Is Trigonometry done differently in the US?

Question

I'm Italian and I've watched some videos from Americans and noticed a weird thing. Let's talk about a linear trigonometric equation like this:

$$\sin x+\cos x+\sqrt3=0.$$

I've seen Americans solving it by just doing an irrational equation, which takes more time and is pointless.

Here we are taught that there are three ways to solve it:

1. We let $Y = \sin x$ and $X = \cos x$ and do a system with the equation of the trigonometric circle, $X^2 + Y^2 = 1$, then we find the values of the sine and cosine and find the related angles, we call this the graphic method.
2. We let $a= \sin x$ and $b=\cos x$, calculate $r=\sqrt{a^2+b^2}$ and $\alpha =\arctan \frac{b}{a}$ and use the formula $r \cdot \sin(x+\alpha)=c$ where $c$ in this case is $-\sqrt 3$. We call this method of the auxiliary angle.
3. We use the formulae $\sin x=\frac{2t}{1-t^2}$, $\cos x=\frac{1-t^2}{1+t^2}$ where $t= \tan(x/2)$ and solve the equation.

Isn't this taught in the US?

Answer 1

As you've tagged this as precalculus, I'll say that there is no standard precalculus curriculum in the United States and very little "official" guidance about what such a curriculum should contain. I think we've always struggled with the puzzle of what to teach in the fourth year of high school to students who mastered the "essential" math in three years. Over the next generation, I predict we'll just toss up our hands and teach introductory Calculus to everyone. As we stand at the moment, the class is a hodgepodge of interesting topics that don't require calculus (like matrices and conic sections) and preparation for calculus (like convergent sequences and the definition of the limit).

The "typical" trigonometry curriculum (and, again, there is no uniformity in the United States) is spread out over several years. Typically, second-year students are learning plane geometry, and learn about using trigonometric functions on acute angles to solve right triangles. Third year students then study the unit circle to extend the trig functions to have a domain of all real numbers. This goes off in two directions: the development of the trig identities (specifically the Law of Sines and Law of Cosines to allow solving oblique triangles) and an analysis of the general sinusoidal function family .$y=a+b\sin(cx+d)$. In practice, there are too many useful trig identities and too many other topics to cover them all in a single year for a non-gifted class, so some of the topics are pushed into the precalculus stewpot or abandoned altogether.

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Just for lulz, I tried solving this problem with my American education. My first attempt was essentially your method 1, because the Girard-Newton identities are a bug that has bitten me recently. But that led me to needing to find the roots of a quadratic equation with irrational coefficients and I ran out of patience with following that trail further. Also, it's not how I would have solved this problem when I was a student anyways.

So here is my second attempt, which is how I would teach students to solve equations of this sort if it were mandated that they learn it (and to be frank, I'm not sure why it should be). It uses the Pythagorean identity and the sine double angle identity that are part of the Common Core. So while an Algebra 2 student may not be taught specifically how to solve equations of this sort, they should have little difficulty verifying my work.

$$\sin x+\cos x=-\sqrt3\\(\sin x+\cos x)^2 =3\\\sin^2x+\cos^2x+2\sin x\cos x=3\\1+\sin 2x=3\\\sin 2x=2$$ which has no solution (in the reals) because the values of sine only range from -1 to 1.

Answer 2

First, I think there is much more to trigonometry overall, than just this sort of problem (which is on a little bit the trickier side). If you look at a book, there will be plenty of trig that is much more foundational. So I think just saying "trig courses are different in 'murica...oooh la la" is not the right overall assessment, given the probable majority of content that's the same.

My copy of Schaum's precalculus says there's no hard and fast rule on how to solve different trig equations, but advocates trying the following three approaches:

A. Factoring (if already possible) into equation equal to zero.

B. Converting into a single function (e.g. getting rid of some cossq to have all sinsq in an equation).

C. Squaring both sides. They have the specific example sinx + cosx = 1, which is almost exactly yours (but ends up easy to factor). You "throw" cosx functions "onto the other side" and then square both sides, then replace sinsq with 1- cossq. Then group terms and you have a quadratic eqn in cosx. Really, it's not that different than if you substituted the radical sq(1-sinsqx) immediately for cosx and then cleaned it up by squaring after.

P.s. I kind of get the impression method 1 ends up the same as what Schaum's talked about. Or those (unlinked) videos you mentioned. I do think the methods 2 and 3 you showed are interesting. But unless you're deep in a strong trig course, you're not going to be jamming that sort of stuff all the time. Yeah...good to have it at the time. But as memory fades, not something you use all the time in higher courses.

P.s.s. I solved the problem using the Schaum's method and the quadratic method. Ans: x =arccos((-sqrt(3)+/-i)/2). Whatever the heck that means. I'm assuming the complex argument results for same reason as Daly not finding any answer. But if you'd given a different constant and the solution was in the reals, the method would have gotten there fine. And I didn't find it at all laborious to use. Maybe seven lines (and I like to show all the steps).