(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
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared 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
To simplify such expressions, it helps to express all trigonometric functions in terms of sines and cosines. That is, convert tan, cot, sec or csc to their equivalent in terms of sin and cos.
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
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, cos squared...you removed the x in the equation.
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.
cos = sqrt(1 - sin^2)
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.
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x