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

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

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Popular Trigonometry >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

Expand

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

1 - 1 = 0

= 2 sin^{2} (x)

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

sin (x)

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

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

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

Manipulating right side

2 csc (2 x) - tan (x)

Express with sin, cos

- tan (x) + 2 csc (2 x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

2 \cdot \frac{1}{sin ( 2 x )} = \frac{2}{sin ( 2 x )}

2 \cdot \frac{1}{sin ( 2 x )}

Multiply fractions:

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

= \frac{1 \cdot 2}{sin ( 2 x )}

Multiply the numbers:

1 \cdot 2 = 2

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

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

Least Common Multiplier of

cos (x), sin (2 x) : cos (x) sin (2 x)

cos (x), sin (2 x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (x)

sin (2 x)

= cos (x) sin (2 x)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos (x) sin (2 x)

For

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

multiply the denominator and numerator by

sin (2 x)

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

For

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

multiply the denominator and numerator by

cos (x)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

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

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

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

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

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

cos (x) \cdot 2 sin (x) cos (x)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

Factor

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

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

Rewrite as

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

Factor out common term

2 cos (x)

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

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

Cancel

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

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

cos (x)

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

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

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (x)

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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