7
$\begingroup$

My precalculus test asked me to determine which was greater: $\tan (53)$ or $\sec (38)$.

I looked at it like this, but it seems so close that it's difficult to imagine that they would ask this:

$\tan (45)$ is 1 and $\tan (60)$ is $\sqrt{3}$, so since 53 is approx between 45 and 60, I took a value in-between $1$ and $1.73\ldots$ say $1.36$

$\dfrac{1}{\cos (30)}$ is about $1.2$, and $\dfrac{1}{\cos (45)}$ is about $1.4.$ Taking a value in-between I chose $1.3$

So, obviously I was correct to choose the tangent value as being larger, but it is actually larger by about $.057$. HOW AM I SUPPOSED TO DO THIS without a calculator?

$\endgroup$
11
  • $\begingroup$ Asking which of the two is bigger doesn't necessarily require actually evaluating them. $\endgroup$ Aug 27 '14 at 21:45
  • $\begingroup$ Can you elaborate on that? $\endgroup$
    – user163862
    Aug 27 '14 at 21:49
  • 4
    $\begingroup$ . . . . . for example, knowing that $\sin22^\circ<\sin23^\circ$ doesn't require evaluating them. $\endgroup$ Aug 27 '14 at 21:49
  • $\begingroup$ Using a product-to-sum formula and rearranging, this inequality turns out to be the same as asking whether $\frac{1}{2}\left(\sin 89^\circ+\sin 17^\circ\right)\gtrless \sin 37^\circ$. That may be easier to analyze... $\endgroup$ Aug 27 '14 at 21:56
  • 2
    $\begingroup$ How about using addition formulas to reduce the problem to estimates for $53 - 38 = 15$ and $53 + 38 = 91$ degrees? It's not hard to evaluate $\sin 15^\circ, \cos 15^\circ$ using double angle formulas, and it should be easy to estimate their values at $91^\circ$. $\endgroup$
    – anomaly
    Aug 27 '14 at 22:29
2
$\begingroup$

Not an answer, but a discussion of the closeness of the situation.

enter image description here

Since $53^\circ$ and $38^\circ$ are very nearly complementary, we have that $\sec 38^\circ \approx \csc 53^\circ$ ... with the left-hand side being ever-so-slightly larger than the right-hand side.

As the first diagram suggests, for big enough (first-quadrant) angles $\theta$, we have that $\tan\theta$ exceeds $\csc\theta$; and, according to that first diagram, $53^\circ$ seems to be one of those "big enough" angles ... but just barely. Is it big enough that the $\tan 53^\circ$ also exceeds the slightly-larger value, $\sec 38^\circ$? Well, the middle diagram confirms that it is (though again: just barely), but of course having a computer program draw an accurate diagram is really no better than using a calculator compute the values.

What makes the approximations especially-tricky here is that $53^\circ$ is very close to the magic (or, should I say, "golden"?) angle, $\theta_\star = 51.8...^\circ$, marking the threshold of those "big enough" angles. If the problem had been to compare, say, $\tan 70^\circ$ with $\sec 21^\circ$, then we would have had more confidence in our ability to fiddle with the numbers.


All things considered, this seems like a bad exercise for a test. I wonder if there was an error in the test question.

$\endgroup$
2
  • $\begingroup$ I so agree that it's a bad question for a test, especially given that this was a no-calculator (obviously), and the actual difference is only .05. I know this teacher is big on concepts. $\endgroup$
    – user163862
    Aug 28 '14 at 18:24
  • $\begingroup$ @user163862: I'm really big on concepts myself, so when you get a chance, please edit your question (or post an answer!) with your teacher's explanation of what techniques you were expected to apply to the problem. If there's a conceptual approach that's able to resolve a 0.058 variation in the values, then I want to know what it is! :) $\endgroup$
    – Blue
    Aug 28 '14 at 18:44
1
$\begingroup$

[All angles are measured in degrees]

The (3 – 4 – 5) right-angled triangle gives the closest approximation of angle equal to 53 degrees (53.1xxxxx degrees to be exact, slightly larger than the required 53 degree angle).

Thus, we construct a right angled triangle of sides (3 + 4d, 4 – 3d, 5); where d is a very small positive increment in the range 50d (at the most) is equal to 1). [The reason of using 4d and 3d will be clear when simplification of the following is performed.]

Then, $\tan 52 = \dfrac {4 – 3d}{3 + 4d}$ and $\sec 38 = \dfrac {5}{4 – 3d}$

$\tan 52 – \sec 38 = …. = \dfrac {1 – 49d + 9d^2}{(3+4d)(4 – 3d)} = ... > 0$ (since d is small as assumed above)

Therefore, $\tan 52 > \sec 38$

Since tan is an increasing function in the range $[0, 90), therefore, \tan 53 > \sec 38$

Remark: Although $(3 + 4d, 4 – 3d, 5)$ is not exactly right-angled, $(3 + 4d, 4 – 3d, \sqrt(25 + 25d^2))$ is. Since d is small enough, we can use that as a close estimate.

$\endgroup$
1
$\begingroup$

Here's one way to do this without protractors and such, with some algebra and basic trigonometric identities. First, notice that the sum and difference of the two angles are 91° and 15°, and the values of trigonometric functions at both these angles are either known or very easy to evaluate. This gives one the idea of transforming the inequality $\tan53° ≶ \sec38°$ in such a way that the awkward angles disappear and the sums and differences of angles come out instead. Since $\tan53°$ and $\sec38°$ are both positive, their squares are related in the same way as the quantities themselves:$$\tan^2 53° ≶ \sec^2 38°,$$or$$\tan^2 53° - \sec^2 38° ≶ 0.$$Transforming the left-hand side with the goal of obtaining sums and differences of the angles,$$LHS=\tan^2 53°-\frac{1}{\cos^2 38°}=\tan^2 53°-(\tan^2 38°+1)=\frac{\sin^2 53°}{\cos^2 53°}-\frac{\sin^2 38°}{\cos^2 38°}-1=$$$$=\frac{\sin^2 53°\cos^2 38°-\sin^2 38°\cos^2 53°}{\cos^2 53°\cos^2 38°}-1=\frac{\sin(53°-38°)\sin(53°+38°)}{((\cos(53°+38°)+\cos(53°-38°))/2)^2}-1=$$ $$=\frac{4\sin15°\sin91°}{(\cos15°+\cos91°)^2}-1$$ This expression is much easier to evaluate, because $\sin91°$ is very close to $1$ and $\cos91°$ is rather close to $0$ (less than 0.02 by absolute value). Neglecting these small quantities for the moment, we find that$$LHS\approx\frac{4\sin15°}{\cos^2 15°}-1=\frac{4\sin15°\cos15°}{\cos^3 15°}-1=\frac{2\sin 30°}{\cos^3 15°}-1=\frac{1}{\cos^3 15°}-1.$$Since the cosine is always smaller than $1$ this quantity is obviously positive, and 15° is a large enough angle that the small quantities we neglected above will not be able to affect the sign of $LHS$. Hence, the correct sign in the original inequality is $>$:$$\tan53°>\sec38°.$$

$\endgroup$
0
$\begingroup$

enter image description here

I think what this all boils down to is that we have to show that $$\cos 54^\circ<\cos 53^\circ<1-\phi\approx \cos 51.8\dots^\circ$$

If we calculate the value of the derivative at $54^\circ$ of the cosine function (differentiating with respect of radians of course) and draw a line whose slope is that value, passing through $(54^\circ,\cos 54^\circ)$, we can use that line to estimate the value of $\cos 53^\circ$. Because of the concavity of the cosine function, we know that the the estimate is greater than the actual value of $\cos 53^\circ$. Thus, if we can show that the estimate is less than $1-\phi$, then the problem is solved.

The actual value of the arccosine of the estimate is approx. $52.99\dots^\circ$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.