I was reading Math Educators stack exchange and one person mentioned he’s appalled year after year that his Calc I students were, without fail, unable to execute this problem:
.8sin(x) + cos(x) = 1.2
Well I tried.
For an hour.
And I still can’t do it. If it wasn’t for that coefficient on the sine I could probably do it. My best hunch involves squaring both sides, but I’ve tried every identity I can think of and look up and it’s not coming out. I’m fairly sure there are infinitely many values for x that satisfy this equation, as is usually the case for periodic functions, but that’s it. At the very least it’ll be one or more functions with acos or atan or asin or something.
I looked back at the page and in the comments his hint was " Since cos^2(x)+sin^2(x)=1 it follows that you can eliminate either sine or cosine to give an algebraic equation which when squared gives you a quadratic which when solved reveals the two solutions for θ. They do have a scientific calculator for the numerics here."
As far as I can tell, this means replacing, e.g. cos^2(x) with sqrt(1-sin^2(x)), then squaring both sides. But I already tried that and hit a dead end, as far as I got it ends up:
It’s not a quadratic equation, you can transform it to
ax^2 + bx*sqrt((1-x)(1+x)) + c = 0
But it doesn’t change the fact that if you’re trying to treat it as
ax^2 + bx + c, then b is a function on x, not a constant. Unless there’s some identity I’m missing that allows you to transform sqrt(1-x^2) I’m at a loss.
The trick is to get the sqrt on one side by itself before you square both sides.
.8sin(x) + cos(x) = 1.2
Substitute cos(x)=sqrt(1-sin2(x)).
.8sin(x) + sqrt(1-sin2(x)) = 1.2
Subtract .8sin(x) from both sides so we isolate the sqrt.
sqrt(1-sin2(x)) = 1.2 - .8sin(x)
Now square both sides.
1-sin2(x) = 1.44 - 1.92sin(x) - .64sin2(x)
Substitute w = sin(x).
1 - w2 = 1.44 - 1.92w - .64w2
Looks like a quadratic equation to me.
Apply the quadratic formula to get w, then use x = arcsin(w).
First solution is approximately .318 ; second is approximately 1.031 .
In what context? At the beginning of the class, as a test of their trigonometry skills? At the end of class? Using Newton’s method, with which an equation like this is pretty straightforward?
Either I’m weird or everyone else is. I can follow this ‘replace cos(x) by sqrt(1-sin^2(x)) and then do a bunch of squaring’ method, but it doesn’t seem a very natural way to look at it. The very first thing I thought was ‘rewrite the LHS as a single sine function with amplitude and phase to be determined’. I.e. use the identity acos(x)+bsin(x) = Rsin(x+k). (you need to know compound angle formulae)
You’d get R=sqrt(0.8^2+1^2) ~= 1.281, k = arctan(1/0.8) = 0.896 rad
so 1.281 sin(x+0.896) = 1.2, rearrange to get x = arcsin(1.2/1.281)-0.896 = 0.317 - principal value, of course.
Probably both. The way “normal” people approach this problem is to throw up their hands and say “I don’t know!”. Or if they’re resourceful, maybe to graph it and look at the intersection, or to ask a computer system like Wolfram Alpha.
I’m probably crazy since I tried to solve the more general case first (of y = 0.8*sin(x) + cos(x)).
My first observation is that since both functions have the same period, they must be equivalent to:
y = A*cos(x - B)
B is the x value for which (0.8sin(x) + cos(x)) is at a peak (since cos(0) is at a peak). We know that cos(x) is falling and 0.8sin(x) is rising (or vice versa), and that at the peak the derivatives must be equal and opposite. So we have:
(0.8sin(x))’ = 0.8cos(x)
(cos(x))’ = -sin(x)
0.8*cos(x) = sin(x)
0.8 = sin(x)/cos(x) = tan(x)
B = x = atan(0.8)
For A, we just substitute something in; 0 works:
0.8sin(0) + cos(0) = Acos(0 - B)
1 = A*cos(-atan(0.8))
A = 1/cos(-atan(0.8)) =~ 1.2806
To solve the problem, we just substitute:
A*cos(x - B) = cos(x - atan(0.8))/cos(-atan(0.8)) = 1.2
x = acos(1.2 * cos(-atan(0.8))) + atan(0.8) =~ 1.0315
ETA: Clearly I did the same thing as Nancarrow but without knowing the compound angle formula
I did ask Wolfram, before I made this thread, so I knew the answer. But I don’t have pro right now so it wouldn’t show me its work. It was bugging me that I couldn’t figure out how to do it.
Beginning of the class.
That’s more intuitive to me, and along the lines of what I originally tried to do, but the page of identities I looked up didn’t deal with the case when the trig functions had coefficients.
If this was really important I probably would have ended up rederiving them from definition, but eh…
Another way to think of Nancarrow’s approach is like this:
The original equation amounts to <1, 0.8> DOT <cos(x), sin(x)> = 1.2.
As you may know, A DOT B = |A| |B| cos(angle between A and B). And the angle to A from B is as good as the difference between the angle to A from <1, 0> and the angle to B from <1, 0>.
In this case, that tells us |<1, 0.8>| cos((angle to <1, 0.8> from <1, 0>) - x) = 1.2.
At this point, solving the problem in terms of standard operations is straightforward; we get that x = (angle to <1, 0.8> from <1, 0>) - arccos(1.2/|<1, 0.8>|) , whatever that comes out to…
Substitute in b = 1.2 - 0.8*a in the second equation and slap it in the ol’ quadratic formula. Basically the same as everyone else is saying, but leaving the square root part to the quadratic formula.
That’s a more elegant approach that the one implied by the professor’s hint in the OP. I like it very much… which makes me wonder if the professor had a reason for doing it the other way, substituting the second equation into the first rather than the first into the second. And we are still stuck with the enigma of why the professor complained that his students had a zero percent success rate in solving the problem.
So far in this thread, we’ve found at least three, arguably four or five, methods for doing this problem. It looks like the professor knew one of them, but was ignorant of the other 2-4. This doesn’t place him much above his students, who were ignorant of 3-5.
But I’m very surprised that literally none of his students was able to do it. What I fear might have happened, is that some of his students did come up with valid solutions (possibly one of the ones mentioned here, possibly yet another one), and the professor rejected them as wrong because they weren’t the specific method he knew.
That would be terrible, but one would hope the teacher (or his graders) could at least recognize a numerically correct answer when seeing one (his comment about scientific calculators suggests he did want the answer ultimately in numeric form), and at that point would proceed to investigate the validity of its derivation.
[Also, for clarification, the professor was apparently complaining about the performance of students in a physics course with a calculus prerequisite, rather than performance within a Calc I course (see here). Though now I’m even more surprised that no one solved it…]
I can’t see why the professor would prefer that approach; in fact, that approach is more dangerous, because in a step such as from “cos(x) = 1.2 - .8sin(x)” to “1 - sin[sup]2/sup = cos[sup]2/sup = (1.2 - .8sin(x))[sup]2[/sup]” [or the analog of this isolating and then squaring sin(x) instead], one introduces spurious solutions (in this case, those for which .8sin(x) - cos(x) = 1.2, which are the supplementary angles to the appropriate solutions; in the analogous case, those for which -.8sin(x) + cos(x) = 1.2, which are the negations of the appropriate solutions); to counter this, one must not at the end simply solve for x from the derived values for sin(x), but rather rederive cos(x) from sin(x) by the posed linear relationship, and then solve for x from <cos(x), sin(x)> (or, with less finesse but amounting to the same thing, apply zut’s step 4 of checking all putative solutions against the original constraint).
Now, if you ignore the possibility of cos(x) [analogously, sin(x)] being negative, as use of the arcsin [analogously, arccos] button on your calculator is liable to lead you to conclude without justification, you will happen to luck out in this case. This would be an example of the right answer by less-than-correct means which the professor would be justified in deducing points for…
But this difficulty is entirely avoided by doing the substitution Hogarth’s way, as well as by appropriately pursuing Nancarrow’s approach (there is a sense in which Hogarth’s and Nancarrow’s approaches are at heart the same, but they of course present quite differently).