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Mathematics 16 Online
OpenStudy (anonymous):

Verify Cos(x)[sec(x)-Cos(x)]=Sin(x)

OpenStudy (anonymous):

(cosx/cosx) - cos^2(x) = sinx 1 - cos^2(x) = sinx cos^2(x) =1 - sin^2(x) sin^2(x) =sin(x) Nope.

OpenStudy (anonymous):

It works if the equation should be Cos(x)[sec(x)-Cos(x)]=Sin^2(x) If it's just a normal Sin(x), it doesn't work. Cos(x) x Sec(x) is 1 Cos(x) x -Cos(x) is -Cos^2(x) so 1 - Cos^2(x) = Sin^2(x) Pythagorean Identity: sin^2(x) + cos^2(x) = 1 Subtract cos so 1 - Cos^2(x) = Sin^2(x) therefore Sin^2(x) = Sin^2(x)

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