How does one prove cot10 + tan5 = csc10? No combination of 3 of the "basic" trigonometric numbers (30, 45, 60, 90 etc.) seems to add up to either 5 or 10.
+129845 In general, we have
cot(2u) = [1 - tan^2(u)]/ [2tanu]
csc(2u) = 1/[2sinucosu]
So we have
[1 - tan^2(u)]/ [2tanu] + tan(u) =
[1 - tan^2(u) + 2tan^2(u)] / [2tan(u)] =
[1 + tan'^2(u)]/[2tan(u)] =
sec^2(u) / [2tan(u)] =
[1 / cos^2(u)] * (1/2)cos(u)/sin(u) =
(1/2)* 1/ (sin(u)cos(u) =
1/[2sin(u)cos(u)] = csc(2u)
For some reason I can't seem to edit my post, so I'll post this as an answer, just to give more info about my question.
We've only learned about the 3 basic formulas, these:
sin(x-y)=sin(x)cos(y)-cos(x)sin(y)
cos(x-y)=cos(x)cos(y)-sin(x)sin(y)
tan(x-y)=[tan(x)+tan(y)]/[1-tan(x)tan(y)].
(And their x+y counterparts)
Fairly certain I'm supposed to use these, with x-y=5 or x-y=10, x being one of the basic trigonometric numbers (0, 30, 45, 60, 90, 120, 135, 150, 180, 210, 225, 240, 270, 300, 315, 330, 360), and y=a+b or y=a-b in which both a and b are also one of those numbers. Is there a way to easily find out what x, a, and b would be, considering they have to be included in the basic trigonometric numbers?
Could be wrong about this entire theory, though.
+129845 In general, we have
cot(2u) = [1 - tan^2(u)]/ [2tanu]
csc(2u) = 1/[2sinucosu]
So we have
[1 - tan^2(u)]/ [2tanu] + tan(u) =
[1 - tan^2(u) + 2tan^2(u)] / [2tan(u)] =
[1 + tan'^2(u)]/[2tan(u)] =
sec^2(u) / [2tan(u)] =
[1 / cos^2(u)] * (1/2)cos(u)/sin(u) =
(1/2)* 1/ (sin(u)cos(u) =
1/[2sin(u)cos(u)] = csc(2u)
