# Differentiation Of Trigonometric Functions Homework

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See the Proof of Trig Limits section of the Extras chapter to see the proof of these two limits. Students often ask why we always use radians in a Calculus class. The proof of the formula involving sine above requires the angles to be in radians.

If the angles are in degrees the limit involving sine is not 1 and so the formulas we will derive below would also change.

When doing a change of variables in a limit we need to change all the $$x$$’s into $$\theta$$’s and that includes the one in the limit.

Doing the change of variables on this limit gives, $\begin\mathop \limits_ \frac & = 6\mathop \limits_ \frac\hspace\theta = 6x\\ & = 6\mathop \limits_ \frac\\ & = 6\left( 1 \right)\\ & = 6\hspace\end$ And there we are.

Therefore, after doing the change of variable the limit becomes, $\mathop \limits_ \frac = \mathop \limits_ \frac = 1$ The previous parts of this example all used the sine portion of the fact.

However, we could just have easily used the cosine portion so here is a quick example using the cosine portion to illustrate this.

A change of variables, in this case, is really only needed to make it clear that the fact does work.

In this case we appear to have a small problem in that the function we’re taking the limit of here is upside down compared to that in the fact.

$\mathop \limits_ \frac = \mathop \limits_ \frac = 6\mathop \limits_ \frac$ Note that we factored the 6 in the numerator out of the limit.

At this point, while it may not look like it, we can use the fact above to finish the limit.

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It may not be obvious, but this problem can be viewed as a differentiation problem. Recall that. If, then, and letting it follows that. Click HERE to return to the list of problems. SOLUTION 9 Differentiate. Apply the chain rule to both functions. If necessary, review the section on the chain rule. Then Recall that.…