Why Was The Angle Assumed To Be In Degrees

calculusdefinite integralsintegrationtrigonometric-integralstrigonometry

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.

Best Answer

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.