Is 1-cos(2x)=tan(x)sin(2x) ?

The answer to whether 1-cos(2x)=tan(x)sin(2x) is True

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prove 1-cos(2x)=tan(x)sin(2x)

prove

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

True

Solution steps

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

Manipulating right side

tan (x) sin (2 x)

Express with sin, cos

sin (2 x) tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Simplify

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

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

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

2 sin (x) cos (x) sin (x)

Apply exponent rule:

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

sin (x) sin (x) = sin^{1 + 1} (x)

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

Add the numbers:

1 + 1 = 2

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

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

Cancel the common factor:

cos (x)

= 2 sin^{2} (x)

= 2 sin^{2} (x)

Use the Double Angle identity:

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

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

= - (cos (2 x) - 1)

Simplify

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

- (cos (2 x) - 1)

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - cos (2 x) + 1

= - cos (2 x) + 1

= - cos (2 x) + 1

= 1 - cos (2 x)

We showed that the two sides could take the same form

\Rightarrow True

Is 1-cos(2x)=tan(x)sin(2x) ?
The answer to whether 1-cos(2x)=tan(x)sin(2x) is True