"The fundamental trigonometric identities" are the basic identities: •The reciprocal identities •The pythagorean identities

2015年4月15日 · How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? How do you prove that #sec xcot x = csc x#? How do you prove that #cos 2x(1 + tan 2x) = 1#?

2016年5月4日 · x=npi+pi/4 cosx=sinx means sinx/cosx=1 or tanx=1=tan(pi/4) Hence x=npi+pi/4

2017年12月9日 · Let: # f(x) = cos(sinx) #..... [A] The Maclaurin series is given by # f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ...

2016年9月18日 · Please see below. (1-cosx)/sinx = (1-cosx)/sinx xx(1+cosx)/(1+cosx) = (1-cos^2x)/(sinx(1+cosx) = sin^2x/(sinx(1+cosx) = sinx/(1+cosx)

2018年3月25日 · How do you verify #cosx/(1-tanx) + sinx/(1-cotx) = sinx + cosx#? Trigonometry Trigonometric Identities and Equations Solving Trigonometric Equations 1 Answer

2015年4月14日 · #f(x) = cos 2x + sin x = 1 - 2sin^2 x + sin x = 0# Call #sin x = t#. This is a quadratic equation in #t#: #f(t) = -2t^2 + t + 1 = 0# Solve this quadratic equation. Because #a + …

2016年5月27日 · cot(x) cos(x)/sin(x) = cot(x)

2015年4月15日 · Solve the equation: f(x) = cos^2 x - sin^2 x - sin x = 0. Replace cos^2 x by (1 - sin^2 x) f(x) = 1 - sin^2 x - sin^2 x - sin x = 0.

2015年7月28日 · y^' = -2/(sinx - cosx)^2 Start by taking a look at your function y = (sinx + cosx)/(sinx - cosx) Notice that this function is actually the quotient of two other functions, let's …