SOLUTION: Hi, please help me solve this problem: Suppose sec alpha = -7/4 and that the terminal side of alpha is in quadrant III. find sin alpha and tan alpha. Thank you!

Question 398188: Hi, please help me solve this problem:
Suppose sec alpha = -7/4 and that the terminal side of alpha is in quadrant III. find sin alpha and tan alpha.
Thank you!
Found 2 solutions by stanbon, lwsshak3:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Suppose sec alpha = -7/4 and that the terminal side of alpha is in quadrant III. find sin alpha and tan alpha.
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sec(theta) = r/x
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If theta is in the 3rd Quadrant, x is negative.
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So, x = -4, r = 7
Solve for "y", which is negative in QIII
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y^2 = r^2-x^2
y^2 = 49-16
y^2 = 33
y = -sqrt(33)
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sin(alpha) = y/r = -sqrt(33)/7
tan(alpha) = y/x = -sqrt(33)/-4 = sqrt(33)/4
=================================================
cheers,
Stan H.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Suppose sec alpha = -7/4 and that the terminal side of alpha is in quadrant III. find sin alpha and tan alpha.
To start, it might help to visualize a picture of the right triangle in question.
Picture a circle with radius 7 and moving the terminal side to the reference angle (alpha) in quadrant III until the cos (horizonal)leg is equal to 4. You now have a right triangle with hypotenuse =7, and the adjacent leg=4. As you can see, both the horizontal (cos) and vertical (sin) legs are negative with respect to the origin.
Use the pythagorean theorem to find the vertical leg, call it x
x=sqrt(7^2-4^2)=sqrt(33)
sin alpha=(opposite/hypotenuse)=sqrt(33)/7
tan alpha = (opposite/adjacent)=sqrt(33)/4
In summary, this is what you have.
Sec alpha = -7/4 (given)
cos alpha = -4/7
sin alpha = sqrt(33)/7
tan alpha = sqrt(33)/4

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