( 2 x )}{cos ( 2 x )}

Multiply

2 sin^{2} (x) \frac{sin ( 2 x )}{cos ( 2 x )} : \frac{2 sin ^{2} ( x ) sin ( 2 x )}{cos ( 2 x )}

2 sin^{2} (x) \frac{sin ( 2 x )}{cos ( 2 x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sin ( 2 x ) \cdot 2 sin ^{2} ( x )}{cos ( 2 x )}

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

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

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

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

Rewrite using trig identities

\frac{sin ( 2 x ) - 2 sin ( 2 x ) sin ^{2} ( x )}{cos ( 2 x )}

Use the Double Angle identity:

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

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

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

Simplify

\frac{sin ( 2 x ) + ( cos ( 2 x ) - 1 ) sin ( 2 x )}{cos ( 2 x )} : sin (2 x)

\frac{sin ( 2 x ) + ( cos ( 2 x ) - 1 ) sin ( 2 x )}{cos ( 2 x )}

Factor

sin (2 x) + (cos (2 x) - 1) sin (2 x) : sin (2 x) cos (2 x)

sin (2 x) + (cos (2 x) - 1) sin (2 x)

Factor out common term

sin (2 x)

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

Refine

= sin (2 x) cos (2 x)

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

Cancel the common factor:

cos (2 x)

= sin (2 x)

= sin (2 x)

= sin (2 x)

We showed that the two sides could take the same form

\Rightarrow True

cot^{2} (θ) csc^{2} (θ) - cot^{2} (θ) = cot^{4} (θ)

sin (x - π) = - sin (x)

cos^{2} (x) + tan^{2} (x) cos^{2} (x) = 1

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

cos (x - \frac{π}{3}) + sin (\frac{π}{6} - x) = cos (x)

Is tan(2x)-2tan(2x)sin^2(x)=sin(2x) ?
The answer to whether tan(2x)-2tan(2x)sin^2(x)=sin(2x) is True