Math

|(sin(x + p), sin(x + q), sin(x + r)), (sin(y + p), sin(y + q), sin(y + r)), (sin(z + p), sin(z + q), sin(z + r))|

Please help me with full solution have done all i could

No hint please just help me please please

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asked by Dema
  1. Oh yeah? What is all this you have done? I don't see any work.
    I get zero for the result. The full determinant value is

    sin(x+p)sin(y+q)sin(z+r)+sin(y+p)sin(z+q)sin(x+r)+sin(z+p)sin(x+q)sin(y+r)
    -sin(z+p)sin(y+q)sin(x+r)-sin(y+p)sin(x+q)sin(z+r)-sin(x+p)sin(z+q)sin(y+r)

    Did you try using the product-to-sum formulas?

  2. actually am still getting into these....i tried it but the result look scary for me plz help me sir

    should i give you my email,so that you can said me the solution to study,because jishka here is a bit restricted,i seriously need help......

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    posted by dema
  3. Recall that
    sinA sinB = 1/2 (cos(A-B) - cos(A+B))
    it's not too hard to expand that into
    sinA sinB sinC = 1/4 (-sin(A-B-C) + sin(A+B-C) + sin(A-B+C) - sin(A+B+C))

    Now, I'm afraid this little problem is just tedious algebra. At least, I don't know of a tricky identity to make things any easier. Get out a nice wide sheet of paper, and write small. You can start with

    sin(x+p) sin(y+q) sin(z+r) =
    1/4 (-sin(x+p-y-q-z-r) + sin(x+p+y+q-z-r) + sin(x+p-y-q+z+r) - sin(x+p+y+q+z+r))

    Now you have to do that same expansion for the other five terms of the determinant. If you are careful, you will find that all those x,y,z,p,q,r expressions appear both as plus and minus terms, cancelling each other out, making the result zero.

    I find very little use of such a problem assignment, as it doesn't really illustrate useful trigonometry. It's just a tricky and unexpected result.

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