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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.
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How do you differentiate sin squared x plus cos squared x?
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
Sin squared theta plus cos squared theta barabar Kya Hota Hai?
1
Is 1- cos 2 x 1 plus cos 2 x equals sin squared x cos squared x an identity?
No, (sinx)^2 + (cosx)^2=1 is though
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
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
How do you prove cot squared theta plus cos squared theta plus sin squared theta scs squared theta?
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
Factor sin cubed plus cos cubed?
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
If cos and theta 0.65 what is the value of sin and theta?
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
How do you differentiate sin squared x plus cos squared x?
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
If a cos theta plus b sin theta equals 8 and a sin theta - b cos theta equals 5 show that a squared plus b squared equals 89?
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.
What is sin squared x minus cos squared x?
Sin squared, cos squared...you removed the x in the equation.
How do you prove that the sin over one minus the cosine minus one plus the cosine over the sine equals zero?
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.
Sin squared theta plus cos squared theta barabar Kya Hota Hai?
1
What is sin squared equal to?
Sin squared is equal to 1 - cos squared.
What is (1 plus cos x)(1- cos x)?
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
What is 1 minus cos squared?
sin squared
Is 1- cos 2 x 1 plus cos 2 x equals sin squared x cos squared x an identity?
No, (sinx)^2 + (cosx)^2=1 is though
Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
