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To solve 2 cos2x + sin x - 1 make use of the identity cos2x + sin2x = 1, then:
cos2x = 1 - sin2x
and
2 cos2x + sin x - 1 = 2 (1 - sin2x) + sin x - 1
= 2 - 2 sin2x + sin x - 1
= 1 + sin x - 2 sin2 x
if you let y = sin x, then you can see you have a quadratic:
1 + sin x - 2 sin2 x = 1 + y - 2y2
which can be solved for 1 + y - 2y2 = 0:
1 + y - 2y2 = 0
⇒ 2y2 - y - 1 = 0
⇒ (2y + 1)(y - 1) = 0
⇒ y = -1/2 or 1
but y = sin x, so:
sin x = -1/2 or sin x = 1
and so solve these for x.
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
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No, (sinx)^2 + (cosx)^2=1 is though
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.