Is (sin(2x))/(tan(2x))=1-2sin^2(x) ?

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

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prove (sin(2x))/(tan(2x))=1-2sin^2(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{sin ( 2 x )}{\frac{s i n ( 2 x )}{c o s ( 2 x )}}

Simplify

\frac{sin ( 2 x )}{\frac{s i n ( 2 x )}{c o s ( 2 x )}} : cos (2 x)

\frac{sin ( 2 x )}{\frac{s i n ( 2 x )}{c o s ( 2 x )}}

Apply the fraction rule:

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

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

Cancel the common factor:

sin (2 x)

= cos (2 x)

= cos (2 x)

= cos (2 x)

Rewrite using trig identities

cos (2 x)

Use the Double Angle identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

tan^{2} (x) = csc^{2} (x) \cdot tan^{2} (x) - 1

\frac{1 + tan ( θ )}{sin ( θ )} - sec (θ) = csc (θ)

cot (a) = tan (\frac{π}{2} - a)

cos (- x) + \frac{sin ( - x )}{cot ( - x )} = sec (x)

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