WEBVTT
00:00:00.980 --> 00:00:11.060
In this video, we will learn how to solve trigonometric equations involving secant, cosecant, and cotangent over different intervals.
00:00:12.070 --> 00:00:24.190
Reciprocal trigonometric equations have several real-world applications in fields such as physics, engineering, architecture, robotics, and navigation.
00:00:25.400 --> 00:00:36.540
In physics, they can be used in projectile motion, when analyzing currents, and finding the trajectory of a mass around a massive body under the force of gravity.
00:00:37.880 --> 00:00:44.480
Letβs begin by recalling the trigonometric functions whose reciprocals we will examine in this video.
00:00:45.230 --> 00:00:52.870
By considering the right triangle drawn, we know that the sin of angle π is equal to the opposite over hypotenuse.
00:00:53.340 --> 00:00:57.820
The cos of angle π is the adjacent over the hypotenuse.
00:00:58.120 --> 00:01:03.310
And the tan of angle π is equal to the opposite over the adjacent.
00:01:04.170 --> 00:01:11.430
This also leads us to the trigonometric identity tan π is equal to sin π over cos π.
00:01:12.800 --> 00:01:19.180
We note that these ratios are defined for acute angles between zero and 90 degrees.
00:01:19.900 --> 00:01:26.480
In order to define these functions for all values of π, we need to consider the unit circle.
00:01:27.590 --> 00:01:32.560
Suppose that a point moves along the unit circle in the clockwise direction.
00:01:33.810 --> 00:01:44.840
At any point with coordinates π₯, π¦ on the unit circle with angle π, we know that π₯ is equal to cos π and π¦ is equal to sin π.
00:01:45.770 --> 00:01:54.450
We know that these trig functions are periodic such that the sin of 360 degrees plus π is equal to sin π.
00:01:55.260 --> 00:02:01.890
Likewise, the cos of 360 degrees plus π is equal to the cos of π.
00:02:02.840 --> 00:02:11.150
The periodicity of the tangent function tells us that tan of 180 degrees plus π is equal to tan π.
00:02:11.860 --> 00:02:28.810
We also know from the symmetry of the sine and cosine graphs that the sin of 180 degrees minus π is equal to sin π and the cos of 360 degrees minus π is equal to cos π.
00:02:29.780 --> 00:02:33.500
These can be seen on a CAST diagram as follows.
00:02:34.420 --> 00:02:40.910
This CAST diagram helps us to remember the signs of the trig functions for each quadrant.
00:02:41.750 --> 00:02:45.490
In the first quadrant, all of the functions are positive.
00:02:46.340 --> 00:02:53.590
In the second quadrant, the sine function is positive, whereas the cosine and tangent functions are negative.
00:02:54.580 --> 00:02:58.580
In the third quadrant, only the tangent function is positive.
00:02:59.390 --> 00:03:07.650
And finally, in the fourth quadrant, the cosine function is positive, whereas the sine and tangent functions are negative.
00:03:08.840 --> 00:03:14.430
Letβs now consider how these relate to the reciprocal trigonometric functions.
00:03:15.440 --> 00:03:21.000
The csc of angle π is equal to one over the sin of angle π.
00:03:21.400 --> 00:03:24.460
It is the reciprocal of the sine function.
00:03:25.610 --> 00:03:31.030
The sec of angle π is equal to one over the cos of angle π.
00:03:31.990 --> 00:03:39.320
And finally, the cot of angle π is equal to one over the tan of angle π.
00:03:40.210 --> 00:03:52.770
This last function, together with the fact that tan π is equal to sin π over cos π, leads us to the fact that cot π is equal to cos π over sin π.
00:03:53.670 --> 00:04:02.980
We will now use these reciprocal functions together with our CAST diagram to solve trigonometric equations in given intervals.
00:04:04.290 --> 00:04:17.090
Find the set of values satisfying root three cot π equals one given π is greater than zero and less than 360 degrees.
00:04:18.100 --> 00:04:23.700
We can solve this problem using our knowledge of the reciprocal trigonometric functions.
00:04:24.520 --> 00:04:30.670
We recall that the cot of angle π is equal to one over the tan of π.
00:04:31.990 --> 00:04:38.140
We can rearrange the equation weβre given by firstly dividing both sides by root three.
00:04:39.130 --> 00:04:44.260
The cot of π is therefore equal to one over root three.
00:04:45.470 --> 00:04:51.910
As tan π is the reciprocal of this, the tan of π equals root three.
00:04:52.960 --> 00:04:58.030
We can find the principal angle here using our knowledge of special angles.
00:04:58.770 --> 00:05:03.590
We know that the tan of 60 degrees is equal to root three.
00:05:04.520 --> 00:05:13.060
This means that one solution to the equation tan π equals root three is π equals 60 degrees.
00:05:13.970 --> 00:05:19.420
We are asked to find all the solutions between zero and 360 degrees.
00:05:20.240 --> 00:05:23.730
We can do this by sketching a CAST diagram.
00:05:24.620 --> 00:05:31.700
As the tan of angle π is positive, there will be solutions in the first and third quadrants.
00:05:32.510 --> 00:05:38.060
We have already found that the solution in the first quadrant is equal to 60 degrees.
00:05:38.900 --> 00:05:49.030
Using the periodicity of the tangent function, we know that the tan of 180 degrees plus π is equal to the tan of angle π.
00:05:50.020 --> 00:05:57.490
Our second solution can therefore be calculated by adding 60 degrees to 180 degrees.
00:05:58.240 --> 00:06:01.870
This is equal to 240 degrees.
00:06:03.140 --> 00:06:12.100
Any further solutions found by adding or subtracting multiples of 180 degrees will be outside the required interval.
00:06:12.640 --> 00:06:18.310
So the solution set is 60 degrees and 240 degrees.
00:06:19.300 --> 00:06:25.520
In our next example, we will consider a problem where the interval is given in radians.
00:06:27.180 --> 00:06:40.190
Find the value of π that satisfies the csc of π minus root two equals zero where π lies on the open interval from zero to π over two.
00:06:41.060 --> 00:06:47.470
We begin by recalling that the csc of angle π is the reciprocal of sin π.
00:06:48.480 --> 00:06:54.700
In order to solve the equation weβre given, we begin by adding root two to both sides.
00:06:55.400 --> 00:06:59.710
This means that the csc of π is equal to root two.
00:07:00.880 --> 00:07:05.580
The sin of angle π is therefore equal to one over root two.
00:07:06.690 --> 00:07:16.200
Using our knowledge of special angles, we know that the sin of 45 degrees is equal to one over root two or root two over two.
00:07:17.250 --> 00:07:22.040
It is important to note at this point that the interval was given in radians.
00:07:22.710 --> 00:07:26.840
We know that π radians is equal to 180 degrees.
00:07:27.510 --> 00:07:33.610
This means that weβre only interested in solutions between zero and 90 degrees.
00:07:34.410 --> 00:07:49.130
We can therefore conclude that the value of π that satisfies the csc of π minus root two equals zero between zero and 90 degrees is 45 degrees.
00:07:50.780 --> 00:07:59.080
In our next example, we will solve a trigonometric equation by changing the interval over which solutions exist.
00:08:00.930 --> 00:08:20.140
Find the solution set of π that satisfies root three multiplied by the csc of 90 degrees minus π minus two equals zero, where π lies on the closed interval from zero degrees to 180 degrees.
00:08:21.020 --> 00:08:26.320
In this question, we begin by considering the reciprocal trigonometric functions.
00:08:27.050 --> 00:08:33.930
We know that the csc of any angle πΌ is equal to one over the sin of angle πΌ.
00:08:34.890 --> 00:08:41.070
Before using this identity, we will let πΌ equal 90 degrees minus π.
00:08:41.970 --> 00:08:50.060
This enables us to solve the simpler equation root three csc of πΌ minus two equals zero.
00:08:50.930 --> 00:09:00.640
Adding two to both sides of this equation and then dividing through by root three, we have the csc of πΌ equals two over root three.
00:09:01.900 --> 00:09:12.600
Using the reciprocal identity, we have sin πΌ is equal to one over two over root three, which is equal to root three over two.
00:09:13.530 --> 00:09:20.110
From our knowledge of special angles, we know that the sin of 60 degrees is equal to root three over two.
00:09:21.090 --> 00:09:29.970
This means that a solution to the equation sin πΌ equals root three over two is πΌ equals 60 degrees.
00:09:30.720 --> 00:09:36.610
We are looking for solutions of π between zero and 180 degrees inclusive.
00:09:37.830 --> 00:09:51.820
To write this interval in terms of πΌ, we subtract 90 degrees from the inequality such that πΌ is greater than or equal to negative 90 degrees and less than or equal to 90 degrees.
00:09:52.490 --> 00:09:59.620
Using a CAST diagram, we can see weβre only looking for solutions in the first or fourth quadrants.
00:10:00.410 --> 00:10:06.930
As the sin of angle πΌ is positive, there will therefore only be a solution in the first quadrant.
00:10:07.670 --> 00:10:13.440
This is the solution we have already found: πΌ is equal to 60 degrees.
00:10:14.600 --> 00:10:22.290
As πΌ is equal to 90 degrees minus π, then π must be equal to 90 degrees minus πΌ.
00:10:23.150 --> 00:10:34.170
Substituting in our value of πΌ, we have π is equal to 90 degrees minus 60 degrees, which gives us an answer of 30 degrees.
00:10:35.020 --> 00:10:48.980
The solution set that satisfies the equation root three multiplied by the csc of 90 degrees minus π minus two equals zero is 30 degrees.
00:10:49.960 --> 00:10:55.760
In our final example, we will look at a more complicated trigonometric equation.
00:10:57.490 --> 00:11:17.350
Find the set of values satisfying two sin π csc π plus sec π cot π equals zero given that π is greater than or equal to zero degrees and less than or equal to 360 degrees.
00:11:18.590 --> 00:11:26.660
We will begin this question by recalling the definition of the cosecant, secant, and cotangent functions.
00:11:27.190 --> 00:11:31.760
csc of angle π is equal to one over sin π.
00:11:32.520 --> 00:11:36.210
The sec of angle π is equal to one over cos π.
00:11:36.210 --> 00:11:42.870
And the cot of π is equal to one over the tan of π.
00:11:44.020 --> 00:11:54.800
We also recall that since tan π is equal to sin π over cos π, the cot of π is equal to cos π over sin π.
00:11:55.810 --> 00:12:12.060
Substituting these identities into our equation, we have two sin π multiplied by one over sin π plus one over cos π multiplied by cos π over sin π is equal to zero.
00:12:13.190 --> 00:12:22.430
The first term on the left-hand side simplifies to two, and the second term simplifies to one over sin π.
00:12:23.440 --> 00:12:29.450
Subtracting two from both sides, we have one over sin π equals negative two.
00:12:30.390 --> 00:12:41.360
This means that the csc of π equals negative two, which in turn tells us that sin of angle π is equal to negative one-half.
00:12:42.360 --> 00:12:46.750
We can solve this equation by firstly sketching a CAST diagram.
00:12:47.760 --> 00:12:54.370
As the sign of angle π is negative, there will be solutions in the third and fourth quadrants.
00:12:55.290 --> 00:13:01.840
From our knowledge of special angles, we know that the sin of 30 degrees is equal to a half.
00:13:02.900 --> 00:13:10.950
We can then use our knowledge of the symmetry of the sine function to calculate the solutions in the third and fourth quadrants.
00:13:12.090 --> 00:13:18.580
In the third quadrant, we have π is equal to 180 degrees plus 30 degrees.
00:13:19.350 --> 00:13:22.520
This is equal to 210 degrees.
00:13:23.690 --> 00:13:30.920
In the fourth quadrant, we have π is equal to 360 degrees minus 30 degrees.
00:13:31.760 --> 00:13:35.680
This is equal to 330 degrees.
00:13:36.770 --> 00:13:55.040
The set of values that satisfy two sin π csc π plus sec π cot π equals zero where π lies between zero and 360 degrees inclusive are 210 and 330 degrees.
00:13:55.040 --> 00:14:00.170
We will now summarize the key points from this video.
00:14:00.290 --> 00:14:12.250
In order to solve reciprocal trigonometric equations, we use the definition of the cosecant, secant, and cotangent functions.
00:14:12.250 --> 00:14:19.220
The csc of angle π is equal to one over sin π.
00:14:20.490 --> 00:14:25.700
The sec of angle π is equal to one over the cos of angle π.
00:14:26.880 --> 00:14:33.180
And the cot of angle π is equal to one over the tan of angle π.
00:14:34.010 --> 00:14:38.950
This is also equal to the cos of angle π over the sin of angle π.
00:14:38.950 --> 00:14:58.350
After finding the principal value solution, using our knowledge of special angles or the inverse trig functions, we can find all solutions in a given interval using the CAST diagram and the periodicity of the trigonometric functions.
00:14:59.230 --> 00:15:09.120
Using the fact that the trigonometric functions are periodic, we can develop this further to find a general solution to trigonometric equations.
00:15:09.460 --> 00:15:12.900
However, this is outside of the scope of this video.