Finding other Trig Values from one

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To find the other five trigonometric function values from theta = arcsin(-5/8), the sine value is established as sin(theta) = -5/8, leading to csc(theta) = -8/5. The Pythagorean theorem is used to determine the adjacent side of the triangle as sqrt(39). The discussion highlights the importance of identifying the correct quadrant, as sine is negative in both the third and fourth quadrants, but arcsin restricts theta to the fourth quadrant where cosine is positive. Therefore, the values of cosine and tangent must be adjusted accordingly based on this quadrant determination.
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The problem gave that:

theta= arcsin(-5/8)

It wants us to find the other five trig function values.

In order to do that, I turned the original function to
sin(theta)= - 5/8

From there I was able to find csc(theta) to be -8/5.

Using the pythagorian theorem I found the other side of the triangle that can be formed to be sqrt(39)

But because sin was negative from the original function, how are you supposed to figure out the quadrant the function lies in, because sin is negative in the 3rd and 4th quadrant.

I understand how to find the other values, I just don't which quadrant it lies in because it would effect the positive and negative signs.
 
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darshanpatel said:
The problem gave that:

theta= arcsin(-5/8)

It wants us to find the other five trig function values.

In order to do that, I turned the original function to
sin(theta)= - 5/8

From there I was able to find csc(theta) to be -8/5.

Using the pythagorian theorem I found the other side of the triangle that can be formed to be sqrt(39)

But because sin was negative from the original function, how are you supposed to figure out the quadrant the function lies in, because sin is negative in the 3rd and 4th quadrant.

I understand how to find the other values, I just don't which quadrant it lies in because it would effect the positive and negative signs.

theta= arcsin(-5/8) is not equivalent with sin(theta)= - 5/8. The range of the arcsin function is [-pi/2, pi/2]. So you can decide about the quadrant the angle lies in.
Inverse trigonometric functions - Wikipedia, the free encyclopedia

ehild
 
ehild said:
theta= arcsin(-5/8) is not equivalent with sin(theta)= - 5/8. The range of the arcsin function is [-pi/2, pi/2]. So you can decide about the quadrant the angle lies in.
Inverse trigonometric functions - Wikipedia, the free encyclopedia

ehild

Those two are equivalent because if you do sin45 it equals 1/(sqrt2) and if you do arcsin(1/sqrt2) it equals 45

They are inverse functions.
 
darshanpatel said:
Those two are equivalent because if you do sin45 it equals 1/(sqrt2) and if you do arcsin(1/sqrt2) it equals 45

They are inverse functions.

sinx has no inverse in its whole domain for -∞ to ∞, as it is not a monotonous function. It can be inverted between -pi/2 and pi/2 when it gives the principal values of arcsin function.

ehild
 
ehild said:
sinx has no inverse in its whole domain for -∞ to ∞, as it is not a monotonous function. It can be inverted between -pi/2 and pi/2 when it gives the principal values of arcsin function.

ehild

I do not have any idea what you are saying, but from what you are saying, what would the quadrant be? You are talking about a range from -pi/2 to pi/2 but I don't understand what you mean. We are not supposed to be using calculators so I don't know. Also, we need to find answers such as a side over another side, like the answer to costheta, or tantheta, which would require a conversion from arcsin to sin.
 
Can you tell me what the quadrants are? And what are the signs of cosine and sine in them?

ehild
 
ehild said:
Can you tell me what the quadrants are? And what are the signs of cosine and sine in them?

ehild

Quadrant 1: sin and cos are both postive

Quadrant 2: sin is only positive, cos negative

Quadrant 3: sin and cos are negative

Quadrant 4: cos is only positive, sin is negative
 
What angles belong to the individual quadrants? For example 300° belongs to which quadrant?

ehild
 
ehild said:
What angles belong to the individual quadrants?

ehild

What do you mean by angles? Also, why does it matter?

There are multiple angles per quadrant from 30 degrees to 60 degrees to 45 degrees and so on for each quadrant. Also, the -5/8 doesn't have a special angle like 30, 60, 90 triangles or the 45, 45, 90 triangles
 
  • #10
darshanpatel said:
What do you mean by angles? Also, why does it matter?

There are multiple angles per quadrant from 30 degrees to 60 degrees to 45 degrees and so on for each quadrant. Also, the -5/8 doesn't have a special angle like 30, 60, 90 triangles or the 45, 45, 90 triangles

Well, f(x) is a trigonometric function. You speak abut quadrants defined by the sign of sine and cosines. What is the interval for x in the first quadrant, second, third, and so on?

ehild
 
  • #11
ehild said:
Well, f(x) is a trigonometric function. You speak abut quadrants defined by the sign of sine and cosines. What is the interval for x in the first quadrant, second, third, and so on?

ehild

I think we are going about this the wrong way. I know for a fact that it is not going to be in the first or second quadrant because the sin was negative. But I don't know weather it is in quadrant 3 or 4 because they both can have negative sines. It might have to do with ranges as you were talking about before. I am thinking quadrant 4 but my gut is saying quadrant 3 and I just want to know how to be able to tell.
 
  • #12
theta =arssin(x) returns theta values between -pi/2 and pi/2. Only a monotonous function has inverse. sin(theta) is monotonous in the range -pi/2≤theta ≤pi/2. -pi/2 is the same as 3pi/2, so the domain includes the fourth and first quadrants. According to the sign of the sine, theta can belong to the third or fourth quadrant, but theta in the third quadrant can not be the value of arcsin(-5/8). That means, theta is in the fourth quadrant, where the cosine is positive.


http://www.marlenesite.com/math/trigonometry/images/quadrant.gif

ehild
 
  • #13
ehild said:
theta =arssin(x) returns theta values between -pi/2 and pi/2. Only a monotonous function has inverse. sin(theta) is monotonous in the range -pi/2≤theta ≤pi/2. -pi/2 is the same as 3pi/2, so the domain includes the fourth and first quadrants. According to the sign of the sine, theta can belong to the third or fourth quadrant, but theta in the third quadrant can not be the value of arcsin(-5/8). That means, theta is in the fourth quadrant, where the cosine is positive. http://www.marlenesite.com/math/trigonometry/images/quadrant.gif

ehild

Thank you, that makes sense but what do you mean by monotonous functions? I remember learning about things such as sin and tan only go from pi/2 to 3pi/2 and cos is the other way, but I don't really remember.
 
  • #14
darshanpatel said:
Thank you, that makes sense but what do you mean by monotonous functions? I remember learning about things such as sin and tan only go from pi/2 to 3pi/2 and cos is the other way, but I don't really remember.

You can invert a function only in the domain where it is monotonous. Otherwise f(x) can be the same at several different x-es, say f(x1)=f(x2). The inverse function returns the x value where f is given. There are two of them, but a function can have only one value.

So you were taught that the range of arcsin is [pi/2; 3pi/2]? It is all right, sin(x) is monotonous also in that range, although [-pi/2, pi/2] is the usual range. Your calculator will return a negative value for the arcsin of a negative number, and taking the cosine of the result, it will be positive. Try.
[pi/2; 3pi/2] includes the second and third quadrants, so you have to choose the cosine accordingly. Check your notes. Maybe, you need to give both the negative and positive cosines as answer.


ehild
 
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