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Given 2sin s equals cos s find all possible values of sin s?
2sin s = cos s; whence, 4sin2 s = cos2 s = 1 - sin2s, and 5sin2 s = 1. Therefore sin s = ±√0.2 = 0.4472, approx. Check: cos2 s = 1 - 0.2 = 0.8, whence, cos s = ±√0.8 = ±2√0.2 = 2 sin s.
What is the integral of 1 divided by cos x multiplied by sin cubed x?
Given the limitations of this browser, I will use S to denote integral (the stretched S).Int = S cos(x)*sin3(x) dxLet t = sin(x) then dt = cos(x)dxso Int = S t3dt = 1/4*t4 + cSubstituting back for x,Int = sin4(x) / 4 + cGiven the limitations of this browser, I will use S to denote integral (the stretched S).Int = S cos(x)*sin3(x) dxLet t = sin(x) then dt = cos(x)dxso Int = S t3dt = 1/4*t4 + cSubstituting back for x,Int = sin4(x) / 4 + cGiven the limitations of this browser, I will use S to denote integral (the stretched S).Int = S cos(x)*sin3(x) dxLet t = sin(x) then dt = cos(x)dxso Int = S t3dt = 1/4*t4 + cSubstituting back for x,Int = sin4(x) / 4 + cGiven the limitations of this browser, I will use S to denote integral (the stretched S).Int = S cos(x)*sin3(x) dxLet t = sin(x) then dt = cos(x)dxso Int = S t3dt = 1/4*t4 + cSubstituting back for x,Int = sin4(x) / 4 + c
How do you find the area of a cone with only the angle degree and base radius given?
Area of the cone part is pi*r*s (s is the slant height which is the hypotenuse of the triangle formed by the radius and the altitude). cosine(theta) = r/s, so s = r/cos(theta), and A = pi*r*s = pi*r^2 / cos(theta). The area of the circle part is given by pi*r^2. Then just add them together.
Solve sec2x equals secx plus 2?
You must mean, either,(1) sec2 x = sec x + 2, or(2) sec(2x) = sec x + 2.Let's first assume that you mean:sec2 x = sec x + 2;whence,if we let s = sec x, we have,s2 = s + 2,s2 - s - 2 = (s - 2)(s + 1) = 0, ands = 2 or -1; that is,sec x = 2 or -1.As, by definition,cos x = 1/sec x ,this means thatcos x = ½ or -1.Therefore, providing that the first assumption is correct,x = 60°, 180°, or 300°; or,if you prefer,x = ⅓ π, π, or 1⅔ π.Now, let's assume, instead, that you mean:sec(2x) = sec x + 2;whence,1/(cos(2x) = (1/cos x) + 2.If we let c = cos x,then we have the standard identity,2c2 - 1 = cos (2x); and,thus, it follows that1/(cos(2x) = 1/(2c2 - 1)= (1/c) + 2 = (1 + 2c)/c.This gives,1/(2c2 - 1) = (1 + 2c)/c;(2c2 - 1)(2c + 1) = 4c3 + 2c2 - 2c - 1 = c; and4c3 + 2c2 - 3c - 1 = (c + 1)(4c2 - 2c - 1) = 0.As our concern is only with real roots,c = cos x = -1; and,therefore, providing that the second assumption is correct,x = 180°; or,if you like,x = π.
What is the length in inches of a simple pendulum whose period s 1 s?
9.5 inches
