# Thread: need these done by tomorrow - trig

1. ## need these done by tomorrow - trig

i accidentally posted these in the trig help forum.

10) Give that tan = -1/7 and sin > 0, use the x-y-r method to find sin and cos.

11) Sketch a 240° angle in standard position. Label a point on its terminal side, draw the right triangle in use, indicate the direction of opening, and give all six trig function values for 240°.

12) Sketch a 90° angle in standard position. Label a point on its terminal side and use the x-y-r method to compute the six trig function values for 90°.

13) In what quadrant(s) does terminate if sin > 0 and cos < 0?
Q1 Q2 Q3 Q4

14) In what quadrant(s) does terminate if tan < 0?
Q1 Q2 Q3 Q4

15) In what quadrant(s) does terminate if cos < 0?
Q1 Q2 Q3 Q4

17) Use a fundamental identity to find cos if sin = 1/5, that terminates in Q2. Show work.

2. Originally Posted by gabby is awesome
i accidentally posted these in the trig help forum.

10) Give that tan = -1/7 and sin > 0, use the x-y-r method to find sin and cos.
I guess I'll solve this one and others can solve the rest

$\displaystyle tan\theta<0 and sin\theta>0$ so by using CAST rule we know that this is in the second quadrant (S). $\displaystyle tan\theta=opposite/adjacent$ so if you draw a triangle on the second quadrant, the length of the line parallel to the y axis is 1 unit and the length on the x axis is 7 units. You can now find hypotenuse using pythagorean theorem
$\displaystyle \sqrt{1^2 + 7^2} = hyp = \sqrt{50}$
then use $\displaystyle sin\theta=opposite/hypotenuse$
$\displaystyle cos\theta=adjacent/hypotenuse$
$\displaystyle sin\theta= 1/\sqrt{50}$
$\displaystyle cos\theta =7/\sqrt{50}$
You can reduce the 50 under the root if you want