1. Anything divided by itself always equals 1.
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
2 x cosine squared x -1 which also equals cos (2x)
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.
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)
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
According to the Pythagorean identity, it is equivalent to sin2theta.
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.
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
No, but cos(-x) = cos(x), because the cosine function is an even function.
Sin squared, cos squared...you removed the x in the equation.
COS squared Theta + SIN squared Theta = 1; where Theta is the angles measurement in degrees.
Sin 15 + cos 105 = -1.9045
The question contains an expression but not an equation. An expression cannot be solved.
2 cos * cos * -1 = 2cos(square) * -1 =cos(square) + cos(square) *-1 =1- sin(square) +cos(square) * -1 1 - 1 * -1 =0
Cos theta squared
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos θ + b sin θ = 8a sin θ - b cos θ = 5then square both sides of each to get:a² cos² θ + 2ab cos θ sin θ + b² sin² θ = 64a² sin² θ - 2ab sin θ cos θ + b² cos² θ = 25Now add the two together:a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ = 89→ a²(cos² θ + sin² θ) + b² (sin² θ + cos² θ) = 89using cos² θ + sin² θ = 1→ a² + b² = 89
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²