Is cot(-θ)*cos(-θ)+sin(-θ)=-csc(θ) ?

The answer to whether cot(-θ)*cos(-θ)+sin(-θ)=-csc(θ) is True

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Popular Trigonometry >

prove cot(-θ)*cos(-θ)+sin(-θ)=-csc(θ)

Solution

prove

cot (- θ) \cdot cos (- θ) + sin (- θ) = - csc (θ)

Solution

True

Solution steps

cot (- θ) cos (- θ) + sin (- θ) = - csc (θ)

Manipulating left side

cot (- θ) cos (- θ) + sin (- θ)

Use the negative angle identity:

sin (- x) = - sin (x)

= - sin (θ) + cos (- θ) cot (- θ)

Use the negative angle identity:

cos (- x) = cos (x)

= - sin (θ) + cos (θ) cot (- θ)

Use the negative angle identity:

cot (- x) = - cot (x)

= - sin (θ) + cos (θ) (- cot (θ))

Express with sin, cos

- sin (θ) + (- cot (θ)) cos (θ)

Use the basic trigonometric identity:

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

= - sin (θ) + (- \frac{cos ( θ )}{sin ( θ )}) cos (θ)

Simplify

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

- sin (θ) + (- \frac{cos ( θ )}{sin ( θ )}) cos (θ)

Remove parentheses:

(- a) = - a

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

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

\frac{cos ( θ )}{sin ( θ )} cos (θ)

Multiply fractions:

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

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{sin ( θ )}

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

Convert element to fraction:

sin (θ) = \frac{sin ( θ ) sin ( θ )}{sin ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

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

- sin (θ) sin (θ) - cos^{2} (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= - sin^{2} (θ) - cos^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= \frac{- 1}{sin ( θ )}

Apply the fraction rule:

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

= - \frac{1}{sin ( θ )}

= - \frac{1}{sin ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

- \frac{1}{\frac{1}{c s c ( θ )}}

Simplify

- \frac{1}{\frac{1}{c s c ( θ )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= - \frac{csc ( θ )}{1}

Apply rule

\frac{a}{1} = a

= - csc (θ)

- csc (θ)

- csc (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is cot(-θ)*cos(-θ)+sin(-θ)=-csc(θ) ?
The answer to whether cot(-θ)*cos(-θ)+sin(-θ)=-csc(θ) is True