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

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To determine sin, cos, and tan values at point P(x, y) on the terminal arm of angle v, the correct formulas are sin(v) = y/√(x²+y²), cos(v) = x/√(x²+y²), and tan(v) = y/x. For point (3, 4), tan(v) is correctly calculated as 4/3, leading to an angle θ of approximately 53.13 degrees. The confusion arises from the distinction between finding the angle θ and calculating the trigonometric functions at that angle. The textbook's instructions were unclear, causing frustration with the online course. Accurate calculations of sin and cos at the specified points are essential for completing the task.
Tazook
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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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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...

$$\sin{\theta} = \frac{y}{\sqrt{x^2+y^2}}$$

$$\tan{\theta} = \frac{y}{x}$$

$$\cos{\theta} = \frac{x}{\sqrt{x^2+y^2}}$$
 
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...

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 $\tan(\theta)$
 
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