What is the general solution for 146=(8-cos(θ))^2+(-9-sin(θ))^2 ?

The general solution for 146=(8-cos(θ))^2+(-9-sin(θ))^2 is θ=0.72664…+pin

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Popular Trigonometry >

146=(8-cos(θ))^2+(-9-sin(θ))^2

Solution

146 = (8 - cos (θ))^{2} + (- 9 - sin (θ))^{2}

Solution

θ = 0.72664 \dots + πn

Degrees

θ = 41.63353 \dots^{\circ} + 18 0^{\circ} n

Solution steps

146 = (8 - cos (θ))^{2} + (- 9 - sin (θ))^{2}

Switch sides

(8 - cos (θ))^{2} + (- 9 - sin (θ))^{2} = 146

Subtract

146

from both sides

cos^{2} (θ) + sin^{2} (θ) + 18 sin (θ) - 16 cos (θ) - 1 = 0

Rewrite using trig identities

- 1 + cos^{2} (θ) + sin^{2} (θ) - 16 cos (θ) + 18 sin (θ)

Use the Pythagorean identity:

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

= - 1 - 16 cos (θ) + 18 sin (θ) + 1

Simplify

- 1 - 16 cos (θ) + 18 sin (θ) + 1 : 18 sin (θ) - 16 cos (θ)

- 1 - 16 cos (θ) + 18 sin (θ) + 1

Group like terms

= - 16 cos (θ) + 18 sin (θ) - 1 + 1

- 1 + 1 = 0

= 18 sin (θ) - 16 cos (θ)

= 18 sin (θ) - 16 cos (θ)

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{- 16 cos ( θ ) + 18 sin ( θ )}{cos ( θ )} = \frac{0}{cos ( θ )}

Simplify

- 16 + \frac{18 sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

- 16 + 18 tan (θ) = 0

- 16 + 18 tan (θ) = 0

Move

16

to the right side

- 16 + 18 tan (θ) = 0

Add

16

to both sides

- 16 + 18 tan (θ) + 16 = 0 + 16

Simplify

18 tan (θ) = 16

18 tan (θ) = 16

Divide both sides by

18

18 tan (θ) = 16

Divide both sides by

18

\frac{18 tan ( θ )}{18} = \frac{16}{18}

Simplify

tan (θ) = \frac{8}{9}

tan (θ) = \frac{8}{9}

Apply trig inverse properties

tan (θ) = \frac{8}{9}

General solutions for

tan (θ) = \frac{8}{9}

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (\frac{8}{9}) + πn

θ = arctan (\frac{8}{9}) + πn

Show solutions in decimal form

θ = 0.72664 \dots + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 146=(8-cos(θ))^2+(-9-sin(θ))^2 ?
The general solution for 146=(8-cos(θ))^2+(-9-sin(θ))^2 is θ=0.72664…+pin