cos θ=-sin(-θ) solve each equation on the interval 0≤θ<2Π 11/3/2013 | Allison from San Diego, CA | 2 Answers | 0 Votes Mark favorite Subscribe Comment
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 } 11/3/2013 | Jason S. Comment Comments I like Robert's answer better. Since sin (-x) = -sin (x), you can substitute. cos x = - (-sin x) cos x = sin x Divide by cos x: cos x = sin x ----- ----- cos x cos x 1 = tan x My way was the longer and much more difficult way. 11/3/2013 | Jason S. Comment
I like Robert's answer better. Since sin (-x) = -sin (x), you can substitute. cos x = - (-sin x) cos x = sin x Divide by cos x: cos x = sin x ----- ----- cos x cos x 1 = tan x My way was the longer and much more difficult way. 11/3/2013 | Jason S.
cosθ = -sin(-θ) is equivalent to tanθ = 1, since sine is an odd function. Answer: θ = pi/4, 5pi/4 11/3/2013 | Robert J. Comment
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