Is sin(2θ)=cot(θ)(1-cos(2θ)) ?

The answer to whether sin(2θ)=cot(θ)(1-cos(2θ)) is True

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prove sin(2θ)=cot(θ)(1-cos(2θ))

prove

sin (2 θ) = cot (θ) (1 - cos (2 θ))

True

Solution steps

sin (2 θ) = cot (θ) (1 - cos (2 θ))

Manipulating right side

cot (θ) (1 - cos (2 θ))

Rewrite using trig identities

cot (θ) (1 - cos (2 θ))

Use the Double Angle identity:

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

= cot (θ) (1 - (1 - 2 sin^{2} (θ)))

Expand

1 - (1 - 2 sin^{2} (θ)) : 2 sin^{2} (θ)

1 - (1 - 2 sin^{2} (θ))

- (1 - 2 sin^{2} (θ)) : - 1 + 2 sin^{2} (θ)

- (1 - 2 sin^{2} (θ))

Distribute parentheses

= - (1) - (- 2 sin^{2} (θ))

Apply minus-plus rules

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

= - 1 + 2 sin^{2} (θ)

= 1 - 1 + 2 sin^{2} (θ)

1 - 1 = 0

= 2 sin^{2} (θ)

= 2 sin^{2} (θ) cot (θ)

= 2 sin^{2} (θ) cot (θ)

Express with sin, cos

2 cot (θ) sin^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

2 \cdot \frac{cos ( θ )}{sin ( θ )} sin^{2} (θ) : 2 cos (θ) sin (θ)

2 \cdot \frac{cos ( θ )}{sin ( θ )} sin^{2} (θ)

Multiply fractions:

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

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

Cancel the common factor:

sin (θ)

= 2 cos (θ) sin (θ)

= 2 cos (θ) sin (θ)

= 2 cos (θ) sin (θ)

Rewrite using trig identities

2 cos (θ) sin (θ)

Use the Double Angle identity:

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

= sin (2 θ)

= sin (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{cos ( 6 0 ^{\circ} )}{sin ( 6 0 ^{\circ} )} = cot (6 0^{\circ})

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

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

sin (θ + \frac{3 π}{2}) = - cos (θ)

sec (3 0^{\circ}) - sin (3 0^{\circ}) tan (3 0^{\circ}) = cos (3 0^{\circ})

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