MHB How Can I Evaluate Trig Functions Without a Calculator?

AI Thread Summary
To evaluate trigonometric functions without a calculator, understanding the unit circle is essential, particularly for determining the sign of functions like cosine. For example, since 3 radians is in the second quadrant, where cosine is negative, cos(3) is negative. Additionally, using a Maclaurin series can provide an approximation for cos(3), yielding a value close to -0.98999. The discussion emphasizes the importance of recognizing the quadrant and suggests that further practice with the textbook will aid in understanding. Engaging with the material and asking questions is encouraged for deeper comprehension.
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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?
 
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RTCNTC said:
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?
 
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?
 
RTCNTC said:
...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.

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

Thus:

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

If we use the first 5 terms, we obtain:

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

According to W|A, we have:

$$\cos(3)\approx-0.9899924966004454$$
 
MarkFL said:
I would likely use a Maclaurin series to approximate $\cos(3)$ to the desired accuracy, if asked to do so by hand.

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

Thus:

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

If we use the first 5 terms, we obtain:

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

According to W|A, we have:

$$\cos(3)\approx-0.9899924966004454$$
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?
 
RTCNTC said:
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. :)
 
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.
 

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