Leave one side unchanged and use algebra and trigonometric substitutions to make the other side the same as the unchanged side. Verify: cot^2(x)/(sin(x) + cos(x)) =(cos^2(x)sin(x) -cos^3(x))/(2sin^4(x) - …

May 25, 2018 · x=2/3*pi writing your equation in the form csc(x)(3cot^2(x)-1)=0 so we get csc(x)=0 or cot(x)=pm1/sqrt(3)

#f (2) = -3.35528# # (df)/ (dx) = 1/ (\sinx).\cosx = cotx; \qquad a_1 = 1/ (1!)\cot (2) = -0.4576;# #\frac {d^2f} {dx^2} = -\csc^2x; \qquad a_2 = -1/ (2!)\csc (2)=-0. ...

For the first function \ \ \ y=cot [2 (x)]\ \ \ \text {Period}=\frac {\pi } {2} Consider the general form of periodic function: y\ =\ a\cot (bx\pm c)\pm d The period of a periodic function can be calculated by …

If you have: (cot^2 (x)+1)* (1-cos^2 (x)) (I) You can write: cot^2 (x)=cos^2 (x)/sin^2 (x) and from: sin^2 (x)+cos^2 (x)=1 (II) you get: sin^2 (x)=1-cos^2 (x) substituting in (I): (cos^2 (x)/sin^2 (x)+1)*sin^2 (x)= …

Explanation: Use trig identity: #sin^2 x = 1/ (1 + cot^2 x)# In this case: Call x the arctan (1/2) #tan x = 1/2# --> #cot x = 1/ (tan) = 2# #sin^2 x = 1/ (1 + 4) = 1/5# #sin x = +- 1/ (sqrt5) = +- sqrt5/5# Because …

Prove: 2cot^2 (x)sin^2 (x)+cos^2 (x)tan^2 (x)=1+cos^2 (x) Substitute cot^2 (x) = cos^2 (x)/sin^2 (x) 2cos^2 (x)/sin^2 (x)sin^2 (x)+cos^2 (x)tan^2 (x)=1+cos^2 (x ...

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