Determining the sin theta, tan theta and cos theta at P (x,y)

In summary: which is the inverse function of $\arctan$ $\cos(\theta) = \frac{x}{\sqrt{x^2+y^2}}$ is also correct because you correctly did $\cos(\theta) = x/\sqrt{x^2+y^2}$ but instead you took $\arccos(x/\sqrt{x^2+y^2})$ instead of giving $\cos(\theta)$ which is the inverse function of $\arccos$
  • #1
Tazook
3
0
Determine the values of sin v, cos v, and tan v at each point P(x, y) on the terminal arm of an angle v in standard position.
(b) (3, 4) ( (d) (12, 5)
(f) (7, 24)
for b I was able to do
tan \theta= y/x
tan \theta= 4/3
\theta = 53.13
My textbook says I am wrong... doing an online course... teacher is so lazy that she never posted how to do it but rather read txtbook pg... Did not help...
 
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  • #2
Tazook said:
Determine the values of sin v, cos v, and tan v at each point P(x, y) on the terminal arm of an angle v in standard position.
(b) (3, 4) ( (d) (12, 5)
(f) (7, 24)
for b I was able to do
tan \theta= y/x
tan \theta= 4/3
\theta = 53.13
My textbook says I am wrong... doing an online course... teacher is so lazy that she never posted how to do it but rather read txtbook pg... Did not help...

\(\displaystyle \sin{\theta} = \frac{y}{\sqrt{x^2+y^2}}\)

\(\displaystyle \tan{\theta} = \frac{y}{x}\)

\(\displaystyle \cos{\theta} = \frac{x}{\sqrt{x^2+y^2}}\)
 
  • #3
Tazook said:
Determine the values of sin v, cos v, and tan v at each point P(x, y) on the terminal arm of an angle v in standard position.
(b) (3, 4) ( (d) (12, 5)
(f) (7, 24)
for b I was able to do
tan \theta= y/x
tan \theta= 4/3
\theta = 53.13
My textbook says I am wrong... doing an online course... teacher is so lazy that she never posted how to do it but rather read txtbook pg... Did not help...

Ah -- I see your trouble here. What you have should be almost correct.

What skeeter recommended is what you (sort of) did.

You are right that $\theta = 53.13$ degrees.

But the question is not asking you to find $\theta$ but the values of $\sin \tan \cos$ at the value of $v = \theta$

In my point of view,

$\tan(\theta) = \frac{4}{3}$ is indeed correct because you correctly did

$\tan(\theta) = y/x$ but instead you took $\arctan(y/x)$ instead of giving $\tan(\theta)$
 

FAQ: Determining the sin theta, tan theta and cos theta at P (x,y)

1. How do you determine sin theta at point P?

To determine sin theta at point P, you need to find the ratio of the opposite side of the triangle to the hypotenuse. This can be calculated using the formula sin theta = opposite/hypotenuse.

2. What is the formula for calculating tan theta at point P?

The formula for calculating tan theta at point P is opposite/adjacent. This means you need to find the ratio of the opposite side to the adjacent side of the triangle.

3. Can you explain how to find cos theta at point P?

To calculate cos theta at point P, you need to find the ratio of the adjacent side to the hypotenuse. This can be calculated using the formula cos theta = adjacent/hypotenuse.

4. What information do you need to determine sin, tan, and cos at point P?

To determine sin, tan, and cos at point P, you need to know the coordinates of the point (x,y) and the length of the sides of the triangle connected to that point.

5. How are sin, tan, and cos related to each other at point P?

Sin, tan, and cos are related to each other at point P through the Pythagorean theorem, which states that sin^2 theta + cos^2 theta = 1. This means that the values of sin, tan, and cos are interconnected and can be calculated using each other.

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