Is cot(x)-2cot(2x)=tan(x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove cot(x)-2cot(2x)=tan(x)

Solution

prove

cot (x) - 2 cot (2 x) = tan (x)

Solution

True

Solution steps

cot (x) - 2 cot (2 x) = tan (x)

Manipulating left side

cot (x) - 2 cot (2 x)

Express with sin, cos

cot (x) - 2 cot (2 x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

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

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

Multiply fractions:

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

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

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

Least Common Multiplier of

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

sin (x), sin (2 x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (x)

sin (2 x)

= sin (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

sin (x) sin (2 x)

For

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

multiply the denominator and numerator by

sin (2 x)

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

For

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

multiply the denominator and numerator by

sin (x)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

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

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

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

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

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

Factor

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

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

Rewrite as

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

Factor out common term

2 sin (x)

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

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

Cancel

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

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

sin (x)

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

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

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

Use the Double Angle identity:

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

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

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

Simplify

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

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

Add similar elements:

cos (2 x) - cos (2 x) = 0

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

Cancel the common factor:

sin (x)

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

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

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

Use the basic trigonometric identity:

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

= tan (x)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove sin(x)=sqrt(3)sin(x/2)prove

sin (x) = 3 sin (\frac{x}{2})

prove sin(θ)cos(θ)sec(θ)csc(θ)=1 prove

sin (θ) cos (θ) sec (θ) csc (θ) = 1

prove sin(pi/2-0)=cos(0)prove

sin (\frac{π}{2} - 0) = cos (0)

prove sin(3x)=sin(3-4sin^2(x))prove

sin (3 x) = sin (3 - 4 sin^{2} (x))

prove (tan^2(θ))/(sec(θ)+1)=sec(θ)-1 prove

\frac{tan ^{2} ( θ )}{sec ( θ ) + 1} = sec (θ) - 1

Frequently Asked Questions (FAQ)

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