Give the exact value using the Half Angle Formula

In summary, the conversation is about finding the exact value of sin -7pi/8 without a calculator using the half-angle identities formula. The attempt at a solution involves finding the cos value of -7pi/4 and using it in the formula, which leads to the final answer of -sqrt((1/2)-(sqrt(2)/4)).
  • #1
Joe_K
33
0

Homework Statement



Find the exact value, without a calculator using half-angle identities, of sin -7pi/8

Homework Equations



Half-angle formula of sin: sin(x/2) = +/- sqrt (1-cosx)/(2)

The Attempt at a Solution



I am confused as to how to use this formula. So far I have:

sin (-7pi/8) = sqrt (1-cos (-7pi/4))/2
 
Physics news on Phys.org
  • #2
That seems like a good start. You'll have to figure out the +/- by hand by figuring out what quadrant -7pi/8 is in.
 
  • #3
From what I figured, -7pi/8 would be in the 3rd quadrant, so I will be taking the negative root assuming I am correct. Am I supposed to take the cos value (x-value) of the point of the terminal ray of the angle -7pi/8? I am unsure how I am supposed to find that value to plug into the formula.
 
  • #4
Joe_K said:
From what I figured, -7pi/8 would be in the 3rd quadrant, so I will be taking the negative root assuming I am correct. Am I supposed to take the cos value (x-value) of the point of the terminal ray of the angle -7pi/8? I am unsure how I am supposed to find that value to plug into the formula.

Right on the negative. But all you have to do to finish it is find cos(-7pi/4). That's easy, yes? It's a multiple of 45 degrees.
 
  • #5
Ah, so I will be using the value of sqrt(2)/2 as the value of cos -7pi/8? Since -7pi/4 falls on points (sqrt2/2 [x], sqrt2/2, [y]) on the unit circle.
 
  • #6
BTW you've got the parentheses wrong on sin (-7pi/8) = sqrt (1-cos (-7pi/4))/2, right? It should be sin (-7pi/8) = sqrt ((1-cos (-7pi/4))/2). The /2 is inside the sqrt.
 
  • #7
Joe_K said:
Ah, so I will be using the value of sqrt(2)/2 as the value of cos -7pi/8? Since -7pi/4 falls on points (sqrt2/2 [x], sqrt2/2, [y]) on the unit circle.

Yes, sqrt(2)/2. But that's cos(-7pi/4), right? Not cos(-7pi/8). Let's not hash this up completely.
 
  • #8
Sorry, I typed that incorrectly about the sqrt(2)/2 being the value of cos -7pi/8, when it is really the value of cos -7pi/4.

So, after working through the problem, I ended up with:

sin -7pi/8= - sqrt(1-(sqrt(2)/2)/2)

Which I believe simplifies to:

sin -7pi/8= - sqrt(1-sqrt(2)/4) as the final answer?
 
  • #9
You've got all the right ideas. But I've got -sqrt((1-sqrt(2)/2)/2). That's just a LITTLE bit different. Try this. Get a calculator and punch in sin(-7pi/8) and then put in your answer. Then figure out where you fluffed a parenthesis. I've been known to do that, just to make sure I'm right.
 
Last edited:
  • #10
Ok, I see where my mistake was now. When I used you solution, it matched what my calculator displayed, which is -.38...

Perhaps simplest form would be something like:

-sqrt((1/2)-(sqrt(2)/4)

but I am not too sure on that. By the way, thank you very much for your help, I appreciate you taking the time to help me.
 
  • #11
Joe_K said:
Ok, I see where my mistake was now. When I used you solution, it matched what my calculator displayed, which is -.38...

Perhaps simplest form would be something like:

-sqrt((1/2)-(sqrt(2)/4)

but I am not too sure on that. By the way, thank you very much for your help, I appreciate you taking the time to help me.

You're welcome. And I'm really sure that's right. My 'calculator' says so, even though you left out the last parenthesis. Tricky aren't they?
 

Related to Give the exact value using the Half Angle Formula

What is the half angle formula?

The half angle formula is a mathematical equation used to find the exact value of trigonometric functions for half angles. It is derived from the double angle formula and can be used to simplify complex trigonometric expressions.

How do you use the half angle formula?

To use the half angle formula, you need to know the original trigonometric function and the value of the angle. Then, you can plug the half angle into the formula and solve for the exact value. It is important to remember that the half angle must be in radians, not degrees.

What is the most common mistake when using the half angle formula?

The most common mistake when using the half angle formula is forgetting to convert the angle to radians. Since the formula requires the angle to be in radians, forgetting to convert from degrees can lead to incorrect results.

Can the half angle formula be used for all trigonometric functions?

Yes, the half angle formula can be applied to all six trigonometric functions: sine, cosine, tangent, cotangent, secant, and cosecant. However, the formula may look slightly different depending on the function being used.

What are some real-life applications of the half angle formula?

The half angle formula has many practical applications in fields such as engineering, physics, and navigation. It is commonly used in calculating forces, distances, and angles in various structures and machines. It is also used in GPS systems to determine exact locations based on satellite data.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
1
Views
962
  • Precalculus Mathematics Homework Help
Replies
14
Views
365
  • Precalculus Mathematics Homework Help
Replies
9
Views
2K
  • Precalculus Mathematics Homework Help
Replies
1
Views
598
  • Precalculus Mathematics Homework Help
Replies
25
Views
641
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
2
Views
959
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
Back
Top