Recent content by Elijah the Wood

Jeez...that was incredible. Thanks a lot...but I don't suppose there was an easier way, was there? haha

can anyone help?

1) tan(x)-sin(x)/2tan(x) = sin^2(x/2) 1-sin(x)/2 = sin^2(x/2) (That's as far as I've gotten on this one...but i don't know if I'm in the right direction or what to do next?) 2) sin2x = 1 / tanx + cot2x 2sin(x)cos(x) = '' sin(x) * cos(x) * sin(x) = '' (The same from #1 applies to...

since sin(x/2)=root17+-15/34 would the final answer be sin (x/2)=root+-16/17? side note: why is sin2x not equal to 2sinx? Is it because in sin2x you are doubling the angle and in 2sinx you're doubling the whole answer?

If sin(x) = 8/17, what would sin(x/2) = ? I knew how to get 8/17, but i have no idea where to go from here Would you just times the denominator by 2?

1) tan(x)-sin(x)/2tan(x) = sin^2(x/2) 1-sin(x)/2 = sin^2(x/2) (That's as far as I've gotten on this one...but i don't know if I'm in the right direction or what to do next?) 2) sin2x = 1 / tanx + cot2x 2sin(x)cos(x) = '' sin(x) * cos(x) * sin(x) = '' (The same from #1 applies to...

I'm retarded...any help would be much congratulated...and i have two problems that are bugging me fiercing Prove the given identity: 1) tan(x)-sin(x)/2tan(x) = sin^2(x/2) 2) sin2x = 1 / tanx + cot2x