Consider the following.
1 answer:
Answer:
Anything in the form x = pi+k*pi, for any integer k
These are not removable discontinuities.
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Explanation:
Recall that tan(x) = sin(x)/cos(x).
The discontinuities occur whenever cos(x) is equal to zero.
Solving cos(x) = 0 will yield the locations when we have discontinuities.
This all applies to tan(x), but we want to work with tan(x/2) instead.
Simply replace x with x/2 and solve for x like so
cos(x/2) = 0
x/2 = arccos(0)
x/2 = (pi/2) + 2pi*k or x/2 = (-pi/2) + 2pi*k
x = pi + 4pi*k or x = -pi + 4pi*k
Where k is any integer.
If we make a table of some example k values, then we'll find that we could get the following outputs:
- x = -3pi
- x = -pi
- x = pi
- x = 3pi
- x = 5pi
and so on. These are the odd multiples of pi.
So we can effectively condense those x equations into the single equation x = pi+k*pi
That equation is the same as x = (k+1)pi
The graph is below. It shows we have jump discontinuities. These are <u>not</u> removable discontinuities (since we're not removing a single point).
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Answer: Angle A = 60°, angle B = 120°, angle C = 60°, angle D = 120°
Step-by-step explanation: Please refer to the attached diagram for further details.
The question has given the parallelogram ABCD as having side AB measuring 6 units, side AK measuring 3 units and line BK has been drawn perpendicular to line AD. This means line AK forms a right angle at the point where it meets line AD. From the information provided and the figure derived from that, we have right angled triangle ABK with the right angle at point K, line AB equals 6 units and line AK equals 3 units.
Using angle A as the reference angle, line AK is the opposite (side facing the reference angle), line AB is the hypotenuse (line facing the right angle) and line AK is the adjacent (side that lies between the right angle and the reference angle).
Hence, using the trigonometric ratios,
Cos A = adjacent/hypotenuse
Cos A = 3/6
Cos A = 0.5
Checking with the calculator, (that is second function of Cos 0.5)
A = 60
The opposite sides of a parallelogram are equal (that is line AD is parallel to line BC) and therefore opposite angles are equal (that is angle A is equal to angle C).
Similarly, angle B is equal to angle D. However, angles in a quadrilateral equals 360 degrees.
Therefore, angles in parallelogram ABCD is derived as;
A + B + C + D = 360
60 + B + 60 + D = 360
120 + B + D = 360
Subtract 120 from both sides of the equation
B + D = 240
Having known that angle B equals angle D, angles B and D is derived as each half of the value 240
Angle B = 240/2
Angle B = 120
and angle D = 120
Therefore, the angles are;
A = 60, B = 120, C = 60, D = 120
