SOLUTION: Can you solve for theta if cos0= negative square root of 2 by 2. Also, i am limited to the third and fourth quadrant. I figured the answer had to be either 7pi by 6 or 5 pi by 4. M
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Question 1006799: Can you solve for theta if cos0= negative square root of 2 by 2. Also, i am limited to the third and fourth quadrant. I figured the answer had to be either 7pi by 6 or 5 pi by 4. My book said 5 pi by 4 but how would I know which special triangle to use?
Answer by HaundC(1) (Show Source): You can put this solution on YOUR website!
For that case, theta WOULD be 5pi/4.
For 30 degree triangles (pi/6, 5pi/6, 7pi/6, 11pi/6), your sine will always be +/-(sqrt(3))/2, your cos will always be +/-1/2, and your tan will always be +/-(sqrt(3))/3.
For a 45 degree triangle (pi/4, 3pi/4, 5pi/4, 7pi/4), your sin and cos will always be +/-(sqrt(2))/2, and your tan will always be +/-1.
For a 60 degree triangle (pi/3, 2pi/3, 4pi/3, 5pi/3), your sin will always be +/-(sqrt(3)/2, your cos will always be +/-1/2, and your tan will always be (sqrt(3)).
(whether they are positive or negative depend on what quadrant they are in)
It's a matter of just memorizing.
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