2015年5月23日 · (1 - sin^2 30) = (1 - sin 30)(1 + sin 30) = (1 - 1/2)(1 + 1/2) = 3/4. How do you apply the fundamental identities to values of #theta# and show that they are true?

2018年3月13日 · #f(t) = (1 - sin^2 t)(1 + cot^2 t)# Transform the first factor: #(1 - sin^2 t) = cos^2t# Transform the second factor:

2018年4月6日 · Or, 1/sin^2x-1/tan^2x=csc^2x-cot^2x=1. Trigonometry . Science Anatomy & Physiology Astronomy ...

2018年1月4日 · int\\ 1/sin^2(x)+cos(2x)\\ dx=1/2sin(2x)-cot(x)+C First, I will split up the integral into two parts: int\\ 1/sin^2(x)+cos(2x)\\ dx=int\\ 1/sin^2(x)\\ dx + int\\ cos(2x)\\ dx I will call the …

Let's begin by simplifying the denominator of the fraction. using #cscx=1/(sinx)" we obtain"#. #sinx-1/sinx# which we require to express as a single fraction.

2017年1月22日 · y' = -sinx By the pythagorean identity sin^2x + cos^2x = 1: y = sqrt(cos^2x) y= cosx This can be differentiated as -sinx, but here's proof using limits. y' = lim_(h ...

2015年5月22日 · Based on this answer, it seems good in the denominator of the original integrand to write (using the double-angle formula for sine after writing #x=2\cdot x/2# and then using …

2018年3月1日 · = (1+tan^2A)(1-sin^2A) = sec^2A xx cos^2A = 1/cancel(cos^2A)xxcancel(cos^2A) = 1

2016年9月11日 · lnabs(sqrt(sin^2t+1)+sint)+C We have: intcost/sqrt(1+sin^2t)dt We will use the substitution sint=tantheta. Thus, costdt=sec^2thetad theta. Substituting, this yields: …

2016年12月26日 · d/(dx) (-1/sin^2x) = (2cosx)/(sin^3x) Based on the chain rule: d/(dx) (-1/sin^x) = d/(dsinx) (-1/sin^2x) * (dsinx)/(dx)= 2/(sin^3x)cosx