Mar 5, 2018 · Take csc^2(x) common then it becomes: csc^2(x)(1+ cot^2(x)) which is equal to csc^4(x) Basic trigonometric identities To prove (csc^2x cot^2x + csc^2 x) = csc^4 x Taking common term csc^2 x on L H S out, => csc^2 x ( 1 + cot^2 x) But csc^2 x = 1 + cot^2 x (Trigonometric identity.) Hence => csc^2 x * csc^2 x = csc^4 x = R H S Q E D.

Jul 22, 2015 · How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? ... How do you prove that #sec ...

Nov 3, 2015 · First term of left side: #((cos x/(sin x))/(1/(cos^2 x))) = (cos^3 x)/(sin x) # (1) Second term of left side: ...

Jun 23, 2018 · #"differentiate using the "color(blue)"chain rule"# #"given "y=f(g(x))" then"# #dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

Mar 28, 2018 · Verify the identity.? cot^2x + csc^2x = 2csc^2x - 1. Show your work. Trigonometry. 1 Answer Jim G.

Nov 2, 2017 · This is why I always put parenthesis around the arguments of fuctions; cos2x+csc2x=2csc2x-1, is not an identity but cot^2(x) + csc^2(x) = 2csc^2(x) - 1 is an identity.

Apr 10, 2018 · We have: #csc^2x(secx- 1) = secx/(1 + cosx)# #1/sin^2x(1/cosx- 1) = (1/cosx)/(1+ cosx)# #1/(sin^2xcosx) - 1/sin^2x = 1/(cosx(1 + cosx))#

May 22, 2018 · int csc^2x/cot^3x "d"x=1/2tan^2x+"c" We want to find intcsc^2x/cot^3x "d"x. Note that csc^2x/cot^3x=1/sin ...

Mar 24, 2017 · How do you find all the solutions for #2 \sin^2 \frac{x}{4}-3 \cos \frac{x}{4} = 0# over the

Apr 18, 2018 · The 2x makes the period pi, the -1 compared to 2 in 2x makes the phase shift 1/2 radian, and the divergent nature of cosecant makes the amplitude infinite. [My tab crashed and I lost my edits. One more try.] Graph of 2csc(2x - 1) graph{2 csc(2x - 1) [-10, 10, -5, 5]} The trig functions like csc x all have period 2\\pi. By doubling the coefficient on x, …