All posts tagged double angle trig identities

Engaging students: Deriving the double angle formulas for sine, cosine, and tangent

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Daniel Adkins. His topic, from Precalculus: deriving the double angle formulas for sine, cosine, and tangent.

How does this topic extend what your students should have already learned?

A major factor that simplifies deriving the double angle formulas is recalling the trigonometric identities that help students “skip steps.” This is true especially for the Sum formulas, so a brief review of these formulas in any fashion would help students possibly derive the equations on their own in some cases. Listed below are the formulas that can lead directly to the double angle formulas.

A list of the formulas that students can benefit from recalling:

Sum Formulas:
- sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
- cos(a+b) = cos(a)cos(b) – sin(a)sin(b)
- tan(a+b) = [tan(a) +tan(b)] / [1-tan(a)tan(b)]

Pythagorean Identity:
- Sin² (a) + Cos²(a) = 1

This leads to the next topic, an activity for students to attempt the equation on their own.

How could you as a teacher create an activity or project that involves your topic?

I’m a firm believer that the more often a student can learn something of their own accord, the better off they are. Providing the skeletal structure of the proofs for the double angle formulas of sine, cosine, and tangent might be enough to help students reach the formulas themselves. The major benefit of this is that, even though these are simple proofs, they have a lot of variance on how they may be presented to students and how “hands on” the activity can be.

I have an example worksheet demonstrating this with the first two double angle formulas attached below. This is in extremely hands on format that can be given to students with the formulas needed in the top right corner and the general position where these should be inserted. If needed the instructor could take this a step further and have the different Pythagorean Identities already listed out (I.e. Cos²(a) = 1 – Sin²(a), Sin²(a) = 1 – Cos²(a)) to emphasize that different formats could be needed. This is an extreme that wouldn’t take students any time to reach the conclusions desired. Of course a lot of this information could be dropped to increase the effort needed to reach the conclusion.

A major benefit with this also is that even though they’re simple, students will still feel extremely rewarded from succeeding on this paper on their own, and thus would be more intrinsically motivated towards learning trig identities.

How can Technology be used to effectively engage students with this topic?

When it comes to technology in the classroom, I tend to lean more on the careful side. I know me as a person/instructor, and I know I can get carried away and make a mess of things because there was so much excitement over a new toy to play with. I also know that the technology can often detract from the actual math itself, but when it comes to trigonometry, and basically any form of geometric mathematics, it’s absolutely necessary to have a visual aid, and this is where technology excels.

The Wolfram Company has provided hundreds of widgets for this exact purpose, and below, you’ll find one attached that demonstrates that sin(2a) appears to be equal to its identity 2cos(a)sin(a). This is clearly not a rigorous proof, but it will help students visualize how these formulas interact with each other and how they may be similar. The fact that it isn’t rigorous may even convince students to try to debunk it. If you can make a student just irritated enough that they spend a few minutes trying to find a way to show you that you’re wrong, then you’ve done your job in that you’ve convinced them to try mathematics for a purpose.

After all, at the end of the day, it doesn’t matter how you begin your classroom, or how you engage your students, what matters is that they are engaged, and are willing to learn.

Wolfram does have a free cdf reader for its demonstrations on this website: http://demonstrations.wolfram.com/AVisualProofOfTheDoubleAngleFormulaForSine/

References

How I Impressed My Wife: Index

Some husbands try to impress their wives by lifting extremely heavy objects or other extraordinary feats of physical prowess.

That will never happen in the Quintanilla household in a million years.

But she was impressed that I broke an impasse in her research and resolved a discrepancy between Mathematica 4 and Mathematica 8 by finding the following integral by hand in less than an hour:

$\displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

Yes, I married well indeed.

In this post, I collect the posts that I wrote last summer regarding various ways of computing this integral.

Part 1: Introduction

Part 2a, 2b, 2c, 2d, 2e, 2f: Changing the endpoints of integration, multiplying top and bottom by $\sec^2 x$ , and the substitution $u = \tan x$ .

Part 3a, 3b, 3c, 3d, 3e, 3f, 3g, 3h, 3i: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and the magic substitution $u = \tan \theta/2$ .

Part 4a, 4b, 4c, 4d, 4e, 4f, 4g, 4h: Double-angle trig identity, combination into a single trig function, changing the endpoints of integration, and contour integration using the unit circle

Part 5a, 5b, 5c, 5d, 5e, 5f, 5g, 5h, 5i, 5j: Independence of the parameter $a$ , the magic substitution $u = \tan \theta/2$ , and partial fractions.

Part 6a, 6b, 6c, 6d, 6e, 6f, 6g:Independence of the parameter $a$ , the magic substitution $u = \tan \theta/2$ , and contour integration using the real line and an expanding semicircle.

Part 7: Concluding thoughts… and ways that should work that I haven’t completely figured out yet.

How I Impressed My Wife: Part 5h

Earlier in this series, I gave three different methods of showing that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Using the fact that $Q$ is independent of $a$ , I’ll now give a fourth method.

Since $Q$ is independent of $a$ , I can substitute any convenient value of $a$ that I want without changing the value of $Q$ . As shown in previous posts, substituting $a =0$ yields the following simplification:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + 2 \cdot 0 \cdot \sin x \cos x + (0^2 + b^2) \sin^2 x}$

$= \displaystyle \int_{0}^{2\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\pi}^{\pi} \frac{dx}{\cos^2 x + b^2 \sin^2 x}$

$= \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{u^4 + (4 b^2 - 2) u^2 + 1}$

The four roots of the denominator satisfy

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| \sqrt{b^2 - 1}$

So far, I’ve handled the cases $|b| = 1$ and $|b| > 1$ . In today’s post, I’ll start considering the case $|b| < 1$ .

Factoring the denominator is a bit more complicated if $|b| < 1$ . Using the quadratic equation, we obtain

$u^2 = \displaystyle 1 - 2b^2 \pm 2|b| i \sqrt{1-b^2}$

However, unlike the cases $|b| \ge 1$ , the right-hand side is now a complex number. So, To solve for $u$ , I’ll use DeMoivre’s Theorem and some surprisingly convenient trig identities. Notice that

$(1-2b^2)^2 + (2|b| \sqrt{1-b^2})^2 = 1 - 4b^2 + b^4 + 4b^2 (1 - b^2) = 1 - 4b^2 + b^4 + 4b^2 - b^4 = 1$ .

Therefore, the four complex roots of the denominator satisfy $|u^2| = 1$ , or $|u| = 1$ . This means that all four roots can be written in trigonometric form so that

$u^2 = \cos 2\phi + i \sin 2\phi$ ,

where $2\phi$ is some angle. (I chose the angle to be $2\phi$ instead of $\phi$ for reasons that will become clear shortly.)

I’ll begin with solving

$u^2 = \displaystyle 1 - 2b^2 + 2|b| i \sqrt{1-b^2}$ .

Matching the real and imaginary parts, we see that

$\cos 2\phi = 1-2b^2$ ,

$\sin 2\phi = 2|b| \sqrt{1-b^2}$

This completely matches the form of the double-angle trig identities

$\cos 2\phi = 1 - 2\sin^2 \phi$ ,

$\sin 2 \phi = 2 \sin\phi \cos \phi$ ,

and so the problem reduces to solving

$u^2 = \cos 2\phi + i \sin 2\phi$ ,

where $\sin \phi = |b|$ and $\cos \phi = \sqrt{1-b^2}$. By De Moivre’s Theorem, I can conclude that the two solutions of this equation are

$u = \pm(\cos \phi + i \sin \phi)$ ,

$u = \pm( \sqrt{1-b^2} + i |b|)$ .

I could re-run this argument to solve $u^2 = \displaystyle 1 - 2b^2 - 2|b| i \sqrt{1-b^2}$ and get the other two complex roots. However, by the Conjugate Root Theorem, I know that the four complex roots of the denominator $u^4 + (4 b^2 - 2) u^2 + 1$ must come in conjugate pairs. Therefore, the four complex roots are

$u = \pm \sqrt{1-b^2} \pm i |b|$ .

Therefore, I can factor the denominator as follows:

$u^4 + (4 b^2 - 2) u^2 + 1 = (u - [\sqrt{1-b^2} + i|b|])(u - [\sqrt{1-b^2} - i|b|])$

$\qquad \times (u - [-\sqrt{1-b^2} + i|b|])(u - [-\sqrt{1-b^2} + i|b|])$

$= (u - \sqrt{1-b^2} - i|b|)(u - \sqrt{1-b^2} + i|b])$

$\qquad \times (u +\sqrt{1-b^2} + i|b|)(u +\sqrt{1-b^2} + i|b|)$

$= ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)$

To double-check my work, I can directly multiply this product:

$([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)$

$= (u^2 - 2u \sqrt{1-b^2} + 1 - b^2 + b^2) (u^2 + 2u \sqrt{1-b^2} + 1 - b^2 + b^2)$

$= ([u^2 +1] - 2u\sqrt{1-b^2})([u^2+1] + 2u\sqrt{1-b^2})$

$= [u^2+1]^2 - [2u\sqrt{1-b^2}]^2$

$= u^4 + 2u^2 + 1 - 4u^2 (1-b^2)$

$= u^4 + u^2 (2 - 4[1-b^2]) + 1$

$= u^4 + u^2 (4b^2 - 2) + 1$ .

So, at last, I can rewrite the integral $Q$ as

$Q = \displaystyle \int_{-\infty}^{\infty} \frac{ 2(1+u^2) du}{ ([u - \sqrt{1-b^2}]^2 +b^2)([u + \sqrt{1-b^2}]^2 +b^2)}$

I’ll continue with this fourth evaluation of the integral, continuing the case $|b| < 1$ , in tomorrow’s post.

1 Comment

by John Quintanilla on August 20, 2015 • Permalink

Posted in Calculus, Precalculus

Tagged complex numbers, cosine, De Moivre's Theorem, double angle trig identities, integral, polar coordinates, sine, square root, trigonometry

Posted by John Quintanilla on August 20, 2015

https://meangreenmath.com/2015/08/20/how-i-impressed-my-wife-part-5h/

How I Impressed My Wife: Part 5b

Amazingly, the integral below has a simple solution:

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

Even more amazingly, the integral $Q$ ultimately does not depend on the parameter $a$ . For several hours, I tried to figure out a way to demonstrate that $Q$ is independent of $a$ , but I couldn’t figure out a way to do this without substantially simplifying the integral, but I’ve been unable to do so (at least so far).

So here’s what I have been able to develop to prove that $Q$ is independent of $a$ without directly computing the integral $Q$ .

Earlier in this series, I showed that

$Q = 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= 2 \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{1 + 2 a \tan x + (a^2 + b^2) \tan^2 x}$

$= 2 \displaystyle \int_{-\infty}^{\infty} \frac{du}{1 + 2 a u + (a^2+b^2) u^2}$

$= \displaystyle \frac{2}{a^2+b^2} \int_{-\infty}^{\infty} \frac{dv}{v^2 + \displaystyle \frac{b^2}{(a^2+b^2)^2} }$

$= \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+ b^2) dv}{(a^2 + b^2) v^2 + b^2 }$

Yesterday, I showed used the substitution $w = (a^2 + b^2) v$ to show that $Q$ was independent of $a$ . Today, I’ll use a different method to establish the same result. Let

$Q(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{(a^2+b^2) dv}{(a^2+b^2)^2 v^2 + b^2 }$ .

Notice that I’ve written this integral as a function of the parameter $a$ . I will demonstrate that $Q'(a) = 0$ , so that $Q(c)$ is a constant with respect to $a$ . In other words, $Q(a)$ does not depend on $a$ .

To do this, I differentiate under the integral sign with respect to $a$ (as opposed to $x$ ) using the Quotient Rule:

$Q'(a) = \displaystyle 2 \int_{-\infty}^{\infty} \frac{ 2a \left[ (a^2+b^2)^2 v^2 + b^2\right] - 2 (a^2+b^2) \cdot (a^2+b^2) v^2 \cdot 2a }{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$

$Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{(a^2+b^2)^2 v^2 + b^2- 2 (a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$

$Q'(a) = \displaystyle 4a \int_{-\infty}^{\infty} \frac{b^2-(a^2+b^2)^2 v^2}{\left[ (a^2+b^2)^2 v^2 + b^2 \right]^2} dv$

I now apply the trigonometric substitution $v = \displaystyle \frac{b}{a^2+b^2} \tan \theta$ , so that

$(a^2+b^2)^2 v^2 = (a^2+b^2)^2 \displaystyle \left[ \frac{b}{a^2+b^2} \tan \theta \right]^2 = b^2 \tan^2 \theta$

and

$dv = \displaystyle \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta$

The endpoints of integration change from $-\infty < v < \infty$ to $-\pi/2 < \theta < \pi/2$ , and so

$Q'(a) = \displaystyle 4a \int_{-\pi/2}^{\pi/2} \frac{b^2- b^2 \tan^2 \theta}{\left[ b^2 \tan^2 \theta + b^2 \right]^2} \frac{b}{a^2+b^2} \sec^2 \theta \, d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta] \sec^2 \theta}{\left[ \tan^2 \theta +1 \right]^2} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\left[ \sec^2 \theta \right]^2} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1-\tan^2 \theta] \sec^2 \theta}{\sec^4 \theta} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \frac{[1- \tan^2 \theta]}{\sec^2 \theta} d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [1- \tan^2 \theta] \cos^2 \theta \, d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} [\cos^2 \theta -\sin^2 \theta] d\theta$

$= \displaystyle \frac{4ab^3}{a^2+b^2} \int_{-\pi/2}^{\pi/2} \cos 2\theta \, d\theta$

$= \displaystyle \left[ \frac{2ab^3}{a^2+b^2} \sin 2\theta \right]^{\pi/2}_{-\pi/2}$

$= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ \sin \pi - \sin (-\pi) \right]$

$= \displaystyle \frac{2ab^3}{a^2+b^2} \left[ 0- 0 \right]$

$= 0$ .

I’m not completely thrilled with this demonstration that $Q$ is independent of $a$ , mostly because I had to do so much simplification of the integral $Q$ to get this result. As I mentioned in yesterday’s post, I’d love to figure out a way to directly start with

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

and demonstrate that $Q$ is independent of $a$ , perhaps by differentiating $Q$ with respect to $a$ and demonstrating that the resulting integral must be equal to 0. However, despite several hours of trying, I’ve not been able to establish this result without simplifying $Q$ first.

1 Comment

by John Quintanilla on August 13, 2015 • Permalink

Posted in Calculus, Precalculus

Tagged derivative, double angle trig identities, integral, Pythagorean trig identities, ratio trig identities, reciprocal trig identities, secant, tangent

Posted by John Quintanilla on August 13, 2015

https://meangreenmath.com/2015/08/13/how-i-impressed-my-wife-part-5b-2/

How I Impressed My Wife: Part 3f

Previously in this series, I showed that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x} = \displaystyle \frac{2\pi}{|b|}.$

My wife had asked me to compute this integral by hand because Mathematica 4 and Mathematica 8 gave different answers. At the time, I eventually obtained the solution by multiplying the top and bottom of the integrand by $\sec^2 x$ and then employing the substitution $u = \tan x$ (after using trig identities to adjust the limits of integration).

But this wasn’t the only method I tried. Indeed, I tried two or three different methods before deciding they were too messy and trying something different. So, for the rest of this series, I’d like to explore different ways that the above integral can be computed.

So far, I have shown that

$Q = \displaystyle \int_0^{2\pi} \frac{dx}{\cos^2 x + 2 a \sin x \cos x + (a^2 + b^2) \sin^2 x}$

$= \displaystyle \int_0^{2\pi} \frac{2 \, dx}{1+\cos 2x + 2 a \sin 2x + (a^2 + b^2)(1-\cos 2x)}$

$= 2 \displaystyle \int_0^{2\pi} \frac{d\theta}{(1+a^2+b^2) + 2 a \sin \theta + (1 - a^2 - b^2) \cos \theta}$

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\theta}{S + R \cos (\theta - \alpha)}$

$= 2 \displaystyle \int_{0}^{2\pi} \frac{d\phi}{S + R \cos \phi}$

$= 2 \displaystyle \int_{-\pi}^{\pi} \frac{d\phi}{S + R \cos \phi}$

where $R = \sqrt{(2a)^2 + (1-a^2-b^2)^2}$ and $S = 1 + a^2 + b^2$ (and $\alpha$ is a certain angle that is now irrelevant at this point in the calculation).

There are actually a couple of ways for computing this last integral. Today, I’ll lay the foundation for the “magic substitution”

$u = \tan \displaystyle \frac{\phi}{2}$

With this substitution, the above integral will become a rational function, which can then be found using standard techniques.

First, we use some trig identities to rewrite $\cos 2x$ in terms of $\tan x$ :

$\cos 2x = 2\cos^2 x - 1$

$= \displaystyle \frac{ \sec^2 x (2 \cos^2 x - 1)}{\sec^2 x}$

$= \displaystyle \frac{ 2 - \sec^2 x)}{\sec^2 x}$

$= \displaystyle \frac{ 2 - [ 1 + \tan^2 x])}{1 + \tan^2 x}$

$= \displaystyle \frac{1- \tan^2 x}{1 + \tan^2 x}$

Next, I’ll replace $x$ by $\phi/2$ :

$\cos \phi = \displaystyle \frac{1- \tan^2 (\phi/2)}{1 + \tan^2 (\phi/2)} = \displaystyle \frac{1-u^2}{1+u^2}$ .

Second, for the sake of completeness (even though it isn’t necessary for this particular integral), I’ll rewrite $\sin 2x$ in terms of $\tan x$ :

$\sin 2x = 2\sin x \cos x$

$= \displaystyle \frac{2\sin x \cos x \sec^2 x}{\sec^2 x}$

$= \displaystyle \frac{ ~ \displaystyle \frac{2 \sin x}{\cos x} ~ }{\sec^2 x}$

$= \displaystyle \frac{ 2 \tan x }{\sec^2 x}$

$= \displaystyle \frac{ 2 \tan x }{1 + \tan^2 x}$

$= \displaystyle \frac{2 \tan x}{1 + \tan^2 x}$

Next, I’ll replace $x$ by $\phi/2$ :

$\sin \phi = \displaystyle \frac{2 \tan^2 (\phi/2)}{1 + \tan^2 (\phi/2)} = \displaystyle \frac{2u}{1+u^2}$ .

Third, again for the sake of completeness,

$\tan \phi = \displaystyle \frac{\sin u}{\cos u} = \displaystyle \frac{ ~ \displaystyle \frac{2u}{1+u^2} ~ }{ ~ \displaystyle \frac{1-u^2}{1+u^2} ~ } = \displaystyle \frac{2u}{1-u^2}$ .

Finally, I need to worry about what happens to the $d\phi$ :

$u = \tan \displaystyle \frac{\phi}{2}$

$du = \displaystyle \frac{1}{2} \sec^2 \displaystyle \frac{\phi}{2} \, d\phi$

$du = \displaystyle \frac{1}{2} \left[ 1 + \tan^2 \displaystyle \frac{\phi}{2} \right] d\phi$

$du = \displaystyle \frac{1}{2} (1+u^2) d\phi$

$\displaystyle \frac{2 du}{1+u^2} = d\phi$

These four substitutions can be used to convert trigonometric integrals into some other integral. Usually, the new integrand is pretty messy, and so these substitutions should only be used sparingly, as a last resort.

I’ll continue this different method of evaluating this integral in tomorrow’s post.

1 Comment

by John Quintanilla on July 31, 2015 • Permalink

Posted in Calculus, Precalculus

Tagged cosine, double angle trig identities, integral, Pythagorean trig identities, ratio trig identities, reciprocal trig identities, trigonometry

Posted by John Quintanilla on July 31, 2015

https://meangreenmath.com/2015/07/31/how-i-impressed-my-wife-part-3e/

Two ways of doing an integral (Part 2)

A colleague placed the following problem on an exam, expecting the following solution:

$\displaystyle \int \frac{dx}{\sqrt{4x-x^2}} = \sin^{-1} \left( \displaystyle \frac{x-2}{2} \right) + C$

However, one student produced the following solution (see yesterday’s post for details):

$\displaystyle \int \frac{dx}{\sqrt{4x-x^2}} = 2 \sin^{-1} \left( \displaystyle \frac{\sqrt{x}}{2} \right) + C$

As he couldn’t find a mistake in the student’s work, he assumed that the two expressions were equivalent. Indeed, he differentiated the student’s work to make sure it was right. But he couldn’t immediately see, using elementary reasoning, why they were equivalent. So he walked across the hall to my office to ask me if I could help.

Here’s how I showed they are equivalent.

Let $\alpha = \displaystyle \sin^{-1} \left( \frac{x-2}{2} \right)$ and $\beta = \sin^{-1} \left( \displaystyle \frac{\sqrt{x}}{2} \right)$ . Then

$\displaystyle \sin(\alpha - 2\beta) = \sin \alpha \cos 2\beta - \cos \alpha \sin 2\beta$ .

Let’s evaluate the four expressions on the right-hand side.

First, $\sin \alpha$ is clearly equal to $\displaystyle \frac{x-2}{2}$ .

Second, $\cos 2\beta = 1 - 2 \sin^2 \beta$ , so that

$\cos 2\beta = \displaystyle 1 - 2\left( \frac{\sqrt{x}}{2} \right)^2 = \displaystyle 1 - \frac{x}{2} = \displaystyle -\frac{x-2}{2}$ .

Third, to evaluate $\cos \alpha$, I’ll use the identity $\displaystyle \cos \left( \sin^{-1} x \right) = \sqrt{1 - x^2}$ :

$\cos \alpha = \displaystyle \sqrt{1 - \left( \frac{x-2}{2} \right)^2 } = \displaystyle \frac{\sqrt{4x-x^2}}{2}$

Fourth, $\sin 2\beta = 2 \sin \beta \cos \beta$ . Using the above identity again, we find

$\sin 2\beta = \displaystyle 2 \left( \frac{ \sqrt{x} }{2} \right) \sqrt{ 1 - \left( \frac{ \sqrt{x} }{2} \right)^2 }$

$= \sqrt{x} \sqrt{1 - \displaystyle \frac{x}{4}}$

$= \displaystyle \frac{\sqrt{4x-x^2}}{2}$

Combining the above, we find

$\sin(\alpha - 2 \beta) = \displaystyle -\left( \frac{x-2}{2} \right)^2 - \left( \frac{\sqrt{4x-x^2}}{2} \right)^2$

$\sin(\alpha - 2 \beta) = \displaystyle \frac{-(x^2 - 4x + 4) - (4x - x^2)}{4}$

$\sin(\alpha - 2 \beta) = -1$

$\alpha - 2 \beta = \displaystyle -\frac{\pi}{2} + 2\pi n$ for some integer $n$

Also, since $-\pi/2 \le \alpha \le \pi/2$ and $0 \le -2\beta \le \pi$ , we see that $-\pi/2 \le \alpha - 2 \beta \le 3\pi/2$ . (From its definition, $\beta$ is the arcsine of a positive number and therefore must be nonnegative.) Therefore, $\alpha - 2\beta = -\pi/2$ .

In other words,

$\sin^{-1} \left( \displaystyle \frac{x-2}{2} \right)$ and $2 \sin^{-1} \left( \displaystyle \frac{\sqrt{x}}{2} \right)$

differ by a constant, thus showing that the two antiderivatives are equivalent.

1 Comment

by John Quintanilla on June 10, 2014 • Permalink

Posted in Calculus

Tagged completing the square, double angle trig identities, integral, sum and difference trig identities, trigonometry

Posted by John Quintanilla on June 10, 2014

https://meangreenmath.com/2014/06/10/two-ways-of-doing-an-integral-part-2/

Why do we teach students about radians?

Throughout grades K-10, students are slowly introduced to the concept of angles. They are told that there are $90$ degrees in a right angle, $180$ degrees in a straight angle, and a circle has $60$ degrees. They are introduced to $30-60-90$ and $45-45-90$ right triangles. Fans of snowboarding even know the multiples of $180$ degrees up to $1440$ or even $1620$ degrees.

Then, in Precalculus, we make students get comfortable with $\pi$ , $\displaystyle \frac{\pi}{2}$ , $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{\pi}{4}$ , $\displaystyle \frac{\pi}{6}$ , and multiples thereof.

We tell students that radians and degrees are just two ways of measuring angles, just like inches and centimeters are two ways of measuring the length of a line segment.

Still, students are extremely comfortable with measuring angles in degrees. They can easily visualize an angle of $75^o$ , but to visualize an angle of $2$ radians, they inevitably need to convert to degrees first. In his book Surely You’re Joking, Mr. Feynman!, Nobel-Prize laureate Richard P. Feynman described himself as a boy:

I was never any good in sports. I was always terrified if a tennis ball would come over the fence and land near me, because I never could get it over the fence – it usually went about a radian off of where it was supposed to go.

Naturally, students wonder why we make them get comfortable with measuring angles with radians.

The short answer, appropriate for Precalculus students: Certain formulas are a little easier to write with radians as opposed to degrees, which in turn make certain formulas in calculus a lot easier.

The longer answer, which Precalculus students would not appreciate, is that radian measure is needed to make the derivatives of $\sin x$ and $\cos x$ look palatable.

Source: http://mathworld.wolfram.com/CircularSector.html

1. In Precalculus, the length of a circle arc with central angle $\theta$ in a circle with radius $r$ is

$s = r\theta$

Also, the area of a circular sector with central angle $\theta$ in a circle with radius $r$ is

$A = \displaystyle \frac{1}{2} r^2 \theta$

In both of these formulas, the angle $\theta$ must be measured in radians.

Students may complain that it’d be easy to make a formula of $\theta$ is measured in degrees, and they’d be right:

$s = \displaystyle \frac{180 r \theta}{\pi}$ and $A = \displaystyle \frac{180}{\pi} r^2 \theta$

However, getting rid of the $180/\pi$ makes the following computations from calculus a lot easier.

2a. Early in calculus, the limit

$\displaystyle \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

is derived using the Sandwich Theorem (or Pinching Theorem or Squeeze Theorem). I won’t reinvent the wheel by writing out the proof, but it can be found here. The first step of the proof uses the formula for the above formula for the area of a circular sector.

2b. Using the trigonometric identity $\cos 2x = 1 - 2 \sin^2 x$ , we replace $x$ by $\theta/2$ to find

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = \displaystyle \lim_{\theta \to 0} \frac{2\sin^2 \displaystyle \left( \frac{\theta}{2} \right)}{ \theta}$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = \displaystyle \lim_{\theta \to 0} \sin \left( \frac{\theta}{2} \right) \cdot \frac{\sin \displaystyle \left( \frac{\theta}{2} \right)}{ \displaystyle \frac{\theta}{2}}$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} =0 \cdot 1$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} =0$

3. Both of the above limits — as well as the formulas for $\sin(\alpha + \beta)$ and $\cos(\alpha + \beta)$ — are needed to prove that $\displaystyle \frac{d}{dx} \sin x = \cos x$ and $\displaystyle \frac{d}{dx} \cos x = -\sin x$ . Again, I won’t reinvent the wheel, but the proofs can be found here.

So, to make a long story short, radians are used to make the derivatives $y = \sin x$ and $y = \cos x$ easier to remember. It is logically possible to differentiate these functions using degrees instead of radians — see http://www.math.ubc.ca/~feldman/m100/sinUnits.pdf. However, possible is not the same thing as preferable, as calculus is a whole lot easier without these extra factors of $\pi/180$ floating around.

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