Is cos(2θ)csc^2(θ)=csc^2(θ)-2 ?

The answer to whether cos(2θ)csc^2(θ)=csc^2(θ)-2 is True

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prove cos(2θ)csc^2(θ)=csc^2(θ)-2

Solution

prove

cos (2 θ) csc^{2} (θ) = csc^{2} (θ) - 2

Solution

True

Solution steps

cos (2 θ) csc^{2} (θ) = csc^{2} (θ) - 2

Manipulating left side

cos (2 θ) csc^{2} (θ)

Express with sin, cos

cos (2 θ) csc^{2} (θ)

Use the basic trigonometric identity:

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

= cos (2 θ) (\frac{1}{sin ( θ )})^{2}

Simplify

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

cos (2 θ) (\frac{1}{sin ( θ )})^{2}

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

(\frac{1}{sin ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( θ )} cos (2 θ)

Multiply fractions:

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

= \frac{1 \cdot cos ( 2 θ )}{sin ^{2} ( θ )}

Multiply:

1 \cdot cos (2 θ) = cos (2 θ)

= \frac{cos ( 2 θ )}{sin ^{2} ( θ )}

= \frac{cos ( 2 θ )}{sin ^{2} ( θ )}

= \frac{cos ( 2 θ )}{sin ^{2} ( θ )}

Rewrite using trig identities

\frac{cos ( 2 θ )}{sin ^{2} ( θ )}

Use the Double Angle identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1 - 2 ( \frac{1}{c s c ( θ )} ) ^{2}}{( \frac{1}{c s c ( θ )} ) ^{2}}

Simplify

\frac{1 - 2 ( \frac{1}{c s c ( θ )} ) ^{2}}{( \frac{1}{c s c ( θ )} ) ^{2}}

(\frac{1}{csc ( θ )})^{2} = \frac{1}{csc ^{2} ( θ )}

(\frac{1}{csc ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{csc ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( θ )}

= \frac{1 - 2 ( \frac{1}{c s c ( θ )} ) ^{2}}{\frac{1}{c s c ^{2} ( θ )}}

2 (\frac{1}{csc ( θ )})^{2} = \frac{2}{csc ^{2} ( θ )}

2 (\frac{1}{csc ( θ )})^{2}

(\frac{1}{csc ( θ )})^{2} = \frac{1}{csc ^{2} ( θ )}

(\frac{1}{csc ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{csc ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( θ )}

= 2 \cdot \frac{1}{csc ^{2} ( θ )}

Multiply fractions:

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

= \frac{1 \cdot 2}{csc ^{2} ( θ )}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{csc ^{2} ( θ )}

= \frac{1 - \frac{2}{c s c ^{2} ( θ )}}{\frac{1}{c s c ^{2} ( θ )}}

Apply the fraction rule:

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

= \frac{( 1 - \frac{2}{c s c ^{2} ( θ )} ) csc ^{2} ( θ )}{1}

Join

1 - \frac{2}{csc ^{2} ( θ )} : \frac{csc ^{2} ( θ ) - 2}{csc ^{2} ( θ )}

1 - \frac{2}{csc ^{2} ( θ )}

Convert element to fraction:

1 = \frac{1 csc ^{2} ( θ )}{csc ^{2} ( θ )}

= \frac{1 \cdot csc ^{2} ( θ )}{csc ^{2} ( θ )} - \frac{2}{csc ^{2} ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot csc ^{2} ( θ ) - 2}{csc ^{2} ( θ )}

Multiply:

1 \cdot csc^{2} (θ) = csc^{2} (θ)

= \frac{csc ^{2} ( θ ) - 2}{csc ^{2} ( θ )}

= \frac{\frac{c s c ^{2} ( θ ) - 2}{c s c ^{2} ( θ )} csc ^{2} ( θ )}{1}

Apply the fraction rule:

\frac{a}{1} = a

= \frac{csc ^{2} ( θ ) - 2}{csc ^{2} ( θ )} csc^{2} (θ)

Multiply fractions:

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

= \frac{( csc ^{2} ( θ ) - 2 ) csc ^{2} ( θ )}{csc ^{2} ( θ )}

Cancel the common factor:

csc^{2} (θ)

= csc^{2} (θ) - 2

csc^{2} (θ) - 2

csc^{2} (θ) - 2

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is cos(2θ)csc^2(θ)=csc^2(θ)-2 ?
The answer to whether cos(2θ)csc^2(θ)=csc^2(θ)-2 is True