3cos(y) = 3/(sqrt(1+x^2)
2 cos * cos * -1 = 2cos(square) * -1 =cos(square) + cos(square) *-1 =1- sin(square) +cos(square) * -1 1 - 1 * -1 =0
Provided that any denominator is non-zero, sin = sqrt(1 - cos^2)tan = sqrt(1 - cos^2)/cos sec = 1/cos cosec = 1/sqrt(1 - cos^2) cot = cos/sqrt(1 - cos^2)
cos(195) = -0.965925826289
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
-1
x = 45 degrees sin(x) = cos(x) = 1/2 sqrt(2)
it equals 4
cos2x = 1 - sin2x = 1 - 0.64 = 0.36 So cos x = +/- 0.6 Since x is acute, cos x is +ve, so cos x = 0.6
sin45=1/root 21/root 2 = x/ycos45= 1/root 2x/y= 1/root 2x=y/root 2
No.
The value of sine and cosine varies between -1 and 1, it never can be 12. If you mean .12, then we have: sin x = .12 x = sin-1 .12 x = 6.89 degrees So that, cos x = cos 6.89 = .99
Cos(-155) = cos(155) = 1 - cos(180-155) = 1-cos(25).
The answer to the math question Cos 5t cos 3t -square root 3 2 - sin 5t cos 3t equals 0. In order to find this answer you will have to find out what each letter is.
a = 0, b = 0.
y = arcsin( cos 48 ); arcsin may be seen as sin-1 on your calculator.
Try to write everything in terms of sines and cosines:1 / cos B - cos B = (sin B / cos B) sin B1 / cos B - cos B = sin2B / cos BMultiply by the common denominator, cos B:1 - cos2B = sin2BUse the pithagorean identity on the left side:sin2B + cos2B - cos2B = sin2Bsin2B = sin2B
To find the hypotenuse with angle a and side b, we use the identity below:cos(a) = b/cWe have a and b, and to find c, we multiply both sides by c and divide both sides by cos(a):c = b/cos(a)c = 5/cos(30)c = 32.41460617mm