Question 2.1.1
Answer: 5 units
Reason:
Use the distance formula to find the distance from the origin (0,0) to the terminal point (-3, 4)
[tex](x_1,y_1) = (0,0) \text{ and } (x_2, y_2) = (-3,4)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(0-(-3))^2 + (0-4)^2}\\\\d = \sqrt{(0+3)^2 + (0-4)^2}\\\\d = \sqrt{(3)^2 + (-4)^2}\\\\d = \sqrt{9 + 16}\\\\d = \sqrt{25}\\\\d = 5\\\\[/tex]
The pythagorean theorem is a similarly related alternative path you can take.
This triangle is a 3-4-5 right triangle.
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Question 2.1.2
Answer: 1/5
Reason:
r = distance from origin to terminal point
r = 5, calculated back in the previous problem above.
sin(alpha) = y/r = 4/5
cos(alpha) = x/r = -3/5
sin(alpha)+cos(alpha) = (4/5)+(-3/5) = 1/5
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Question 2.1.3
Answer: 1
Reason:
Recall that tangent is the ratio of sine over cosine
tan = sin/cos
Therefore,
[tex]\tan(\alpha) *\frac{\cos(\alpha)}{\sin(\alpha)}\\\\\frac{\sin(\alpha)}{\cos(\alpha)}*\frac{\cos(\alpha)}{\sin(\alpha)}\\\\\frac{\sin(\alpha)*\cos(\alpha)}{\cos(\alpha)*\sin(\alpha)}\\\\1[/tex]
This assumes that neither sin(alpha) nor cos(alpha) are zero. Otherwise, we have a division by zero error.
