This is a problem that requires solving a trigonometric equation.

Also, remember that sin(theta)=sqrt(1/2) can be simplified to sin(theta)=1/sqrt(2), which is equivalent to sin(theta)=sqrt(2)/2. In summary, the conversation is about solving equations involving trigonometric functions, specifically finding values of theta that satisfy given equations. The first problem involves using basic trigonometric identities and solving for theta given the values of sine, while the second problem involves using quadratic equations and substituting back in for theta to find the values that make the equations true. The book provides answers in radians, indicating that basic knowledge of trigonometric functions is expected. The conversation also mentions using trigonometric identities and simplifying expressions.
  • #1
arizona_cards_11
17
0
Trig Functions...

I don't really understand how my book wants me to approach this problem. And I know that you appreciate work...because this is for my benefit after all...but how exactly would this be worked?


Direction: Solve the equation for (theta)... 0 is less than or equal to (theta) < 2*pi

2sin^2(theta) = 1

The answers in the back of the book given in radians:

(theta) = (pi/4) , ((3*pi)/(4)) , ((5*pi)/(4)) , ((7*pi)/(4))
 
Physics news on Phys.org
  • #2
This should be a pretty simple problem, what is the trouble you're having exactly? First solve for sin(theta), and then find all theta that satisfy that equation.
 
  • #3
2sin^2 (theta) = 1

sin^2(theta) = 1/2

sin(theta) = sqrt(1/2

? Those are my first few steps...are there any problems?
 
  • #4
Nope that is perfectly correct, now you just need to find the values of theta with sines of sqrt(1/2).
 
  • #5
I don't have a calculator with me so I'll have to wait until school...

The next question using same directions:

tan^2(theta) - tan(theta) = 0
 
  • #6
You might notice that this is a quadratic equation in tan(theta) so let x=tan(theta) and solve the resulting quadratic equation. Then you will have 2 equations to solve for tan(theta).
 
  • #7
okay...

x^2 - x = 0

(x-1)(x+0) = 0

x = 1 or x = 0
 
  • #8
Am I on the right track here?
 
  • #9
arizona_cards_11 said:
Am I on the right track here?

Yep so now you can substitute x=tan(theta) back in and find the values of theta such that those equations are satisfied.
 
  • #10
I'm a little bit confused on this point...

I plug tan(theta) back into (x-1)(x+0) ?

Thus, making... tan(theta) = 1 and tan(theta) = 0
 
  • #11
arizona_cards_11 said:
I'm a little bit confused on this point...

I plug tan(theta) back into (x-1)(x+0) ?

Thus, making... tan(theta) = 1 and tan(theta) = 0

No you had x=1 or x=0, from there put x=tan(theta) and then find teh values of theta that will satisfy that.
 
  • #12
Is there any way to show your work besides plugging in...as my teacher is a stickler for descriptive work?
 
  • #13
Well I'm not really sure there are many wasy to show your work for this kind of a problem, but once you have it down to solving for theta, if you've memorized the important angles then it should be pretty simple to find what angles satisfy these conditions and then just explain that these are the angles satisfying the equations.
 
  • #14
The answers in the book are...

0 , (pi/4) , pi , ((5pi)/(4))
 
  • #15
You are clearly expected to know the trig functions for some basic angles, not just use a calculator.
 
  • #16
hey this stuff is easy
cant we use the identities
like

cos ( A-B) = cosA.cosB + sinA.sinB
and the other identities?
 
  • #17
arizona_cards_11 said:
2sin^2 (theta) = 1

sin^2(theta) = 1/2

sin(theta) = sqrt(1/2

? Those are my first few steps...are there any problems?

If sin^2(theta)=1/2 then you should have two possibilities for sin(theta), not just the one you have.
 

What is a trigonometric equation?

A trigonometric equation is an equation that involves trigonometric functions such as sine, cosine, and tangent. These functions are used to relate the side lengths of a right triangle to its angles.

Why is solving a trigonometric equation important?

Solving a trigonometric equation is important because it allows us to find missing information in a triangle, such as side lengths or angles. This is useful in many real-world applications, such as surveying, navigation, and engineering.

What are the different methods for solving a trigonometric equation?

There are several methods for solving a trigonometric equation, including using inverse trigonometric functions, using the unit circle, and using trigonometric identities. The best method to use will depend on the specific equation and its complexity.

Can a trigonometric equation have more than one solution?

Yes, a trigonometric equation can have multiple solutions. This is because trigonometric functions are periodic, meaning they repeat themselves after a certain interval. Therefore, an equation may have multiple solutions within that interval.

How can I check if my solution to a trigonometric equation is correct?

You can check your solution by substituting it back into the original equation and simplifying. If both sides of the equation are equal, then your solution is correct. You can also use a calculator to graph the equation and see if your solution aligns with the points where the graph intersects the x-axis.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
15
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
8
Views
999
  • Precalculus Mathematics Homework Help
Replies
4
Views
530
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
1K
Replies
3
Views
956
  • Calculus
Replies
29
Views
703
  • Precalculus Mathematics Homework Help
Replies
11
Views
1K
  • New Member Introductions
Replies
1
Views
71
Back
Top