Question Video: Solving Trigonometric Equations Using Trigonometric Values of Special Angles Mathematics

Find the set of values satisfying cos (𝜃 − 105) = −1/2 where 0° < 𝜃 < 360°.

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Video Transcript

Find the set of values satisfying the cos of 𝜃 minus 105 is equal to negative a half, where 𝜃 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 𝜃 minus 105 such that the cos of 𝛼 is equal to negative one-half and 𝜃 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 𝜃. 120 plus 105 is equal to 225, and 240 plus 105 is 345. The set of values that satisfy cos of 𝜃 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 𝜃 is equal to the cos of 360 degrees minus 𝜃.

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 𝜃 equals 𝐶 is 𝜃 is equal to 𝜃 sub one plus 360𝑛 and 𝜃 is equal to 360 minus 𝜃 sub one plus 360𝑛 for all integer values of 𝑛. Once again, if 𝜃 is measured in radians, we replace 360 degrees with two 𝜋 radians.

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