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In the following problem: $$F(x) = \int_x^{x+2} \sin t \ dt$$ we are required to find $x$ when $F(x)$ is a maximum. If we differentiate and equate to zero, we will get the result:

$$\sin x = \frac{\sin 2}{\sqrt{2(1–\cos 2)}}$$

Now, I didn't exactly know whether the angles of $\sin 2$ and $\cos 2$ were degrees or radians, because the result was obtained by a substitution of a length function. But neither are radians, are a length quantity (because they're dimensionless), nor degrees are a length quantity. So I just tried evaluating it assuming it was degrees, which gave me $x = 89$. This was exactly what I expected graphically. (Assuming $89$ is in degrees). When I then tried evaluating the result using radians, I got approximately $0.5707$, which is again exactly what I was expecting to get, for:

$$0.5707 = \frac{\pi}{2} – 1 \ .$$

Does it then not matter, whatever way I evaluate a trigonometric expression (either using radians or degrees)?

Why is this so?

Thank you in advance.

Brendan Darrer
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Camelot823
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    In mathematics (as opposed to geography) it is usual to assume $\sin(x)$ in radians unless otherwise stated: you then get simple results such as $\frac{d}{dx} \sin(x)=\cos(x)$. Similarly in mathematics (as opposed to computer science) it is usual to assume $\log(x)$ is base $e$ unless otherwise stated: you then get simple results such as $\frac{d}{dx} \log(x) = \frac1x$. – Henry Aug 03 '23 at 13:58
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    @Henry however, why did both give the correct result? Was this expected? – Camelot823 Aug 03 '23 at 14:02
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    To be clear and to emphasize Henry's point, $\frac{d}{dx}\sin_\circ(x)$ where $\sin_\circ$ is the version of $\sin$ who takes its arguments in degrees and $x$ is in degrees is not equal to $\cos_\circ(x)$, but is rather equal to $\frac{\pi}{180}\cos_\circ(x)$. See here. Integration is similarly altered. – JMoravitz Aug 03 '23 at 14:11
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    Arguments to trig functions should be pure numbers. Some programming languages or calculators may take degrees but those functions do the conversion for you. – John Douma Aug 03 '23 at 14:11
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    Wait, now I'm more confused. Shouldn't $\frac{\pi}{2} – 1$ be $89°$? I'm having such a confusion episode right now. I believe I have mistooken the result with $\frac{\pi}{2} rad – 1°$ (Am I correct?). However, I expected the result must be $89°$ if you think of what the problem means geometrically. $(\frac{\pi}{2} – 1) rad ≈ 32.70°$ – Camelot823 Aug 03 '23 at 14:13
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    @Camelot823 No. You can't convert $\frac{\pi}{2}$ to degrees and then subtract $1$. You must convert $\frac{\pi}{2}-1$. – John Douma Aug 03 '23 at 14:21

1 Answers1

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You are essentially asking why they both gave you "a right angle minus $1$" despite the ambiguity of "$1$".

It is essentially a "half-angle formula": $$\cos \frac{y}2 = \frac{\sin y}{\sqrt{2(1–\cos y)}}$$ with $y=2$ , so you get: $$\cos 1 = \frac{\sin 2}{\sqrt{2(1–\cos 2)}} \ .$$

Because you are taking cosines and sines, the identity works in positive radians and degrees up towards four right angles. You do not want $\cos y =1$.

Then you change the first $\cos$ to $\sin$ so instead of the cosine of $1$ whatever, you have the sine of the complementary angle, a right angle minus $1$ whatever. Since the measurement of a right angle changes but you are still subtracting $1$, so too does the result.

You really had two different equations to solve:

  • In radians: with $y=2$ radians in the half angle formula (an obtuse angle), $\sin x = \frac{\sin 2}{\sqrt{2(1–\cos 2)}}$ has the solution $x=\frac\pi 2 - 1$
  • In degrees: with $y=2^\circ$ in the half angle formula (a very acute angle), $\sin x^\circ = \frac{\sin 2^\circ}{\sqrt{2(1–\cos 2^\circ)}}$ has the solution $x=90^\circ - 1^\circ=89^\circ$

The two $y$s are different angles here, so the equations are different and it is no surprise that the two solutions for the $x$s are also different angles.

Henry
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    I think I understand it now. If I am correct, a trigonometric identity then works for both radians and degrees, however, this is only not the case with trig identities pertaining derivatives and integrals. So it is best to make sure the angle is converted into radians before operating with those restricted identities. In this question, though, no trig identities pertaining derivates or integrals were used. The last identity was simply an identity between sines and cosines, hence substituting degrees, or radians must provide the appropriate result. – Camelot823 Aug 03 '23 at 14:37
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    However, one slight problem I'm still having is that: according to your answer in the comments, the correct result to the question is $x =( \frac{\pi}{2} – 1$rad)$≈ 32.70°$. However, since substituting either degrees, or radians should give the appropriate result, we get $x = 89°$ when we use degrees. Which is supposed to be correct. But that does not agree with the previous result. This is my only concern right now. Please help me out. – Camelot823 Aug 03 '23 at 14:41
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    I have added to the answer saying that you get different results because you are essentially solving two different equations – Henry Aug 03 '23 at 14:51
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    I see now how they are regarded as two different equations: but which equation is the one we desire to solve? In the comments, you said that usually radians are assumed. I don't see the objective reason for that (especially since there are no trig-calc identities involved here so we're not simplifying anything). Moreover, if we geometrically interpret the problem, we are being asked to find the value $x$ at which the area of the strip in the interval $[x, x + 2]$ is greatest. Obviously, the greatest value is at $x = 90° ≈ 1.571$ rad. Taking the interval $[89°, 91°]$ must be the greatest. – Camelot823 Aug 03 '23 at 15:06
  • You started with a calculus question, so I would expect this to be based on $\int\limits_a^b \sin(t), dt = \cos(a)-\cos(b)$, i.e. using radians. Otherwise you end up multiplying or dividing by $\pi/180$ – Henry Aug 03 '23 at 15:24
  • Oh, I didn't know that identity relied on the assumption of us using radians. Thank you very much for clearing that out!! – Camelot823 Aug 03 '23 at 15:42
  • But are you sure that the answer is $≈32.70°$? Because it logically seems off. The real question was to find the interval $[x, x+ 2]$ which makes the area under $y = \sin t$ the greatest. Perhaps I modeled the problem incorrectly? – Camelot823 Aug 03 '23 at 15:45
  • @Camelot823 Personally I would give the answer $\frac \pi 2 -1 \approx 0.570796$ and leave it like that – Henry Aug 03 '23 at 17:28
  • I just understood exactly what I was doing wrong. The values of $x$ in $[x, x + 2]$ must be measured by the same unit that the $t$ axis is measured in (because it takes on only those values on the $t$ axis), which is in radians. However, I wasn't seeing this so I assumed that I could impose any meaning on $x$, which I chose to be degrees. So I was expecting $x$ in $[89°, 90°]$. However since $[x, x + 2]$ is in radians, that means that the answer $x ≈ 0.570796$ rad would be in $[32.70°,147.29°]$ if we were to measure in degrees. The result is now logically correct. – Camelot823 Aug 03 '23 at 19:41
  • Thank you very much for your answer, truly, it has helped greatly. – Camelot823 Aug 03 '23 at 19:42