How Can I Evaluate Trig Functions Without a Calculator?

[mathdad]

I am in the trigonometry section of my precalculus textbook by David Cohen. In Section 6.2, David explains how to evaluate trig functions without using a calculator but it is not clear to me.

Sample:

Is cos 3 positive or negative?

How do I determine if cos 3 is positive or negative without using a calculator?

Is the unit circle needed in this case?

 

[tkhunny]

I am in the trigonometry section of my precalculus textbook by David Cohen. In Section 6.2, David explains how to evaluate trig functions without using a calculator but it is not clear to me.

Sample:

Is cos 3 positive or negative?

How do I determine if cos 3 is positive or negative without using a calculator?

Is the unit circle needed in this case?

I don't see any "evaluate" in your examples. All I see is a suggestion that we can gain SOMETHING rather easily.

3 is pretty close to $pi$. It's a little less. Is cosine positive or negative over on the $pi$ side of the Unit Circle?

 

[mathdad]

Cosine is negative in quadrant 2.

So, cos 3 is negative.

- - - Updated - - -

What if the question is EVALUATE cos 3 without using a calculator? Do we need the unit circle?

 

[MarkFL]

...What if the question is EVALUATE cos 3 without using a calculator? Do we need the unit circle?

I would likely use a Maclaurin series to approximate $\cos(3)$ to the desired accuracy, if asked to do so by hand.

\(\displaystyle \cos(x)=\sum_{k=0}^{\infty}\left(\frac{(-1)^k}{(2k)!}x^{2k}\right)\)

Thus:

\(\displaystyle \cos(3)=\sum_{k=0}^{\infty}\left(\frac{(-1)^k}{(2k)!}9^{k}\right)\)

If we use the first 5 terms, we obtain:

\(\displaystyle \cos(3)\approx1-\frac{9}{2}+\frac{81}{24}-\frac{729}{720}+\frac{6561}{40320}=-\frac{4367}{4480}\approx-0.9748\)

According to W|A, we have:

\(\displaystyle \cos(3)\approx-0.9899924966004454\)

 

[mathdad]

I would likely use a Maclaurin series to approximate $\cos(3)$ to the desired accuracy, if asked to do so by hand.

\(\displaystyle \cos(x)=\sum_{k=0}^{\infty}\left(\frac{(-1)^k}{(2k)!}x^{2k}\right)\)

Thus:

\(\displaystyle \cos(3)=\sum_{k=0}^{\infty}\left(\frac{(-1)^k}{(2k)!}9^{k}\right)\)

If we use the first 5 terms, we obtain:

\(\displaystyle \cos(3)\approx1-\frac{9}{2}+\frac{81}{24}-\frac{729}{720}+\frac{6561}{40320}=-\frac{4367}{4480}\approx-0.9748\)

According to W|A, we have:

\(\displaystyle \cos(3)\approx-0.9899924966004454\)

MarkFL,

You are a true mathematician. I am far from your level of understanding mathematics. I am not familiar with the Maclaurin series. It looks very interesting but I am not there yet. Is my reply concerning cos 3 being negative correct, good or bad?

 

[MarkFL]

MarkFL,

You are a true mathematician. I am far from your level of understanding mathematics. I am not familiar with the Maclaurin series. It looks very interesting but I am not there yet. Is my reply concerning cos 3 being negative correct, good or bad?

Yes, 3 radians is in quadrant II, and so the cosine of that angle is negative. As stated 3, is close to $\pi$, and so we should expect $\cos(3)$ to be close to -1. :)

 

[mathdad]

When I get home tonight, I will work on selected questions from Section 6.2. If I get stuck, like always, I will come back here with my questions, and my math work seeking help. Thank you so much for your help, and your patience as I travel through the wonderful David Cohen textbook.