iF Cos=-4/5 , a lies in second quadrent find the values of sin a
Answers
Answer:
Yes, I can. I suppose you want me to do so, though you did not ask that.
The key to the solution is the trigonometric identity sin² x + cos² x = 1.
Therefore sin² x = 1 − cos² x = 1 − (4/5)² =(25 − 16)/25 = 9/25.
Taking the [principal] square root of both sides yields:
|sin x| = 3/5.
[Remember that √(x²) = |x|, not x, not ±x, and |x| = c means x = ±c for c > 0.]
Therefore, sin x = ±3/5 is the final answer.
Note that cos x > 0 in Quadrants 1 and 4, while sin x > 0 in Quadrant 1 but sin x < 0 in Quadrant 4. The problem did not indicate any restriction on quadrants, so both answers are valid and required to be stated; otherwise, the solution is incomplete. That is a problem with the method of drawing a right triangle, as so many other answers used—it tends to make you focus on Quadrant 1 so much that you forget about the other quadrants, which was indeed the outcome for several answers using that technique.
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Answer:
Answer:
Yes, I can. I suppose you want me to do so, though you did not ask that.
The key to the solution is the trigonometric identity sin² x + cos² x = 1.
Therefore sin² x = 1 − cos² x = 1 − (4/5)² =(25 − 16)/25 = 9/25.
Taking the [principal] square root of both sides yields:
|sin x| = 3/5.
[Remember that √(x²) = |x|, not x, not ±x, and |x| = c means x = ±c for c > 0.]
Therefore, sin x = ±3/5 is the final answer.
Note that cos x > 0 in Quadrants 1 and 4, while sin x > 0 in Quadrant 1 but sin x < 0 in Quadrant