2016年8月26日 · cos^2x Rearrange the pythagorean identity sin^2x + cos^2x = 1 to isolate cos^2x: cos^2x = 1 - sin^2x Hence, 1- sin^2x = cos^2x

2016年8月2日 · Since #sin^2 theta + cos^2 theta =1# #1-sin^2 theta = cos^2 theta# Answer link. Related questions.

2016年10月12日 · How do you prove #(1-\cos^2 x)(1+\cot^2 x) = 1#? 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#?

2018年6月22日 · But to take the derivative this form in terms of sin is more useful. Use the quotient rule for the overall fraction and the chain rule to differentiate #sin^2x# . #d/dx[1/sin^2x]=(sin^2x*0-1*2sinxcosx)/sin^4x#

2016年7月25日 · Note that this is NOT: 1/(sin(sin2)) which is not the same thing. If you did have 1/(sin(sin(2)), it would be equal to (sin(sin2))^(-1). However, even though sin^2(x) = (sinx)^2, it does not mean that sin^(-1)(x) = (sinx)^(-1).

2016年5月27日 · How do I evaluate the indefinite integral #int(tan^2(x)+tan^4(x))^2dx# ? How do I evaluate the indefinite integral #intx*sin(x)*tan(x)dx# ? See all questions in Integrals of Trigonometric Functions

2016年10月20日 · #I=int1/(1+sin^2x)dx# Notice that dividing by #cos^2(x)# will get us working with #tanx# and #secx# functions, which are derivatives of themselves:

2016年2月5日 · To prove trig identity either manipulate the left side into the form of the right side or the right side into the form of the left side or manipulate both sides together until they agree.

2016年10月15日 · The range of sin^-1 or arcsin is between pi/2 and -pi/2. If you are finding sin^-1 of a positive value, the answer will be between 0 and pi/2, or the first quadrant in the unit circle. Do NOT use the second quadrant angle with a sine of 1/2, because it does not fall within the range of sin^-1. Using the unit circle, the angle with a sine of 1/2 ...

2016年12月16日 · How do you solve and find the value of #sin(sin^-1(1/2))#? Trigonometry Inverse Trigonometric Functions Inverse Trigonometric Properties. 1 Answer