Is (cos(4x)+cos(6x))/(sin(4x)-sin(6x))=-cot(x) ?

The answer to whether (cos(4x)+cos(6x))/(sin(4x)-sin(6x))=-cot(x) is True

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Popular >

prove (cos(4x)+cos(6x))/(sin(4x)-sin(6x))=-cot(x)

Solution

prove

\frac{cos ( 4 x ) + cos ( 6 x )}{sin ( 4 x ) - sin ( 6 x )} = - cot (x)

Solution

True

Solution steps

\frac{cos ( 4 x ) + cos ( 6 x )}{sin ( 4 x ) - sin ( 6 x )} = - cot (x)

Manipulating left side

\frac{cos ( 4 x ) + cos ( 6 x )}{sin ( 4 x ) - sin ( 6 x )}

Rewrite using trig identities

\frac{cos ( 4 x ) + cos ( 6 x )}{sin ( 4 x ) - sin ( 6 x )}

Use the Sum to Product identity:

cos (s) + cos (t) = 2 cos (\frac{s + t}{2}) cos (\frac{s - t}{2})

= \frac{2 cos ( \frac{4 x + 6 x}{2} ) cos ( \frac{4 x - 6 x}{2} )}{sin ( 4 x ) - sin ( 6 x )}

Simplify

\frac{2 cos ( \frac{4 x + 6 x}{2} ) cos ( \frac{4 x - 6 x}{2} )}{sin ( 4 x ) - sin ( 6 x )} : \frac{2 cos ( x ) cos ( 5 x )}{sin ( 4 x ) - sin ( 6 x )}

\frac{2 cos ( \frac{4 x + 6 x}{2} ) cos ( \frac{4 x - 6 x}{2} )}{sin ( 4 x ) - sin ( 6 x )}

2 cos (\frac{4 x + 6 x}{2}) cos (\frac{4 x - 6 x}{2}) = 2 cos (\frac{10 x}{2}) cos (- \frac{2 x}{2})

2 cos (\frac{4 x + 6 x}{2}) cos (\frac{4 x - 6 x}{2})

Add similar elements:

4 x + 6 x = 10 x

= 2 cos (\frac{10 x}{2}) cos (\frac{4 x - 6 x}{2})

\frac{4 x - 6 x}{2} = - \frac{2 x}{2}

\frac{4 x - 6 x}{2}

Add similar elements:

4 x - 6 x = - 2 x

= \frac{- 2 x}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 x}{2}

= 2 cos (\frac{10 x}{2}) cos (- \frac{2 x}{2})

= \frac{2 cos ( \frac{10 x}{2} ) cos ( - \frac{2 x}{2} )}{sin ( 4 x ) - sin ( 6 x )}

Use the negative angle identity:

cos (- x) = cos (x)

= \frac{2 cos ( \frac{2 x}{2} ) cos ( \frac{10 x}{2} )}{sin ( 4 x ) - sin ( 6 x )}

Divide the numbers:

\frac{2}{2} = 1

= \frac{2 cos ( x ) cos ( \frac{10 x}{2} )}{sin ( 4 x ) - sin ( 6 x )}

Divide the numbers:

\frac{10}{2} = 5

= \frac{2 cos ( x ) cos ( 5 x )}{sin ( 4 x ) - sin ( 6 x )}

= \frac{2 cos ( x ) cos ( 5 x )}{sin ( 4 x ) - sin ( 6 x )}

Use the Sum to Product identity:

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

= \frac{2 cos ( x ) cos ( 5 x )}{2 sin ( \frac{4 x - 6 x}{2} ) cos ( \frac{4 x + 6 x}{2} )}

Simplify

\frac{2 cos ( x ) cos ( 5 x )}{2 sin ( \frac{4 x - 6 x}{2} ) cos ( \frac{4 x + 6 x}{2} )} : - \frac{cos ( x )}{sin ( x )}

\frac{2 cos ( x ) cos ( 5 x )}{2 sin ( \frac{4 x - 6 x}{2} ) cos ( \frac{4 x + 6 x}{2} )}

2 sin (\frac{4 x - 6 x}{2}) cos (\frac{4 x + 6 x}{2}) = 2 sin (- \frac{2 x}{2}) cos (\frac{10 x}{2})

2 sin (\frac{4 x - 6 x}{2}) cos (\frac{4 x + 6 x}{2})

\frac{4 x - 6 x}{2} = - \frac{2 x}{2}

\frac{4 x - 6 x}{2}

Add similar elements:

4 x - 6 x = - 2 x

= \frac{- 2 x}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 x}{2}

= 2 sin (- \frac{2 x}{2}) cos (\frac{4 x + 6 x}{2})

Add similar elements:

4 x + 6 x = 10 x

= 2 sin (- \frac{2 x}{2}) cos (\frac{10 x}{2})

= \frac{2 cos ( x ) cos ( 5 x )}{2 sin ( - \frac{2 x}{2} ) cos ( \frac{10 x}{2} )}

Divide the numbers:

\frac{2}{2} = 1

= \frac{cos ( x ) cos ( 5 x )}{sin ( - \frac{2 x}{2} ) cos ( \frac{10 x}{2} )}

Use the negative angle identity:

sin (- x) = - sin (x)

= \frac{cos ( 5 x ) cos ( x )}{cos ( \frac{10 x}{2} ) ( - sin ( \frac{2 x}{2} ) )}

Refine

= - \frac{cos ( 5 x ) cos ( x )}{cos ( \frac{10 x}{2} ) sin ( \frac{2 x}{2} )}

Divide the numbers:

\frac{10}{2} = 5

= - \frac{cos ( 5 x ) cos ( x )}{cos ( 5 x ) sin ( \frac{2 x}{2} )}

Divide the numbers:

\frac{2}{2} = 1

= - \frac{cos ( 5 x ) cos ( x )}{cos ( 5 x ) sin ( x )}

Cancel the common factor:

cos (5 x)

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

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

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

Use the basic trigonometric identity:

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

= - cot (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is (cos(4x)+cos(6x))/(sin(4x)-sin(6x))=-cot(x) ?
The answer to whether (cos(4x)+cos(6x))/(sin(4x)-sin(6x))=-cot(x) is True