There are two functions happening here, sin() and x 2. But it is not sin(x), it is sin(the result of x 2) Let's use "u" for x 2 so we can have: dy dx = dy du du dx. Which becomes: d dx sin(x 2) = d du sin(u) d dx x 2. The individual derivatives are: d du sin(u) = cos(u) d dx x 2 = 2x; So: d dx sin(x 2) = cos(u) (2x) Substitute back u = x 2: d
As we know cos(a) = x = x 1 we can label the adjacent leg as x and the hypotenuse as 1. The Pythagorean theorem then allows us to solve for the second leg as √1 −x2. With this, we can now find sin(cos−1(x)) as the quotient of the opposite leg and the hypotenuse. sin (cos^ (-1) (x)) = sqrt (1-x^2) Let's draw a right triangle with an angle
Answer link. Express sin (x/2) in terms of cos x. Ans: sin (x /2) = sqrt ( (1 - cos x)/2) By applying the trig identity: cos 2a = 1 - 2sin^2 a, we get: cos x = 1 - 2sin^2 (x/2) 2sin^2 (x/2) = 1 - cos x sin^2 (x/2) = (1 - cos x)/2 sin (x/2) = +- sqrt ( (1 - cos x)/2) So, if we take the first derivative, if we take the first derivative, derivative of cos(x) = -sin(x) if we take the derivative of that, if we take the derivative of that, derivative of sin(x) is cos(x), and we have the negative there, so it's -cos(x) so if we take the derivative of that, so this is the third derivative of cos(x), now it's just Properties of Trigonometric Functions. The properties of the 6 trigonometric functions: sin (x), cos (x), tan (x), cot (x), sec (x) and csc (x) are discussed. These include the graph, domain, range, asymptotes (if any), symmetry, x and y intercepts and maximum and minimum points. Unit Circle - Part 2. . 347 188 71 84 439 471 76 139