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Old 02-16-2018, 10:54 AM
blood63 blood63 is offline
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Help me solve a math problem

This is not homework. Pinky swear.
I love to try word problems that come my way.
This one simplifies to
300cos(angle) = 5sin(angle) + 3

It works out to around 88.5 degrees but I was only able to solve it by trial and error.
I would appreciate a trig lesson here.

Thanks.
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Old 02-16-2018, 11:20 AM
sbunny8 sbunny8 is offline
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If you square both sides, then apply sin^2 + cos^2 = 1, it becomes a quadratic equation, which you can solve any number of ways but the quadratic formula is probably easiest.

90000cos^2 = 25sin^2 + 30sin + 9

90000(1-sin^2) = 25sin^2 + 30sin + 9

Then substitute x=sin^2 and solve for x

Last edited by sbunny8; 02-16-2018 at 11:24 AM.
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Old 02-16-2018, 12:13 PM
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Chronos Chronos is online now
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I'm curious just what word problem you were doing that ended up reducing to that.

And it should be noted that, although this one is solvable, it's very easy when putting together equations involving sines and cosines to end up with what's called a transcendental equation, for which the only recourse is something like your trial-and-error solution (though of course, there are usually more efficient methods than pure trial-and-error available).
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Old 02-16-2018, 02:25 PM
MikeS MikeS is offline
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Another method is to apply trig identities. Your equation is

300 cos x - 5 sin x = 3.

Now draw a right triangle with legs 300 and 5. The hypotenuse of this triangle will be q = √(3002 + 52). The left-hand side of your equation can be rewritten as

q ( (300/q) cos theta - (5/q) sin alpha) = 3.

But if you go back to your drawing of the triangle, and you define alpha to be the angle between the length-300 side and the hypotenuse, then you can see that 300/q = cos alpha and 5/q = sin alpha. Thus, this is equivalent to

q [(cos alpha)(cos theta) - (sin alpha)(sin theta)] = 3.

The quantity in square brackets is a trigonometric addition identity, meaning that

cos (alpha + theta) = 3/q.

Thus, theta = cos-1 (3/q) - alpha, with alpha and q defined as above. This sort of trick can be used in general to solve for an angle that appears in both a sine and a cosine in an equation.

Last edited by MikeS; 02-16-2018 at 02:27 PM.
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Old 02-16-2018, 02:28 PM
blood63 blood63 is offline
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Quote:
Originally Posted by Chronos View Post
I'm curious just what word problem you were doing that ended up reducing to that.

And it should be noted that, although this one is solvable, it's very easy when putting together equations involving sines and cosines to end up with what's called a transcendental equation, for which the only recourse is something like your trial-and-error solution (though of course, there are usually more efficient methods than pure trial-and-error available).
This is actually from my son's homework in vectors. I wanted to try it as I loved physics in high school. He never got back to me with a solution. I think he has forgotten about it but I get a bit obsessive over these things.
Here is the word problem as I remember.
A river is 0.5 km wide. A boat must cross the river to a point 15 km downstream. The current is 12 km/h and the speed of the boat is 20 km/h. At what angle must the boat head?
My solution involved breaking it up into components and having the t for the vertical and horizontal motions be equal. There must a vector solution as well.
I was able to solve the quadratic that resulted from the hint of squaring as above. I got 88.9 degrees which I found is not correct through a simple check.
Getting closer but still stumped.
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Old 02-16-2018, 04:50 PM
bob++ bob++ is offline
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This might help
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Old 02-16-2018, 05:35 PM
DPRK DPRK is offline
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Quote:
Originally Posted by Chronos View Post
And it should be noted that, although this one is solvable, it's very easy when putting together equations involving sines and cosines to end up with what's called a transcendental equation, for which the only recourse is something like your trial-and-error solution (though of course, there are usually more efficient methods than pure trial-and-error available).
Of course, sin and cos are already transcendental functions, so computing them or their inverses already takes you beyond pure algebra. Similarly, you could have an equation like cos x = x. However, if all you have are sines and cosines of the same angle (or appropriate multiples thereof), you can always apply sbunny8's trick and get rid of them by making a substitution like sin x = 2u/(1+u), cos(x) = (1-u)/(1+u) and all you have to do is solve the resulting polynomial equation for u instead.
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Old 02-17-2018, 10:30 PM
AWB AWB is offline
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A second solution is Θ=-89.618
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Old 02-21-2018, 10:32 AM
blood63 blood63 is offline
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Quote:
Originally Posted by MikeS View Post
Another method is to apply trig identities. Your equation is

300 cos x - 5 sin x = 3.

Now draw a right triangle with legs 300 and 5. The hypotenuse of this triangle will be q = √(3002 + 52). The left-hand side of your equation can be rewritten as

q ( (300/q) cos theta - (5/q) sin alpha) = 3.

But if you go back to your drawing of the triangle, and you define alpha to be the angle between the length-300 side and the hypotenuse, then you can see that 300/q = cos alpha and 5/q = sin alpha. Thus, this is equivalent to

q [(cos alpha)(cos theta) - (sin alpha)(sin theta)] = 3.

The quantity in square brackets is a trigonometric addition identity, meaning that

cos (alpha + theta) = 3/q.

Thus, theta = cos-1 (3/q) - alpha, with alpha and q defined as above. This sort of trick can be used in general to solve for an angle that appears in both a sine and a cosine in an equation.
Just getting to this now. Very cool. Thanks. I get an angle of 88.5 degrees.
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