Find cos theta, given that sin theta=3/5 and theta is in quadrant II
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sinØ = 3/5 , Ø in II
make a sketch of the corresponding right-angled triangle to see that y = 3, r = 5
x^2 + 3^2 = 5^2
x = ± 4, but we are in II , so x = -4
cosØ = -4/5
To find cos(theta) given that sin(theta) = 3/5 and theta is in quadrant II, we can use the Pythagorean identity to find the cosine of the angle.
The Pythagorean identity states that sin^2(theta) + cos^2(theta) = 1.
First, we need to find cos(theta) using the given information. To do so, we can use the fact that theta is in quadrant II and sin(theta) = 3/5. In quadrant II, sine is positive, but cosine is negative.
Since sin(theta) = 3/5 in quadrant II, we can use the Pythagorean identity to find the value of cos(theta).
Start by using the Pythagorean identity:
sin^2(theta) + cos^2(theta) = 1
Plugging in the given value of sin(theta) = 3/5:
(3/5)^2 + cos^2(theta) = 1
Simplifying the equation:
9/25 + cos^2(theta) = 1
Now, subtract 9/25 from both sides:
cos^2(theta) = 1 - 9/25
cos^2(theta) = 25/25 - 9/25
cos^2(theta) = 16/25
Taking the square root of both sides:
cos(theta) = ± √(16/25)
Since we are in quadrant II where cosine is negative, we take the negative square root of 16/25:
cos(theta) = -√(16/25)
Simplifying:
cos(theta) = -4/5
Therefore, cos(theta) is equal to -4/5.