It's hard to use clear notation without a special program for typing Maths, but I take it that sin^3x means "sine cubed x",
i.e. (sin x)^3. In that case, the first two terms of this equation are equal to sin 3x (standard formula), and so the equation is
sin 3x - k = 0.
Since A and B satisfy the equation, then sin 3A and sin 3B must each be equal to k.
i.e. sin 3A = sin 3B
However, 3A > 3B, and so the possibilities are
(1) 3A = pi - 3B, and so
A + B = pi/3
whence
C = pi - (A + B)
= 2pi/3
(2) 3A = 2pi + 3B
A = 2pi/3 + B
whence
C= pi - (A + B)
= pi - (2pi/3 + 2B)
= pi/3 - 2B.
Of course A, B, C must all be positive, and so B must be less than pi/6 (so that pi/3 - 2B will be positive) and C must be less than pi/3 (since B is positive). So any value of C:
0 < C < pi/3, is a correct solution.
(3) 3A = 3pi - 3B i.e. A = pi - B is not really a possibility as it leaves C=0.
So, unfortunately, I disagree with the problem as set, agreeing only with the third answer provided, and finding an infinite set of others.
If C were equal to pi/3, then A + B would be 2pi/3, and so
3A + 3B would be 2pi. But sin (2pi - 3A) = -sin 3A, which can equal sin 3A only if they are both zero. Since B > 0, the only possibility is 3A = pi = 3B, which contradicts A>B.
If C = pi/2, then 3A + 3B = 3pi/2, and so
sin 3B = sin(3pi/2 - 3A)
= - cos 3A, which can equal sin 3A only if 3A =3pi/4, and then
3B = 3pi/2 - 3pi/4
= 3pi/4,
again contradicting A>B.
2006-08-29 13:17:25
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answer #1
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answered by Hy 7
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The equation -4u^3 + 3u - k = 0 has discriminant 4 + k^2, which is positive; therefore there is only one real solution. Therefore the two solutions A and B have sines equal to the same number u, and
A = pi/2 - B
Since A + B + C = pi, we must have C = pi/2
2006-08-29 10:35:13
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answer #2
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answered by dutch_prof 4
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Since angle A and angle B don't seem to show up anywhere in the equation you gave (and k seems to have just 'popped up' out of nowhere), I'd say you're just SOL.
Doug
2006-08-29 10:20:20
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answer #3
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answered by doug_donaghue 7
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