Is ((cos(x)+1))/(sin^3(x))=((csc(x)))/(1-cos(x)) ?

The answer to whether ((cos(x)+1))/(sin^3(x))=((csc(x)))/(1-cos(x)) is True

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Popular >

prove ((cos(x)+1))/(sin^3(x))=((csc(x)))/(1-cos(x))

Solution

prove

\frac{( cos ( x ) + 1 )}{sin ^{3} ( x )} = \frac{( csc ( x ) )}{1 - cos ( x )}

Solution

True

Solution steps

\frac{cos ( x ) + 1}{sin ^{3} ( x )} = \frac{csc ( x )}{1 - cos ( x )}

Manipulating right side

\frac{csc ( x )}{1 - cos ( x )}

Express with sin, cos

\frac{csc ( x )}{1 - cos ( x )}

Use the basic trigonometric identity:

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

= \frac{\frac{1}{s i n ( x )}}{1 - cos ( x )}

Simplify

\frac{\frac{1}{s i n ( x )}}{1 - cos ( x )} : \frac{1}{sin ( x ) ( 1 - cos ( x ) )}

\frac{\frac{1}{s i n ( x )}}{1 - cos ( x )}

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

1 + cot^{2} (x) = csc^{2} (x)

1 = csc^{2} (x) - cot^{2} (x)

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{- ( \frac{c o s ( x )}{s i n ( x )} ) ^{2} + csc ^{2} ( x )}{( 1 - cos ( x ) ) sin ( x )}

Use the basic trigonometric identity:

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

= \frac{- ( \frac{c o s ( x )}{s i n ( x )} ) ^{2} + ( \frac{1}{s i n ( x )} ) ^{2}}{( 1 - cos ( x ) ) sin ( x )}

Simplify

\frac{- ( \frac{c o s ( x )}{s i n ( x )} ) ^{2} + ( \frac{1}{s i n ( x )} ) ^{2}}{( 1 - cos ( x ) ) sin ( x )} : \frac{cos ( x ) + 1}{sin ^{3} ( x )}

\frac{- ( \frac{c o s ( x )}{s i n ( x )} ) ^{2} + ( \frac{1}{s i n ( x )} ) ^{2}}{( 1 - cos ( x ) ) sin ( x )}

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{- \frac{c o s ^{2} ( x )}{s i n ^{2} ( x )} + \frac{1}{s i n ^{2} ( x )}}{sin ( x ) ( - cos ( x ) + 1 )}

Combine the fractions

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

Apply rule

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

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

= \frac{\frac{- c o s ^{2} ( x ) + 1}{s i n ^{2} ( x )}}{sin ( x ) ( - cos ( x ) + 1 )}

Apply the fraction rule:

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

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

Factor

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

- cos^{2} (x) + 1

Factor out common term

- 1

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

Factor

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

cos^{2} (x) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (cos (x) + 1) (cos (x) - 1)

= - (cos (x) + 1) (cos (x) - 1)

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

Factor

sin^{2} (x) (1 - cos (x)) sin (x) : - sin^{3} (x) (cos (x) - 1)

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

Factor

1 - cos (x) : - (cos (x) - 1)

1 - cos (x)

Factor out common term

- 1

= - (cos (x) - 1)

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

Refine

= - sin^{3} (x) (cos (x) - 1)

= \frac{( cos ( x ) + 1 ) ( cos ( x ) - 1 )}{sin ^{3} ( x ) ( cos ( x ) - 1 )}

Cancel the common factor:

cos (x) - 1

= \frac{cos ( x ) + 1}{sin ^{3} ( x )}

= \frac{cos ( x ) + 1}{sin ^{3} ( x )}

= \frac{1 + cos ( x )}{sin ^{3} ( x )}

= \frac{cos ( x ) + 1}{sin ^{3} ( x )}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is ((cos(x)+1))/(sin^3(x))=((csc(x)))/(1-cos(x)) ?
The answer to whether ((cos(x)+1))/(sin^3(x))=((csc(x)))/(1-cos(x)) is True