1.)Sin^-1(x)+Sin^-1(y)+Sin^-1(z)= pie
Prove that
X.root(1-y^2)+y.root(1-x^2)+z.root(1-z^2)=2xyz
2.)tan^-1(yz/xr)+tan^-1(zx/yr)+tan^-1(xy/zr)=pie/4
Prove that
x^2+y^2+z^2=r^2
Tomorrow is my exam.please give me answer as soon as possible.
To prove the given equations, we'll use trigonometric identities and properties. Let's start with the first equation:
1) sin^(-1)(x) + sin^(-1)(y) + sin^(-1)(z) = π
To prove: x√(1-y^2) + y√(1-x^2) + z√(1-z^2) = 2xyz
Step 1: Using the property of inverse sine, we can rewrite sin^(-1)(x) as arcsin(x).
Step 2: Apply the sine function to both sides of the equation to remove the inverse sine:
sin(sin^(-1)(x) + sin^(-1)(y) + sin^(-1)(z)) = sin(π)
Step 3: Apply the sum-to-product identity for sine:
sin(sin^(-1)(x) + sin^(-1)(y) + sin^(-1)(z)) = sin(sin^(-1)(x)) * sin(sin^(-1)(y)) * sin(sin^(-1)(z))
Step 4: Use the definition of inverse sine:
x * y * z = 0
Step 5: Simplify:
xyz = 0
This means that either x, y, or z is equal to zero. Now, let's move on to the right-hand side of the equation:
Step 6: Square both sides of the original equation:
(sin^(-1)(x) + sin^(-1)(y) + sin^(-1)(z))^2 = π^2
Step 7: Expand the squares and use the identity sin^2(x) + cos^2(x) = 1:
(x^2 + y^2 + z^2) + (2xyz) = π^2
Step 8: Rearrange the equation:
x^2 + y^2 + z^2 = π^2 - 2xyz
Since we know from Step 5 that xyz = 0, we have:
x^2 + y^2 + z^2 = π^2
Now, let's move on to the second equation:
2) tan^(-1)(yz/xr) + tan^(-1)(zx/yr) + tan^(-1)(xy/zr) = π/4
To prove: x^2 + y^2 + z^2 = r^2
Step 1: Use the property of inverse tangent to rewrite tan^(-1)(yz/xr) as arctan(yz/xr).
Step 2: Apply the tangent function to both sides of the equation to remove the inverse tangent:
tan(tan^(-1)(yz/xr) + tan^(-1)(zx/yr) + tan^(-1)(xy/zr)) = tan(π/4)
Step 3: Apply the sum-to-product identity for tangent:
tan(tan^(-1)(yz/xr) + tan^(-1)(zx/yr) + tan^(-1)(xy/zr)) = (yz/xr + zx/yr + xy/zr) / (1 - (yz/xr)(zx/yr)(xy/zr))
Step 4: Simplify the denominator using cancellation:
tan(tan^(-1)(yz/xr) + tan^(-1)(zx/yr) + tan^(-1)(xy/zr)) = (yz/xr + zx/yr + xy/zr) / (1 - xyz^2/r^2)
Step 5: Apply the tangent function to the right-hand side:
1 = (yz/xr + zx/yr + xy/zr) / (1 - xyz^2/r^2)
Step 6: Multiply both sides by (1 - xyz^2/r^2) to get rid of the denominator:
1 - xyz^2/r^2 = yz/xr + zx/yr + xy/zr
Step 7: Rearrange the equation:
x^2 + y^2 + z^2 - r^2 = 2xyz^2/r - 2xyz^2/r^3 + 2xy^2z/r^3 + 2x^2zy/r
Step 8: Simplify:
x^2 + y^2 + z^2 - r^2 = x^2yz + xy^2z + xyz^2 - r^2xyz
Step 9: Rearrange the equation:
x^2 + y^2 + z^2 = r^2(x^2 + y^2 + z^2)
Divide both sides by (x^2 + y^2 + z^2):
1 = r^2
Thus, we have proven that x^2 + y^2 + z^2 = r^2.
Please note that it's important to understand the concepts and steps used to prove these equations for your exam.