Michael W. answered • 04/27/15
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Satjayeet,
Let's see if I can get you started, and see if that matches up with the direction you were going...
Think of "arcsin(x)" as "the angle whose sine is x." So, it's an angle. Just to make things easier (and maybe a bit more obvious?), let's call that angle "A." So, A is arcsin(x), and sin(A) = x.
Same thing for the second piece. Let's call angle B "the angle whose sine is 2x." So, B is arcsin(2x), and sin(B) = 2x.
Just using our labels, what we now have is something that looks like this:
A + B = pi/3
That's nothing write home about, but we know about those angles. We know their sines. And, if we know their sines, we could draw out the triangles, and then we'd be able to figure out their cosines, too.
Let's see if I can get you started, and see if that matches up with the direction you were going...
Think of "arcsin(x)" as "the angle whose sine is x." So, it's an angle. Just to make things easier (and maybe a bit more obvious?), let's call that angle "A." So, A is arcsin(x), and sin(A) = x.
Same thing for the second piece. Let's call angle B "the angle whose sine is 2x." So, B is arcsin(2x), and sin(B) = 2x.
Just using our labels, what we now have is something that looks like this:
A + B = pi/3
That's nothing write home about, but we know about those angles. We know their sines. And, if we know their sines, we could draw out the triangles, and then we'd be able to figure out their cosines, too.
So, we can take advantage of that by taking the sine or the cosine of both sides of our equation. We get:
sin(A+B) = sin(pi/3).
Or, we could get:
Or, we could get:
cos(A+B) = cos(pi/3).
Not really thinking about it, the first time I did the problem, I went with sin, because the original problem had arcsin in it. Baaaad idea. :) The fact that we started with arcsin doesn't matter. Cosine is way cleaner (and after you continue a few steps in the problem, you might see how it helps).
So, if we go with the cosine version:
cos(A+B) = cos(pi/3).
The right-hand side we know. That's 1/2.
For the left-hand side,isn't there an identity for something like cos (A+B)? Can't we write it in terms of cos A, cos B, sin A, and sin B? Using the triangles you drew, those are actually four values we can plug in. They all have x's in them, but that's okay, because x is the variable we're trying to solve for.
If you can hunt down the identity, and use the triangles you drew to determine the values you need, then what you get is a big messy expression with x's in it. Perform some algebracadabra, and poof, you can find x. Easier said than done...
...but does that help?
Let us know!
Satyajeet S.
thank you sir very much, for your advice step by step but the real problem comes in when you do algebraic manipulation, (sir i had already done the basic steps before which you told now),that is you get different answer when you do from sin way and when you manipulate it using cos,but even if you get different answer none of those fit in the eqn.Please do it yourself and check if your solution fits in !!
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04/28/15
Michael W.
I was actually worried about my algebra, too, given that it was rather messy. So yes, I did do it myself, and yes, I did check what I got, and yes, one of my answers worked.
cos (A + B) = cos (pi/3)
The cosine of pi/3 is 1/2, and the identity for cos gives:
cos(A)cos(B) - sin(A)sin(B) = 1/2
So, going back to the triangles you drew for angles A and B, you should get:
√(1-x2) * √(1-4x2) - (x)(2x) = 1/2
To solve that, we can move the 2x2 to the other side, and then we square the whole equation to get rid of the square roots. The left-hand side actually gets a little better, because the square roots are multiplied together...so when you square it, you get rid of both roots. The right-hand side, you have to multiply out.
After we square it, it looks like this:
(1 - x2)(1 - 4x2) = (2x2 + 1)2
After you multiply out both sides:
1 - 4x2 - x2 + 4x4 = 4x4 + x2 + x2 + 1/4
Mercifully, the x4 terms cancel. After you combine like terms on each side, all you're left with are x2 terms and constants. Isolate the x2 on one side of the equation, get the constants on the other, and I land here:
7x2 = 3/4
Divide off the 7, which gives 3/28 on the right, and then take the square root, and:
x = ±√(3/28)
You can actually simplify that radical, so you don't end up with the square root of 28 in the denominator, but for the purposes of checking the answer, the radical won't bother our calculators. :) Go ahead and plug it back into the original equation:
arcsin(√(3/28)) + arcsin (2√(3/28)) = pi/3 (or 60 degrees if you want to check it in degrees)
Even the double-check is a little bit messy, so be careful.
When I plugged it in, the positive solution worked, and then the negative solution still gives pi/3, but negative pi/3, so that's no good. Extraneous.
Can you use this to spot your error?
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04/28/15
Satyajeet S.
thank you very much!!!!!!!
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05/02/15
Michael W.
04/27/15