Sum or difference formula (sin, cos, and tan)

Click For Summary

Discussion Overview

The discussion revolves around finding the exact values of sine, cosine, and tangent for the angle ${-13\pi}/{12}$ using sum or difference formulas. The focus is on applying trigonometric identities to simplify the angle into a more manageable form.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant suggests adding $2\pi$ to the angle ${-13\pi}/{12}$ to convert it to a positive angle, resulting in ${11\pi}/{12}$.
  • Another participant proposes expressing ${11\pi}/{12}$ as a sum of angles, specifically ${1\pi}/{2} + {1\pi}/{4} + {1\pi}/{6}$, to utilize the angle-sum formulas.
  • A further suggestion includes alternative angle combinations such as ${3\pi}/{4} + {1\pi}/{6}$ or ${\pi}/{4} + {2\pi}/{3}$ to simplify the calculations without repeated use of compound angle formulas.

Areas of Agreement / Disagreement

Participants appear to agree on the method of converting the angle and expressing it as a sum of known angles, but there are multiple approaches suggested for how to express ${11\pi}/{12}$, indicating a lack of consensus on the best method.

Contextual Notes

There may be limitations regarding the assumptions made about the angles used in the sum, as well as the potential for different interpretations of how to apply the sum or difference formulas effectively.

Who May Find This Useful

Students or individuals studying trigonometry, particularly those looking to understand the application of sum and difference formulas in calculating trigonometric values.

Taryn1
Messages
25
Reaction score
0
So I'm supposed to find the exact values of the sine, cosine, and tangent of an angle by using a sum or difference formula ( i.e. sin(x+y)=sin(x)cos(y)+cos(x)sin(y) ), but this is the angle I was given: ${-13\pi}/{12}$. How do I use a sum or difference formula to get the sin, cos, and tan of that?
 
Last edited:
Physics news on Phys.org
I would first add $2\pi$ to get:

$$-\frac{13}{12}\pi+2\pi=\frac{11}{12}\pi$$

And then write:

$$\frac{11}{12}\pi=\frac{1}{2}\pi+\frac{1}{4}\pi+\frac{1}{6}\pi$$

Now you can use the angle-sum formulas. :)
 
MarkFL said:
I would first add $2\pi$ to get:

$$-\frac{13}{12}\pi+2\pi=\frac{11}{12}\pi$$

And then write:

$$\frac{11}{12}\pi=\frac{1}{2}\pi+\frac{1}{4}\pi+\frac{1}{6}\pi$$

Now you can use the angle-sum formulas. :)

Or even just $\displaystyle \begin{align*} \frac{3\pi}{4} + \frac{\pi}{6} \end{align*}$ or $\displaystyle \begin{align*} \frac{\pi}{4} + \frac{2\pi}{3} \end{align*}$ to avoid multiple uses of the compound angle formulae...
 
Thanks for your help! That makes more sense now.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K