trigonometry: prove that, in a triangle, [sin2a sin2b sin2c]/[sina sinb sinc]=8sina/2sinb/2sinc/2.
Published 1 month ago • 14 plays • Length 7:33Download video MP4
Download video MP3
Similar videos
-
11:01
prove (sin2a sin2b sin2c )/(sina sinb sinc ) = 8 sin(a/2) sin(b/2) sin(c/2)
-
4:02
in deltaabc, prove that:a) (sin2a sin2b sin2c)/(sina sinb sinc) = 8sin(a/2) sin(b/2)sin(c/2)...
-
21:52
trigonometry for beginners!
-
7:12
if a b c = 180, prove that sin2a sin2b sin2c = 4 sina sinb sinc
-
5:51
prove that : sina-sinb-sinc=-4cos a/2.sin b/2.sin c/2
-
3:52
prove that : -sina sinb sinc=4cos a/2.sin b/2.sin c/2
-
3:57
prove that sin²a/2 sin²b/2 - sin²c/2 = 1 - 2cosa/2 cosb/2 sinc/2 | edutainment online
-
6:18
proof of sina/2.sinb/2.sinc/2 less than or equals 1/8, when a, b, c are the angles of a triangle.
-
13:41
if a, b, c are angles in a triangle, sin 2a sin 2b sin 2c = 4 sin a sin b sin c# trigonometry
-
15:14
how to solve two triangle trigonometry problems
-
8:55
prove that sin²a sin²b sin²c = 2 2cosacosbcosc if a b c =π | #trigonometry #mmacademy
-
4:30
if `a b c =pi ,` prove that `sin 2a sin 2b sin 2c=4 sin asin b sin c.`
-
27:57
trig identities
-
7:39
prove that sina/2 sinb/2 sinc/2 = 1 4sin[(π-a)/4]sin[(π-b)/4]sin[(π-c)/4] | edutainment online
-
24:51
verifying trigonometric identities
-
6:24
prove that sin2a sin2b - sin2c = 4cosa cosb sinc
-
4:11
ex: product to sum trigonometric identity involving (sin(a)*sin(b) and cos(a)*cos(b))
-
3:18
solved example : trigonometric proof : problem 1