|(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

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?

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......

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.

I understand that you need help with calculating the determinant of the given matrix:

```
|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)|
```

To solve this, we can use the properties of determinants. Let's denote the given matrix as A:

```
A = |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)|
```

To find the determinant of matrix A, we can use the formula for a 3x3 matrix:

```
det(A) = (a1 * (b2 * c3 - b3 * c2)) - (a2 * (b1 * c3 - b3 * c1)) + (a3 * (b1 * c2 - b2 * c1))
```

Where a1, b1, c1 are the elements of the first row, a2, b2, c2 are the elements of the second row, and a3, b3, c3 are the elements of the third row.

Applying this formula to our given matrix A, we have:

```
det(A) = (sin(x + p) * (sin(y + q) * sin(z + r) - sin(z + q) * sin(y + r))) - (sin(x + q) * (sin(y + p) * sin(z + r) - sin(z + p) * sin(y + r))) + (sin(x + r) * (sin(y + p) * sin(z + q) - sin(z + p) * sin(y + q)))
```

This is the full solution for finding the determinant of the given matrix. Simply substitute the values of x, y, z, p, q, and r into the above equation to get the numerical value of the determinant.