another trig function (1 Viewer)

[jemsta]

prove:
sin2x+cos2x/2cosx+sinx-2(cos^3x+sin^3x)=cosec x

 

[Trev]

Is that:
sin2x+(cos2x/2cosx)+sinx-2(cos^3x+sin^3x)=cosec x
or
(sin2x+cos2x)/(2cosx+sinx) - 2(cos^3x+sin^3x)=cosec x
???

 

[jemsta]

its
(sin2x+cos2x)/(2cosx+sinx) - 2(cos^3x+sin^3x)=cosec x

 

[Trev]

I don't think the question is right.
Trying x=90 you get LHS=-3 and RHS=1.

 

[icycloud]

LHS = [sin(2x) + cos(2x)] / [2cos(x) + sin(x) - 2{cos³(x) + sin³(x)}]

Let s = sin(x), c = cos(x)

Thus, we have:
sin(2x) = 2sc, cos(2x) = 2c² - 1
cos³(x) + sin³(x) = (s+c)(s² - cs + c²) = (s+c)(1-cs)

Thus, LHS = (2sc + 2c² - 1) / (2c + s - 2(s+c)(1-cs))
= (2sc + 2c² - 1) / (2c + s - 2(c- c²s + s - cs²))
= (2sc + 2c² - 1) / (2c + s - 2c + 2c²s - 2s + 2cs²)
= (2sc + 2c² - 1) / (2c²s + 2cs² - s)
= (2sc + 2c² - 1) / (s(2c² + 2cs - 1))
= 1/s
= 1/sin(x)
= cosec(x)
= RHS #

 

[Trev]

So it's all under the denominator... lol *touches nose*