- prove identity
sin^{ 2}x/1-sinx= secx+1/secx
- solve sin2y=cos4y for y, where between 0 degree and 360 degree
- cosx(secx-cscx)= 1-cotx
- (a) solve 11sin^{ 2} y=13-3sin^{ 2} y for y, where 0 to 360 degree
(b) what second-quadrant angle satisfices this equation?
(c) what fouth- quadrant angle satisfices this equation?
Hi,
I assume that L.S. = sin^{ 2}x/(1-sinx) and not sin^{ 2}x/1-sinx which would be
the same as sin^{ 2}x - sinx.
It seems reasonable to start with the right side by changing secx into cosx
and finding a common denominator:
R.S. = secx + 1/secx
= 1/cosx + cosx
= 1/cosx + cos^{ 2}x/cosx
= (1 + cos^{ 2}x)/cosx
The denominators of cos x (R.S) and 1-sinx (L.S.) requires a little trick
to deal with.
cos^{ 2}x=1-sin^{ 2}x=(1-sinx)(1+sinx) suggests we multiply numerator and
denominator of the left side by 1+sinx, and hope things work out (note that
this will have the added benefit of getting rid of a difference in the
denominator of L.S.):
L.S. = (sin^{ 2}x)(1+sinx)/(1-sinx)(1+sinx)
= (1-cos^{ 2}x)(1+sinx)/cos^{ 2}x
= (1-cos^{ 2}x+sinx-sinxcos^{ 2}x)/cos^{ 2}x
I am now starting to think that the identity is false. So I tried 30 degrees.
L.S.(x=30) = 1/2
R.S.(x=3-) = 7*sqr(3)/6
Thus, the identity is false.
I decided not to spend any time with the remaining questions in case there
was a problem with these as well.
Paul