# Let s=(t)=\sin t, where ''t'' is in degrees, and let f(x)=\sin x, where ''x'' is in radians. (a)...

## Question:

Let {eq}s=(t)=\sin t, {/eq} where t is in degrees, and let {eq}f(x)=\sin x, {/eq} where x is in radians.

(a) Draw the graphs of s and f in the same figure, with domain (0, 6) on the horizontal axis. What is the difference in the graphs? Is one graph easier to analyze than the other?

(b) Recall that t degrees {eq}=\frac{\pi }{180}t {/eq} radians. It follows that s(t) in degrees{eq}=\sin \left ( \frac{\pi }{180}t \right) {/eq} in radians. Find both s(t) and f (x), using the derivative of the sine function in the book and the Chain Rule where necessary. Would s be easier to use than f , or vice versa? Why?

(c) Let {eq}c(t)=\cos t, {/eq} where t is in degrees, and let {eq}g(x)=\cos x, {/eq} where x is in radians. Using ideas analogous to those in (b), find c (t) and g (x). Which would be easier to work with

It is important to remember that degrees are arbitrary units, and because they are arbitrary, ultimately meaningless in equations unless we do something to make them meaningful. Radians, in contrast, are meaningful: they equate angle measure to the arc length of the unit circle, making them into sort of scale factors that simply by their very nature work in equations just as they are.

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Part A

Let's look at the plots on the same graph. Here {eq}s (t) = \sin t {/eq} (degree measure) is green, and {eq}f (x) = \sin x {/eq} (radians)...