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Verify tanx = secx/cscx
The general way to solve this type of problem is to go back to the definition of tangent, secant, cosecant, etc. Convert the values into their equivalents involving only sin and cosine using recognized trigonometric identities:
=\frac {sec(x)}{csc(x)})
 = \frac {sin(x)}{cos(x)})
. This comes from the definition of sine, cosine, and tangent on a unit circle.
 = \frac {1}{cos(x)})
(definition of secant)
 = \frac {1}{sin(x)})
(definition of cosecant)
Plug the three equations into the first equation.
}{cos(x)} = \frac {\frac {1}{cos(x)}}{\frac {1}{sin(x)}}=\frac {sin(x)}{cos(x)})
Now, you try the others, one at a time. I'll be around to help you out when you post what you think is the answer.
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Verify tan(pi - A) = -tan(A)
Start with a unit circle (a circle centered on the origin of a Cartesian coordinate system Draw an angle from the origin corresponding to the angle, A. Draw another corresponding to pi-A. With the angles, drop a perpendicular line to the X-axis forming a triangle. Using the definition of the tangent, you can solve the problem.
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Use synthetic Division to find the remainder of (x^4 - 3x^2 + 2x - 1)/(x - 1)
This is just like long division you did in the fourth grade.
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cotx + tanx all over sec^2x
A: cotx
B: sinx
C: cosx
D: tanx
Do this just like the first problem = substitute cot=cos/sin and tan=sin/cos and sec = 1/cos
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State the amplitude, period, phase shift, and vertical shift for y = 3sin(x/4 + pi/2) - 2
a. Remember that sin is periodic. It goes from +1 to -1. The amplitude is always 1 -- unless you multiply it by something.
b. Sin(x) goes from 0 at x=0 to 1 at x=π/2 (π=pi) to 0 at x=π to -1 at x=3π/2 back to zero. Its natural period (peak-to-peak) is 2π. You can vary the period only by multiplying or dividing whatever is inside the sin(x) parentheses.
c. A phase shift is whatever moves the graph from what I stated in b.
d. The vertical shift is anything that moves the entire graph up or down from what I stated in b.
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Given (triangle)ABC and A = 36(degrees), B = 101(degrees), and b = 42. Solve for C, a, and c.
This is very simple trigonometry. Take a piece of graph paper and construct a triangle according to what was specified. Pick an arbitrary angle for A (say 30 degrees) just so you can visualize the problem. Use the definitions of the sine, cosine, and tangent functions of the angles to solve the triangle.
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Solve the following system of equations 2x + 3y = -7 & x - y = 4.
Solve the following system of equations x - 2y + z = 15 & 2x + 3y - 3z = 1 & 4x + 10y - 5z = -3.
In the first, solve one of the two equations for x in terms of y. Put that value for x in the other equation. Solve for y. Take the value you now have for y and put it in either one of the equations to solve for x.
In the second, the procedure is similar except that you have three equations and three unknowns.
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What is the secant of -180(degrees).
Secant is the inverse of the cosine: Sec = 1/Cos. Use your calculator.
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Does cosx + cosx * tan^2x = secx? True or False.
This is solved exactly the same way as the first problem in this list.
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Find the value of k so that the remainder of (x^3 - 3x^2 + kx - 6)/(x + 2) = 0.
A: k = -11
B: k = 11
C: k = 6
D: k = -13
Do synthetic division and you'll have an expression in "k" as the remainder. Solve the equation for k.
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Given f(x) = 3x^2 + 2x -2 and g(x) = 4x + 1. Find f/g.
Just divide one equation by the other (synthetic division).
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Find the x & y intercepts of the equation y = 2/5x -2.
A line in its standard form is Y=mX+b where m is the slope and b is the Y-intercept. You simply figure out in your head what "m" is and what "b" is. If your equation is not in the standard form (yours is), you manipulate using algebra it until it is in the standard form.
