by using rightangle triangle
I'm sorry the question is not correctly displayed. If f(x) = cos(2x).cos(4x).cos(6x).cos(8x).cos(10x) then, find the limit of {1 - [f(x)]^3}/[5(sinx)^2] as x tends to 0 (zero).
Find the measure of this angles m1 equals 123 m8 equals?
cos 2x = cos2 x - sin2 x = 2 cos2 x - 1; whence, cos 2x / cos x = 2 cos x - (1 / cos x) = 2 cos x - sec x.
sec x = 1/cos x sec x cos x = [1/cos x] [cos x] = 1
180-x...... hahaahahaaa
cos(theta) = 0.7902 arcos(0.7902) = theta = 38 degrees you find complimentary angles
If the angles are measured in radians then the answer is -0.2678
yes
Yes. Quadrantal angles have reference angles of either 0 degrees (e.g. 0 degrees and 180 degrees) or 90 degrees (e.g. 90 degrees and 270 degrees).
There is no reason at all. For most angles sin plus cos do not equal one.
sin cos tan -soh cah toa
If you know the length of the sides of a triangle you can find all the angles of the triangle using the Law of cosines such as: Step 1. cos A = (b^2 + c^2 - a^2)/(2bc) cos B = (a^2 + c^2 - b^2)/(2ac) cos C = (a^2 + b^2 - c^2)/(2ab) Step 2. Find the arc cosine A, arc cosine B, and arc cosine C in order to find the angles A, B, and C.
sin θ = cos (90° - θ) cos θ = sin (90° - θ)
sin(x) = cos(90° - x) cos(x) = sin(90° - x)
cos(195) = cos(180 + 15) = cos(180)*cos(15) - sin(180)*sin(15) = -1*cos(15) - 0*sim(15) = -cos(15) = -cos(60 - 45) = -[cos(60)*cos(45) + sin(60)*sin(45)] = -(1/2)*sqrt(2)/2 - sqrt(3)/2*sqrt(2)/2 = - 1/4*sqrt(2)*(1 + sqrt3) or -1/4*[sqrt(2) + sqrt(6)]
There are many. For example, if A and B are the two acute angles, then A + B = 90 degrees or sin(A) = cos(B) or cos(A) = sin(B) or tan(A) = 1/tan(B)
I would start by looking up the formulae for multiple angles, and convert that to simgle angles. In this case, sin 2x = 2 sin x cos x, so your equation becomes:2 sin x cos x sin x = cos x2 sin2x cos x = cos xNext divide both sides by cos x; note that you must consider the possibility that cos x = 0 (this may give additional solutions to the equation).