sin2(1) = 1 - cos2(1) = 1 - [cos(1)]2
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
The answer is 1. sin^2 x cos^2/sin^2 x 1/cos^2 cos^2 will be cancelled =1 sin^2 also will be cancelled=1 1/1 = 1
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
1
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
tan θ = sin θ / cos θ sec θ = 1 / cos θ sin ² θ + cos² θ = 1 → sin² θ - 1 = - cos² θ → tan² θ - sec² θ = (sin θ / cos θ)² - (1 / cos θ)² = sin² θ / cos² θ - 1 / cos² θ = (sin² θ - 1) / cos² θ = - cos² θ / cos² θ = -1
Sin squared, cos squared...you removed the x in the equation.
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 is equal to 1 - cos squared.
(sin x + cos x) / cosx = sin x / cos x + cosx / cos x = tan x + 1
sin squared
22
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
Note that an angle should always be specified - for example, 1 - cos square x. Due to the Pythagorean formula, this can be simplified as sin square x. Note that sin square x is a shortcut of (sin x) squared.
cos = sqrt(1 - sin^2)
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x