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)
22
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.
Cos theta squared
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
sin squared