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Lesson Video: Simple Trigonometric Equations Mathematics • 10th Grade

In this video, we will learn how to find angle measures given interval and function values.

16:54

Video Transcript

In this video, we will learn how to find the general solution of a trigonometric equation or how to solve it over a specified interval. A trigonometric equation is an equation which involves at least one of the following: a trig function such as sine, cosine, and tangent; a reciprocal trig function, cosecant, secant, and cotangent; or an inverse of any of these. Some of the simpler examples of such equations can be solved without using a calculator. In these cases, we will use our knowledge of the special angles together with the symmetry and periodicity of the sine, cosine, and tangent graphs.

We will begin this video by recalling the exact values for the sine, cosine, and tangent of a number of special angles. It is important that we are able to recall the sine, cosine, and tangent of zero, 30, 45, 60, and 90 degrees. It is also important to know the corresponding values of these angles in radians. Whilst we’ll not consider the proofs of these in any detail in this video, it is important to recall that they come from our knowledge of right-angle trigonometry and the Pythagorean theorem. The exact values of sine, cosine, and tangent of the angles given are shown. Recalling the identity tan πœƒ is equal to sin πœƒ over cos πœƒ, we can calculate the tangent of any of these angles by dividing the value of the sine of the angle by the value of the cosine of the angle.

In our first example, we will demonstrate how to use the symmetry of the graph of the sine function alongside this table of values to find all solutions to a simple trig equation.

What is the general solution of sin πœƒ is equal to root two over two?

In order to find the general solution to a trigonometric equation, we begin by finding a particular solution. In this case, the table of exact trig values can help. For any angle πœƒ given in radian measure, the exact values of the sine function are as shown. We observe that the sin of πœ‹ by four radians is equal to root two over two. This means that πœƒ equals πœ‹ over four is a particular solution to the equation sin πœƒ equals root two over two. To find further solutions, we sketch the graph of 𝑦 equals sin πœƒ between zero and two πœ‹. The solutions to sin πœƒ equals root two over two are found by adding the line 𝑦 equals root two over two to the diagram.

We notice that this intersects the curve twice between zero and two πœ‹. The first point of intersection corresponds to the solution πœ‹ over four. Since the sine curve has symmetry about πœ‹ over two in the interval from zero to πœ‹, the second solution is found by subtracting πœ‹ over four from πœ‹. This is equal to three πœ‹ over four. We now have two solutions to the equation sin πœƒ equals root two over two: πœƒ equals πœ‹ over four and πœƒ equals three πœ‹ over four. Recalling that the sine function is periodic with a period of 360 degrees or two πœ‹ radians, we can find the general solution. Firstly, we have πœƒ is equal to πœ‹ over four plus two π‘›πœ‹ β€” we can write it like this because further solutions are found by adding or subtracting multiples of two πœ‹ or 360 degrees β€” and secondly, three πœ‹ over four plus two π‘›πœ‹, where 𝑛 is an integer.

In this question, we demonstrated how to interpret the symmetry of the graph of the sine function to find all solutions to an equation. Another way of extending the domain of the sine function is the unit circle. Recalling that the unit circle is centered at the origin with a radius of one unit, we can work out the sine of any angle πœƒ by starting at the point one, zero and traveling along the circumference of the circle in a counterclockwise direction until the angle that is formed between this point, the origin, and the positive π‘₯-axis is equal to πœƒ. If this point has coordinates π‘₯, 𝑦, then sin πœƒ is equal to the value of 𝑦. The value of the 𝑦-coordinate is positive in both the first and second quadrants. Hence, the value of sin πœƒ will also be positive in these quadrants.

Since the unit circle has reflection symmetry about the 𝑦-axis, we can see that sin πœƒ is equal to sin of 180 degrees minus πœƒ. By continuing to move along the circumference of the unit circle, we see that sin πœƒ is also equal to sin of 360 plus πœƒ for all values of πœƒ. These results can be generalized as shown where the set of all solutions to sin πœƒ equals 𝐢 is πœƒ equals πœƒ sub one plus 360𝑛 and πœƒ equals 180 minus πœƒ sub one plus 360𝑛, where 𝑛 is an integer. Note that if πœƒ is measured in radians, 360 degrees is replaced by two πœ‹ and 180 degrees by πœ‹. Whilst we might feel inclined to memorize these formulae, in practice, it could be much more effective to sketch the graph of the function or the unit circle.

In our next example, we will look at how to use the symmetry of the graph of the cosine function to solve a trigonometric equation.

Find the set of values satisfying the cos of πœƒ minus 105 is equal to negative a half, where πœƒ is greater than zero degrees and less than 360 degrees.

In order to find the solutions to a trig equation in a given interval, we begin by finding a particular solution. In this case, the table of exact trigonometric values can help. We will first redefine the argument of the function by letting 𝛼 equal πœƒ minus 105 such that the cos of 𝛼 is equal to negative one-half and πœƒ is equal to 𝛼 plus 105. We can then amend the interval over which our solutions are valid by adding 105 to each part of the inequality; 𝛼 is greater than 105 degrees and less than 465 degrees. Filling in the table for the exact values of cos 𝛼, we can see that cos 𝛼 equals a half when 𝛼 is 60 degrees. However, there are no values of 𝛼 in the table such that cos 𝛼 is equal to negative a half.

By sketching the graph of the cosine function along with the lines 𝑦 equals one-half and 𝑦 equals negative a half, we can find the associated value of 𝛼. It appears on the graph that there might be three values between 105 and 465 degrees. As the graph has rotational symmetry between zero and 180 degrees about 90 degrees, zero, the first solution is equal to 180 minus 60. This is equal to 120 degrees, which lies in the required interval. Next, using symmetry of the curve, we have 𝛼 is equal to 180 plus 60. This is equal to 240 degrees, which also lies in the given interval. The third solution corresponds to 120 plus 360 degrees. However, this value of 480 degrees lies outside of our interval for 𝛼. Hence, the solutions to cos 𝛼 equals negative one-half are 𝛼 equals 120 degrees and 𝛼 equals 240 degrees.

We can now calculate the corresponding values of πœƒ. 120 plus 105 is equal to 225, and 240 plus 105 is 345. The set of values that satisfy cos of πœƒ minus 105 equals negative one-half are 225 degrees and 345 degrees. An alternative technique to find the particular solution to cos of 𝛼 equals negative one-half is to use the inverse cosine function such that 𝛼 is equal to the inverse cos of negative one-half, which is equal to 120 degrees. From this point, we would use the same steps to find the other solutions. This could also have been done using the unit circle, which would lead us to the general rule: the cos of πœƒ is equal to the cos of 360 degrees minus πœƒ.

Using the symmetry of the unit circle and periodicity of the cosine function, we can quote formulas for the general solution to equations involving this function. In the same way as we have already seen for the sine function, then the set of all solutions to cos πœƒ equals 𝐢 is πœƒ is equal to πœƒ sub one plus 360𝑛 and πœƒ is equal to 360 minus πœƒ sub one plus 360𝑛 for all integer values of 𝑛. Once again, if πœƒ is measured in radians, we replace 360 degrees with two πœ‹ radians.

We saw in the previous question how to solve a trig equation where the argument of the function has been transformed in some way. We will now look at a similar version of this involving the tangent function.

Find the set of values satisfying the tan of two π‘₯ plus πœ‹ over five is equal to negative one, where π‘₯ is greater than or equal to zero and less than or equal to two πœ‹.

To solve this equation, we’ll begin by redefining the argument as this will allow us to use the symmetry of the tangent function. We will let πœƒ equal two π‘₯ plus πœ‹ over five. This means that we need to solve the tan of πœƒ equals negative one where πœƒ is greater than or equal to πœ‹ over five and less than or equal to 21πœ‹ over five as we multiply each part of the inequality by two and then add πœ‹ over five. Next, we recall that for πœƒ measured in radians, the exact values of tan πœƒ are as shown. We see that tan of πœ‹ over four is equal to one. Next, we will sketch the graph of 𝑦 equals the tan of πœƒ. We will then add the horizontal lines where 𝑦 equals one and 𝑦 equals negative one.

Due to the rotational symmetry of the tangent function, the first solution occurs when πœƒ is equal to πœ‹ minus πœ‹ over four. This is equal to three πœ‹ over four. As the function is periodic with a period of πœ‹ radians, we can find the remaining solutions by adding multiples of πœ‹ to this value. Firstly, three πœ‹ over four plus πœ‹ is equal to seven πœ‹ over four. We also have solutions 11πœ‹ over four and 15πœ‹ over four. These are the four points of intersection shown on the graph. Clearing some space and rewriting our four solutions for πœƒ, we can now calculate the values of π‘₯. As πœƒ is equal to two π‘₯ plus πœ‹ over five, two π‘₯ is equal to πœƒ minus πœ‹ over five. Dividing through by two, we have π‘₯ is equal to πœƒ over two minus πœ‹ over 10.

We can now substitute each of our values of πœƒ into this equation. This gives us four values of π‘₯ equal to 11πœ‹ over 40, 31πœ‹ over 40, 51πœ‹ over 40, and 71πœ‹ over 40. This is the set of values that satisfies the equation tan of two π‘₯ plus πœ‹ over five equals negative one where π‘₯ lies between zero and two πœ‹ inclusive.

As we have done for the sine and cosine functions, we can now quote the general solutions to equations involving the tangent function. When πœƒ is measured in degrees, the solutions are πœƒ is equal to πœƒ sub one plus 180𝑛 where 𝑛 is an integer. And if πœƒ is measured in radians, we have πœƒ is equal to πœƒ sub one plus π‘›πœ‹ where, once again, 𝑛 is an integer.

In this video, we’ve only considered the standard trigonometric functions sine, cosine, and tangent. Whilst we’ll not cover them here, it is important to understand that the process holds for the reciprocal functions cosecant, secant, and cotangent.

We will now recap the key points from this video. We can solve simple trigonometric equations using tables of exact values or the inverse trig functions. To help us calculate all solutions to a given equation in a specified range, we can draw the graph of the necessary trig function or use the unit circle. The symmetry and periodicity of the sine, cosine, and tangent functions allow us to calculate further solutions to trigonometric equations or general solutions involving integer multiples of 360 degrees or two πœ‹ radians for sine and cosine and 180 degrees or πœ‹ radians for tangent.

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