This is geometry/trigonometry, not calculus.
That said, cos 2a = .3
The property you need: cos 2u = 1 - 2sin2u = 2cos2u - 1
Using the first equality:
cos(2a) = 1 - 2sin2a Let u = a
(.3) = 1 - 2sin2a Substitute cos(2a) = .3
.7 = sin2a Subtract .3 from both sides, get rid of negative sign
sin a = .71/2 Square root both sides
a = sin-1 .71/2= .991156 Not a special angle, use calculator.
1. Anything divided by itself always equals 1.
No, (sinx)^2 + (cosx)^2=1 is though
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
Since the word 'equals' appears in your questions it might be what is called a trigonometric identity, in other words a statement about a relationship between various trigonometric values.
Given y = tan x: dy/dx = sec^2 x(secant of x squared)
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1. Anything divided by itself always equals 1.
2 x cosine squared x -1 which also equals cos (2x)
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)
No, (sinx)^2 + (cosx)^2=1 is though
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.
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
Sin squared, cos squared...you removed the x in the equation.
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
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.
Cos theta squared
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)