cos θ=-sin(-θ)

solve each equation on the interval 0≤θ<2Π

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cos^2 = sin^2

Since:

cos^2 = 1-sin^2

1 - sin^2 = sin^2

Add sin^2 to both sides:

1 = 2 sin^2

Divide by 2:

1/2 = sin^2

Take the square root:

+/- sqrt(2)/2 = sin (-x)

Where is sine +sqrt(2)/2? pi/4 and 3pi/4

Sine is -sqrt(2)/2 at 5pi/4 and 7pi/4.

Since,

cos(pi/4) = -sin(-pi/4)

cos(5pi/4) = -sin(-5pi/4)

θ = { pi/4, 5pi/4 }

cosθ = -sin(-θ) is equivalent to

tanθ = 1, since sine is an odd function.

Answer: θ = pi/4, 5pi/4

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