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Having trouble getting a start on this problem, any help would be appreciated!

Given point $P = (-3,5)$ is on the terminal arm of angle $\theta$ in standard position. Calculate the exact value of $\sin\theta, \cos\theta$, and $\tan\theta$.

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Compute it directly from the side lengths of the implied rectangular triangle. – Hagen von Eitzen Jun 18 '13 at 19:53
up vote 7 down vote accepted


What quadrant is the point $\;P = ({\bf x, y}) = (-3, 5)$ located?

Draw the right triangle that point $P$ makes with the $x$ axis - the length of the hypotenuse of the right triangle will equal $\;{\bf h} = \sqrt{(-3)^2 + 5^2} = \sqrt{34}$

Use SOH CAH TOA to unpack the definitions of $\tan \theta, \;\sin\theta,\;\cos\theta$:

$$\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac yx = \quad?\;$$ $$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac yh = \quad?\;$$ $$\cos \theta= \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{h} = \quad?$$

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For the graph, which is the proper way to orient the triangle? – ComradeYakov Jun 18 '13 at 20:14
@amWhy: SOHCAHTOA it is! :-) – Amzoti Jun 18 '13 at 20:17
Ok, I have to go now but I'll review this again once I get back home in about an hour or so. – ComradeYakov Jun 18 '13 at 20:22
Ok, let me know what you think. Exact value for sin(theta)= -3/6, cos(theta)=5/6, tan(theta)=-3/5 @amWhy – ComradeYakov Jun 18 '13 at 23:56
Oops, $\sqrt{34} \neq 6.$ Actually, your $\tan \theta = -\frac{5}{3}$ is correct, but your mixing up $x$ and $y$: $P = (x, y) = (-3, 5)$: Cosine correlates with $x$. Sine correlates with $y$: $\cos \theta = \frac{-3}{\sqrt{34}}$, $\sin\theta = \frac{5}{\sqrt{34}}$ – amWhy Jun 18 '13 at 23:59

Current solutions will provide your answer. Something to think about. Plot the point P you are interested in. Draw the line from the point to the origin, O. Also draw the circle, centre (0,0), radius 1 and the line x = 1. The cosine and sine of the angle are the coordinates of the point where the line OP cuts the circle. This is true whatever point you are talking about. In fact this is a definition of sine and cosine. The tangent of your angle is given by the coordinates (1, t) where the line OP (extended in this case) cuts the line x=1 which is a tangent to the unit circle. You can use similar triangles to find all these values. The calculations will be exactly those already provided.

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$\sin \theta = p/h$ means perpendicular/hypotenuse
$\cos \theta = b/h$ means base/hypotenuse
$\tan \theta = p/b$ means perpendicular/base

This all means, if given two values in a right-angle triangle, we can find another third side of triangle. Especially for 10 class, this formula helps.

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