What is the general solution for cos(θ)=-sin(-θ) ?

The general solution for cos(θ)=-sin(-θ) is θ= pi/4+pin

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cos(θ)=-sin(-θ)

cos (θ) = - sin (- θ)

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

cos (θ) = - sin (- θ)

Rewrite using trig identities

cos (θ) = - sin (- θ)

cos (θ) = - (- sin (θ))

Simplify

- (- sin (θ)) : sin (θ)

- (- sin (θ))

Apply rule

- (- a) = a

= sin (θ)

cos (θ) = sin (θ)

cos (θ) = sin (θ)

Subtract

sin (θ)

from both sides

cos (θ) - sin (θ) = 0

Rewrite using trig identities

cos (θ) - sin (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

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

Simplify

1 - \frac{sin ( θ )}{cos ( θ )} = 0

Use the basic trigonometric identity:

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

1 - tan (θ) = 0

1 - tan (θ) = 0

Move

1

to the right side

1 - tan (θ) = 0

Subtract

1

from both sides

1 - tan (θ) - 1 = 0 - 1

Simplify

- tan (θ) = - 1

- tan (θ) = - 1

Divide both sides by

- 1

- tan (θ) = - 1

Divide both sides by

- 1

\frac{- tan ( θ )}{- 1} = \frac{- 1}{- 1}

Simplify

tan (θ) = 1

tan (θ) = 1

General solutions for

tan (θ) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

sin (x) - 3 cos (x) = 3

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

3 csc (x) + csc (x) cot (x) = 0

- 3 cos^{2} (θ) - 10 cos (θ) = - 3 cos (θ) - 6

- 4 sin (x) = - cos^{2} (x) + 4

What is the general solution for cos(θ)=-sin(-θ) ?
The general solution for cos(θ)=-sin(-θ) is θ= pi/4+pin