Video: Using Pythagorean Identities to Evaluate Trigonometric Expressions

Find the value of cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽, given sin 𝛼 = 4/5 where 𝛼 ∈ (πœ‹/2, πœ‹) and 5 cos 𝛽 βˆ’ 3 = 0, where 𝛽 ∈ (3πœ‹/2, 2πœ‹).

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Video Transcript

Find the value of cos 𝛼 cos 𝛽 plus sin 𝛼 sin 𝛽, given sin 𝛼 is equal to four-fifths, where 𝛼 is an element of the open interval πœ‹ by two to πœ‹, and five cos 𝛽 minus three equals zero, where 𝛽 is an element of the open interval three πœ‹ by two to two πœ‹.

We’ve been given the value of sin 𝛼 and an equation involving cos of 𝛽. To answer this question, it’s quite clear we’re going to need to work out the value of cos of 𝛼 and sin of πœ‹. Let’s begin with the information about sin of 𝛼. Let’s begin with sin of 𝛼. In order to work out the value of cos of 𝛼, we’re not going to solve by taking the inverse sin of both sides. Instead, we’re going to recognize that sin of 𝛼 equals four-fifths can be represented using a right-angle triangle.

We know that the sine ratio tells us that sine of some angle is equal to the length of the opposite side divided by the length of the hypotenuse. So if we construct a right-angle triangle with an included angle of 𝛼, the side opposite that must be four units and the hypotenuse must be five. And then, we spot that we have a Pythagorean triple. We know that the numbers three, four, five form a Pythagorean triple. That is, three squared plus four squared equals five squared. And so the side adjacent to our angle 𝛼 must be three units.

Now, since cos of the angle is equal to the adjacent divided by the hypotenuse, in this exact triangle, we can say that cos of 𝛼 must be equal to three-fifths. But we’ve not yet used this information. That is, 𝛼 is an element of the open interval from πœ‹ by two to πœ‹ radians. In our triangle, we’ve assumed that 𝛼 is an acute angle. So to work out the exact value of cos of 𝛼 based on the fact that sin of 𝛼 is equal to four-fifths, we’re going to use the CAST diagram.

Remember, a CAST diagram tells us the sine-tells us the sine of a trigonometric ratio dependent on which quadrant the angle falls in. Our angle is in the open interval from πœ‹ by two to πœ‹ radians. So it must lie in this second quadrant. In this quadrant, sine is positive; cos, however, is not. And therefore, it must be negative. So we get that sin of 𝛼 is equal to four- fifths, but cos of 𝛼 in this quadrant must be equal to negative three-fifths.

Let’s repeat this process with the information about cos of 𝛽. We’re going to manipulate the equation a little bit first by adding three to both sides and then dividing through by five. So this time, we find that cos of 𝛽 is equal to three-fifths. Now, in fact, if we go back to our earlier triangle, but this time label the angle 𝛽 instead of 𝛼, we know that the adjacent is three and the hypotenuse must be five. So the opposite side in this triangle is still four. So sin of 𝛽 would be equal to four-fifths in this triangle. But we’re going to need to consider the CAST diagram.

This time, our angle is in the open interval from three πœ‹ by two to two πœ‹. This is in the fourth quadrant. And over in this quadrant for angles between three πœ‹ by two and two πœ‹, cos of that angle is positive. This means sine of the angle is negative. And so for cos of 𝛽 is equal to three-fifths, where 𝛽 is an element of the open interval three πœ‹ by two to two πœ‹, sin of 𝛽 is negative four-fifths. And so cos of 𝛼 times cos of 𝛽 is negative three-fifths times three-fifths. And sin of 𝛼 times sin of 𝛽 is four-fifths times negative four-fifths.

We multiply fractions by simply multiplying their numerators and then multiplying their denominators. So we get negative nine twenty-fifths plus negative sixteen twenty-fifths. And, of course, since the denominators are equal, we can simply add or subtract their numerators. So we get negative 25 over 25, which is equal to negative one. And so, given the information about sin of 𝛼 and cos of 𝛽, we find cos of 𝛼 times cos of 𝛽 plus sin of 𝛼 times sin of 𝛽 to be equal to negative one.

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