Lesson Video: Solving Reciprocal Trigonometric Equations | Nagwa Lesson Video: Solving Reciprocal Trigonometric Equations | Nagwa

Lesson Video: Solving Reciprocal Trigonometric Equations Mathematics

In this video, we will learn how to solve trigonometric equations involving secant, cosecant, and cotangent over different intervals in degrees and radians.

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Video Transcript

In this video, we will learn how to solve trigonometric equations involving secant, cosecant, and cotangent over different intervals. Reciprocal trigonometric equations have several real-world applications in fields such as physics, engineering, architecture, robotics, and navigation. 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.

Let’s begin by recalling the trigonometric functions whose reciprocals we will examine in this video. By considering the right triangle drawn, we know that the sin of angle 𝜃 is equal to the opposite over hypotenuse. The cos of angle 𝜃 is the adjacent over the hypotenuse. And the tan of angle 𝜃 is equal to the opposite over the adjacent. This also leads us to the trigonometric identity tan 𝜃 is equal to sin 𝜃 over cos 𝜃. We note that these ratios are defined for acute angles between zero and 90 degrees.

In order to define these functions for all values of 𝜃, we need to consider the unit circle. Suppose that a point moves along the unit circle in the clockwise direction. At any point with coordinates 𝑥, 𝑦 on the unit circle with angle 𝜃, we know that 𝑥 is equal to cos 𝜃 and 𝑦 is equal to sin 𝜃. We know that these trig functions are periodic such that the sin of 360 degrees plus 𝜃 is equal to sin 𝜃. Likewise, the cos of 360 degrees plus 𝜃 is equal to the cos of 𝜃. The periodicity of the tangent function tells us that tan of 180 degrees plus 𝜃 is equal to tan 𝜃.

We also know from the symmetry of the sine and cosine graphs that the sin of 180 degrees minus 𝜃 is equal to sin 𝜃 and the cos of 360 degrees minus 𝜃 is equal to cos 𝜃. These can be seen on a CAST diagram as follows. This CAST diagram helps us to remember the signs of the trig functions for each quadrant. In the first quadrant, all of the functions are positive. In the second quadrant, the sine function is positive, whereas the cosine and tangent functions are negative. In the third quadrant, only the tangent function is positive. And finally, in the fourth quadrant, the cosine function is positive, whereas the sine and tangent functions are negative.

Let’s now consider how these relate to the reciprocal trigonometric functions. The csc of angle 𝜃 is equal to one over the sin of angle 𝜃. It is the reciprocal of the sine function. The sec of angle 𝜃 is equal to one over the cos of angle 𝜃. And finally, the cot of angle 𝜃 is equal to one over the tan of angle 𝜃. This last function, together with the fact that tan 𝜃 is equal to sin 𝜃 over cos 𝜃, leads us to the fact that cot 𝜃 is equal to cos 𝜃 over sin 𝜃. We will now use these reciprocal functions together with our CAST diagram to solve trigonometric equations in given intervals.

Find the set of values satisfying root three cot 𝜃 equals one given 𝜃 is greater than zero and less than 360 degrees.

We can solve this problem using our knowledge of the reciprocal trigonometric functions. We recall that the cot of angle 𝜃 is equal to one over the tan of 𝜃. We can rearrange the equation we’re given by firstly dividing both sides by root three. The cot of 𝜃 is therefore equal to one over root three. As tan 𝜃 is the reciprocal of this, the tan of 𝜃 equals root three. We can find the principal angle here using our knowledge of special angles. We know that the tan of 60 degrees is equal to root three. This means that one solution to the equation tan 𝜃 equals root three is 𝜃 equals 60 degrees. We are asked to find all the solutions between zero and 360 degrees. We can do this by sketching a CAST diagram.

As the tan of angle 𝜃 is positive, there will be solutions in the first and third quadrants. We have already found that the solution in the first quadrant is equal to 60 degrees. Using the periodicity of the tangent function, we know that the tan of 180 degrees plus 𝜃 is equal to the tan of angle 𝜃. Our second solution can therefore be calculated by adding 60 degrees to 180 degrees. This is equal to 240 degrees. Any further solutions found by adding or subtracting multiples of 180 degrees will be outside the required interval. So the solution set is 60 degrees and 240 degrees.

In our next example, we will consider a problem where the interval is given in radians.

Find the value of 𝜃 that satisfies the csc of 𝜃 minus root two equals zero where 𝜃 lies on the open interval from zero to 𝜋 over two.

We begin by recalling that the csc of angle 𝜃 is the reciprocal of sin 𝜃. In order to solve the equation we’re given, we begin by adding root two to both sides. This means that the csc of 𝜃 is equal to root two. The sin of angle 𝜃 is therefore equal to one over root two. 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.

It is important to note at this point that the interval was given in radians. We know that 𝜋 radians is equal to 180 degrees. This means that we’re only interested in solutions between zero and 90 degrees. We can therefore conclude that the value of 𝜃 that satisfies the csc of 𝜃 minus root two equals zero between zero and 90 degrees is 45 degrees.

In our next example, we will solve a trigonometric equation by changing the interval over which solutions exist.

Find the solution set of 𝜃 that satisfies root three multiplied by the csc of 90 degrees minus 𝜃 minus two equals zero, where 𝜃 lies on the closed interval from zero degrees to 180 degrees.

In this question, we begin by considering the reciprocal trigonometric functions. We know that the csc of any angle 𝛼 is equal to one over the sin of angle 𝛼. Before using this identity, we will let 𝛼 equal 90 degrees minus 𝜃. This enables us to solve the simpler equation root three csc of 𝛼 minus two equals zero. Adding two to both sides of this equation and then dividing through by root three, we have the csc of 𝛼 equals two over root three. Using the reciprocal identity, we have sin 𝛼 is equal to one over two over root three, which is equal to root three over two.

From our knowledge of special angles, we know that the sin of 60 degrees is equal to root three over two. This means that a solution to the equation sin 𝛼 equals root three over two is 𝛼 equals 60 degrees. We are looking for solutions of 𝜃 between zero and 180 degrees inclusive. To write this interval in terms of 𝛼, we subtract 90 degrees from the inequality such that 𝛼 is greater than or equal to negative 90 degrees and less than or equal to 90 degrees. Using a CAST diagram, we can see we’re only looking for solutions in the first or fourth quadrants. As the sin of angle 𝛼 is positive, there will therefore only be a solution in the first quadrant. This is the solution we have already found: 𝛼 is equal to 60 degrees.

As 𝛼 is equal to 90 degrees minus 𝜃, then 𝜃 must be equal to 90 degrees minus 𝛼. Substituting in our value of 𝛼, we have 𝜃 is equal to 90 degrees minus 60 degrees, which gives us an answer of 30 degrees. The solution set that satisfies the equation root three multiplied by the csc of 90 degrees minus 𝜃 minus two equals zero is 30 degrees.

In our final example, we will look at a more complicated trigonometric equation.

Find the set of values satisfying two sin 𝜃 csc 𝜃 plus sec 𝜃 cot 𝜃 equals zero given that 𝜃 is greater than or equal to zero degrees and less than or equal to 360 degrees.

We will begin this question by recalling the definition of the cosecant, secant, and cotangent functions. csc of angle 𝜃 is equal to one over sin 𝜃. The sec of angle 𝜃 is equal to one over cos 𝜃. And the cot of 𝜃 is equal to one over the tan of 𝜃. We also recall that since tan 𝜃 is equal to sin 𝜃 over cos 𝜃, the cot of 𝜃 is equal to cos 𝜃 over sin 𝜃. Substituting these identities into our equation, we have two sin 𝜃 multiplied by one over sin 𝜃 plus one over cos 𝜃 multiplied by cos 𝜃 over sin 𝜃 is equal to zero. The first term on the left-hand side simplifies to two, and the second term simplifies to one over sin 𝜃.

Subtracting two from both sides, we have one over sin 𝜃 equals negative two. This means that the csc of 𝜃 equals negative two, which in turn tells us that sin of angle 𝜃 is equal to negative one-half. We can solve this equation by firstly sketching a CAST diagram. As the sign of angle 𝜃 is negative, there will be solutions in the third and fourth quadrants.

From our knowledge of special angles, we know that the sin of 30 degrees is equal to a half. We can then use our knowledge of the symmetry of the sine function to calculate the solutions in the third and fourth quadrants. In the third quadrant, we have 𝜃 is equal to 180 degrees plus 30 degrees. This is equal to 210 degrees. In the fourth quadrant, we have 𝜃 is equal to 360 degrees minus 30 degrees. This is equal to 330 degrees. The set of values that satisfy two sin 𝜃 csc 𝜃 plus sec 𝜃 cot 𝜃 equals zero where 𝜃 lies between zero and 360 degrees inclusive are 210 and 330 degrees.

We will now summarize the key points from this video. In order to solve reciprocal trigonometric equations, we use the definition of the cosecant, secant, and cotangent functions. The csc of angle 𝜃 is equal to one over sin 𝜃. The sec of angle 𝜃 is equal to one over the cos of angle 𝜃. And the cot of angle 𝜃 is equal to one over the tan of angle 𝜃. This is also equal to the cos of angle 𝜃 over the sin of angle 𝜃.

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. Using the fact that the trigonometric functions are periodic, we can develop this further to find a general solution to trigonometric equations. However, this is outside of the scope of this video.

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