cos2t=(1-tan^2(t))/(1+tan^2(t))
2007-12-19 19:56:36 · 4 answers · asked by mike m 1 in Science & Mathematics ➔ Mathematics
RHS
(1 - sin ² t / cos ² t) / (1 + sin ² t / cos ² t )
( cos ² t - sin ² t ) ( cos ² t + sin ² t )
cos 2t / 1
cos 2t
LHS
cos 2t
LHS = RHS
2007-12-19 20:33:35 · answer #1 · answered by Como 7 · 2⤊ 0⤋
RHS = (1 -- tan^2t) / (1 + tan^2t)
= (cos^2t -- sin^2t) / (cos^2t + sin^2t)
= cos2t
= LHS
2007-12-20 04:26:43 · answer #2 · answered by sv 7 · 0⤊ 0⤋
LHS
= cos(2t)
= cos²(t) - sin²(t)
= (cos²(t) - sin²(t)) / (cos²(t) + sin²(t)) ---- cos²(t) + sin²(t) = 1
= ((cos²(t) - sin²(t))/cos²(t)) / ((cos²(t) + sin²(t))/cos²(t))
= (1 - tan²(t)) / (1 + tan²(t)) ---- sin(t)/cos(t) = tan(t)
= RHS
2007-12-20 04:13:42 · answer #3 · answered by gudspeling 7 · 2⤊ 0⤋
The Bourne Identity... for sure it is.
2007-12-20 04:04:24 · answer #4 · answered by Anonymous · 0⤊ 1⤋