What is the trig identity for sin^2x + cos^2x = 1?

In summary, This conversation is discussing the use of trigonometric identities to solve for a variable in a circular motion problem. The identity sin(kx)^2 + cos(kx)^2 = 1 is mentioned and it is explained that the symbol for the argument in the identity can be changed without altering the meaning of the equation. It is also mentioned that this is one of the first trig identities taught in a trigonometry course.
  • #1
bobsmith76
336
0

Homework Statement



Screenshot2012-02-25at20521AM.png


If t = pi/2, then that would equal 2, but they don't say that t = pi/2 so how do they get 2?

Maybe since it's circular motion t = pi, that would work too, but I want to be sure before i move on.
 
Last edited:
Physics news on Phys.org
  • #2
It's a simple trignometric identity.

sin2θ + cos2θ = 1
 
  • #3
bobsmith76 said:

Homework Statement



Screenshot2012-02-25at20521AM.png


If t = pi/2, then that would equal 2, but they don't say that t = pi/2 so how do they get 2?

Maybe since it's circular motion t = pi, that would work too, but I want to be sure before i move on.

Note the identity sin(kx)^2 + cos(kx)^2 = 1
|v| becomes sqrt(4(1)) = 2
 
  • #4
What about the 2t? Unless the 2t gets subsumed in the theta and whatever theta is cos^2 and sin^2 add up to 1, is that right?
 
  • #5
bobsmith76 said:
What about the 2t? Unless the 2t gets subsumed in the theta and whatever theta is cos^2 and sin^2 add up to 1, is that right?

Yes.

If you let u = 2t, then we'd have
sin(u)^2 + cos(u)^2

Which equals 1, and we don't have a "u" to plug back into.
 
  • #6
bobsmith76 said:
What about the 2t? Unless the 2t gets subsumed in the theta and whatever theta is cos^2 and sin^2 add up to 1, is that right?

Theta is just a symbol for "the argument (or angle) that you pass into the trigonometric function." In this case, the argument to the function is 2t. In other words, θ = 2t here. The identity holds regardless of what theta is. I could have written the identity as:

sin2x + cos2x = 1

OR

sin2ϕ+ cos2ϕ = 1

OR

sin2:smile:+ cos2:smile: = 1

It doesn't matter. All that's changing here is the SYMBOL that I'm using to represent the argument. It doesn't change the meaning of the identity. All of these equations mean the same thing. Expressed in words, the identity is:

"The square of the sine of an angle, plus the square of the cosine of that same angle, together must equal 1."
 
  • #7
bobsmith76 said:
What about the 2t? Unless the 2t gets subsumed in the theta and whatever theta is cos^2 and sin^2 add up to 1, is that right?
sin2(whatever) + cos2(whatever) [itex]\equiv[/itex] 1.

"Whatever" can be anything, as long as it is the same in both terms.

This is one of the first trig identities presented in a trigonometry course. There aren't many trig identities that you need to memorize, but this is definitely one of them.
 

Related to What is the trig identity for sin^2x + cos^2x = 1?

What is the normal of a curve?

The normal of a curve is a line that is perpendicular to the tangent line at a specific point on the curve.

Why is it important to find the normal of a curve?

Finding the normal of a curve is important because it allows us to determine the direction in which the curve is curving at a specific point. This information is useful in various applications, such as engineering and physics.

How do you find the normal of a curve?

To find the normal of a curve, we first find the slope of the tangent line at the desired point. Then, we take the negative reciprocal of this slope to find the slope of the normal line. Finally, we use the point-slope formula to find the equation of the normal line.

Can the normal of a curve be found at any point?

Yes, the normal of a curve can be found at any point where the curve is differentiable, meaning that it has a well-defined tangent line.

What is the relationship between the normal and tangent of a curve?

The normal and tangent of a curve are always perpendicular to each other at a specific point. This means that the slopes of the two lines are negative reciprocals of each other.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
373
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
336
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
979
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
7K
  • Calculus and Beyond Homework Help
Replies
7
Views
840
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
Replies
9
Views
1K
Back
Top