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.

## Comments Differentiation Of Trigonometric Functions Homework

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## Unit 2 - The Trigonometric Functions - Classwork

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## Differentiation of Inverse Trigonometric Functions

Each of the six basic trigonometric functions have corresponding inverse functions when appropriate restrictions are placed on the domain of the original functions. All the inverse trigonometric functions have derivatives, which are summarized as follows…

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## Solutions to Differentiation of Trigonometric Functions

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.…