2sin2 x
2(1 - cos2 x) = 2 + cos x
2 - 2cos2 x = 2 + cos x
- 2cos2 x - cos x = 0
cos x(-cos x - 1) = 0
cos x = 0 or -cos x - 1 = 0
cos x = -1
Since the cosine has a period of 2pi, in the interval 0 ≤ x ≤ 2pi
when cos x = 0, x = pi/2 and x = 3pi/2, and
when cos x = -1, x = pi.
All values of x that satisfy the given equation are:
x = pi/2 + 2npi, x = pi +2npi, and x = 3pi/2+ 2npi where n is any integer.
No, (sinx)^2 + (cosx)^2=1 is though
1. Anything divided by itself always equals 1.
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
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The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
No, (sinx)^2 + (cosx)^2=1 is though
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1. Anything divided by itself always equals 1.
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
2 x cosine squared x -1 which also equals cos (2x)
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos. Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.) (This is diff of 2 squares because 1 is the square of 1.) And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos squared) - 1. Then from basic trig we know that (sin squared) + (cos squared) = 1, so this is 0.
No. Cos squared x is not the same as cos x squared. Cos squared x means cos (x) times cos (x) Cos x squared means cos (x squared)
Well, darling, if we square the first equation and the second equation, add them together, and do some algebraic magic, we can indeed show that a squared plus b squared equals 89. It's like a little math puzzle, but trust me, the answer is as sassy as I am.
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The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x