a) If angle $\theta\\$ lies in Quadrant II and $\sin \theta ={3 \over {\sqrt {45} }}$. Determine possible coordinates for point $P$ on the terminal arm of angle $\theta$.

b) Determine the Quadrant in which angle $\theta$ is located if $\cos\theta < 0$ and $\tan \theta > 0$.

How would I do such a problem?

  • $\begingroup$ Apply the Pythagorean Theorem sin²t+cos²t=1. Substitute your given sine value and calculate the cosine value. You know the quadrant, hence the sign of the cosine value. $\endgroup$ – imranfat May 29 '13 at 2:49
  • $\begingroup$ I believe you are asking 2 different questions. Also, the tag of calculus is not appropriate. Perhaps trigonometry? $\endgroup$ – Calvin Lin May 29 '13 at 2:49
  • $\begingroup$ SIN^(2)2+COS^2(2)=1? $\endgroup$ – user73122 May 29 '13 at 2:50
  • $\begingroup$ much apolgizes ; $\endgroup$ – user73122 May 29 '13 at 2:50
  • $\begingroup$ If you draw the sine value in a right triangle: with respect to angle theta, the opposite is 3, the hypotenuse is √45 and so through the Pythagorean Theorem you find the adjacent to be 6. Now if you know the definition of cosine, it should not be difficult anymore $\endgroup$ – imranfat May 29 '13 at 2:59

For part a, $\sin\theta=\frac{3}{\sqrt{45}}=\frac{3}{3\sqrt{5}}=\frac{1}{\sqrt{5}}$ and since we know that the sine of an angle is the opposite over the hypotenuse we can imagine a triangle whose hypotenuse is $\sqrt{5}$ and whose opposite side is $1$. The Pythagorean theorem tells us that the adjacent side is then $\sqrt{\sqrt{5}^2-1^2}=\sqrt{5-1}=2$.

Since we are given that the point must be in the second quadrant, this gives a possible point of $(-2,1)$. Note that any positive multiple of this will also work, e.g. $(-4,2)$.

For part b, we use the fact that the cosine function is negative in the second and third quadrants and the tangent is positive in the first and third quadrants. Therefore, $\theta$ must be in the third quadrant for both conditions to be satisfied.


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