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Through out my studies, I had been through a lot of trigonometry and differential equations with trigonometry. Whenever 'pi' appears inside the sin/cos/tan..etc, I assumed it to be 180, and when it appeared outside due to integration or something else, I assumed it to be the irrational number-3.14, sometimes, both at a time. I read a lot about radians and degrees. But Most math problems don't specify if it's radians or degrees. Is it implied that whenever pi appears inside the trigonometric functions it is in degrees and outside in radians? or the other way round? Sorry I'm totally confused. I just wanna have some clarity. Also, calculators seem to take the same 'pi' value inside and outside which is causing ridiculous errors. like sin 180 coming out as sin 22/7*. Please help.

Please give an example: say this: So, if there is an integral which, say x*sin(x), integrated over 0 and pi/5., After the integration is over, how do you apply the limits? Do you use pi=180 inside the -cos and the irrational just number on the plain old x?

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I think your confusion originates from misunderstanding radians and degrees.

$\pi$ is a number. It happens that the circumference of a circle is always $2\pi$ times the radius of the circle. So, in radians, an angle of $\pi$ does correspond to an angle of $180$ degrees.

However, $\pi \neq 180.$ When you see $\pi$ appearing in an integral, it's probably appearing as an area--$\pi$ is the area of the unit circle.

We may define $1^{\circ} = \pi/180$ to define degrees. But that just means 1 degree equals $\pi/180,$ not that $1 = \pi/180.$

Just like $100\text{cm} = 1\text{m}$ doesn't mean $100=1.$ There are different units.

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    $\begingroup$ Very good. Emphasis on $ \pi $ is just a number. $\endgroup$ – Jake Mirra Oct 30 at 20:34
  • $\begingroup$ So, if there is an integral which, say x*sin(x), integrated over 0 and pi/2., After the integration is over, how do you apply the limits? Do you use pi=180 inside the -cos and the irrational just number on the plain old x? $\endgroup$ – Vishwa Mithra Tatta Oct 30 at 20:38
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    $\begingroup$ @VishwaMithraTatta Your calculator is likely finding the value of $\int_{0}^{\pi /2}\sin^\text{o}(x) x \, dx$ rather than $\int_{0}^{\pi /2}\sin^\text{r}(x) x \, dx$. $\sin^\text{o}$ and $\sin^\text{r}$ are different functions. $\endgroup$ – Joe Oct 30 at 21:13
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    $\begingroup$ @VishwaMithraTatta Yes, unfortunately, calculators don't use common sense. When you input $\sin \pi$, they interpret it as $\pi^{\text{o}}$, even though a human would understand from context that you mean $\pi \text{ rad}$. $\endgroup$ – Joe Oct 30 at 21:36
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    $\begingroup$ @VishwaMithraTatta The trouble occurs because there are two functions that use the representation $\sin$. The first one, $\sin^\text{o}(x)$ is the $y$-coordinate when you go $x$ degrees anticlockwise around a unit circle, starting from the point $(0,1)$. $\sin^\text{r}(x)$ is exactly the same, except that you are going $x$ radians around the circle. $\endgroup$ – Joe Oct 30 at 21:40

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