Algebra 2 hw help please
2007-05-16 10:02:52 · 7 answers · asked by Destiny K 2 in Science & Mathematics ➔ Mathematics
cot(THETA) is the same as cos(THETA)/sin(THETA).
Now when you divide sin(THETA) by cot(THETA), you get sin^2(THETA)/cos(THETA) * cos(THETA).
The cosines cancel, and you're left with
sin^2(THETA)
2007-05-16 10:10:04 · answer #1 · answered by Nicknamr 3 · 0⤊ 0⤋
sin²A +sin²B+2 sinA sinB+cos²A+cos²B+2cosAcosB - 2 (1) + 2 sin A sin B + (1) + 2 cos A cos B - 2 2 ( cos A cos B + sin A sin B ) 2 cos ( A - B )
2016-04-01 04:47:49 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Keep in mind: tan(theta) = sin(theta)/cos(theta) and cot(theta) = 1/tan(theta) = cos(theta)/sin(theta)
Thus (sin/cot) X cos = [sin//cos/sin] X cos = sin^2
Also keep in mind: 1 = cos^2 + sin^2; so that sin^2 = 1 - cos^2, which is useful.
2007-05-16 10:17:07 · answer #3 · answered by oldprof 7 · 0⤊ 0⤋
Split the cotangent term into cos(θ) / sin(θ), then simplify:
[sin(θ) / cot(θ)] cos(θ)
[sin(θ) / (cos(θ)/sin(θ)) ] cos(θ)
[sin^2 (θ) / cos(θ) ] cos(θ)
sin^2 (θ)
2007-05-16 10:10:21 · answer #4 · answered by Anonymous · 0⤊ 0⤋
(sin x / cot x)(cos x)
(sin x)(tan x)(cos x)
(sin x)((sin x)/(cos x))(cos x)
sin^2 x
2007-05-16 10:08:20 · answer #5 · answered by hawkeye3772 4 · 0⤊ 0⤋
(sinθ/cotθ)cosθ = (sinθ)(sinθ/cosθ)cosθ = sin^2θ
2007-05-16 10:10:27 · answer #6 · answered by Damian_Anderson 2 · 0⤊ 0⤋
(sin)(tan)(cos)
(sin)[(sin)/(cos)](cos): cos cancels out
sin*sin
2007-05-16 10:10:17 · answer #7 · answered by Cool Nerd At Your Service 4 · 0⤊ 0⤋