Question

Easy Trig Question FAN ANDA MEDAL

Answer 1

\[\sin^2A/2 + \sin^2B/2 + \sin^2C/2 = 1-2sinA/2sinB/2sinC/2\]

Answer 2

@ganeshie8

Answer 3

@zepdrix

Answer 4

Sin[A]^2/2 + Sin[B]^2/2 + Sin[C]^2/2 = 1 - (Sin[A] Sin[B] Sinc[x])/4

Answer 5

Dont know if thats whats your looking for

Answer 6

its sin^2 (A/2) etc.

Answer 7

and on the RHS its 2 sin (A/2) etc

Answer 8

x = sin^(-1)((-2 sin(A/2) sin(B/2) sin(C/2)-sin^2(A/2)-sin^2(B/2)+1)^(1/C))

Answer 9

http://www.wolframalpha.com/input/?i=sin%5E2%28A%2F2%29%2Bsin%5E2%28B%2F2%29%2Bsin%5E%282C%2F2%29%3D1%E2%88%922sin%28A%2F2%29sin%28B%2F2%29sin%28C%2F2%29

Answer 10

This might help, not really sure though.

Answer 11

no theres no x

Answer 12

http://www.wolframalpha.com/input/?i=sin%5E2%28A%2F2%29%2Bsin%5E2%28B%2F2%29%2Bsin%5E2%28C%2F2%29%3D1%E2%88%922sin%28A%2F2%29sin%28B%2F2%29sin%28C%2F2%29

Answer 13

My bad, misplaced the parentheses.

Answer 14

To me, the expression is unclear. I don't get what we are supposed to do. Can you use latex to type it clearer.??

Answer 15

especially the right hand side, which is /.../.../ ??? divide/ divide/ divide. Ha!!!

Answer 16

\[\sin^2\frac{ A }{2 }+\sin^2\frac{ B }{ 2 }+\sin^2\frac{ C }{ 2 }=1-2\sin \frac{ A }{ 2 }\sin \frac{ B }{ 2 }\sin \frac{ C }{ 2 }\]

Answer 17

oh, I got it, it is sin (A/2) sin (B/2) sin(C/2) and we have to prove it, right?

Answer 18

Yeah, now looking at it I'm pretty sure its a proof.

Answer 19

where A+B+C=180 sorry

Answer 20

do you know the half angle formulas? or are you not to use them in this proof

Answer 21

how would the half angle formula help?

Answer 22

wait nevermind, thatsnot applicable.

Answer 23

yeah sorry