Video: Evaluating Determinants Involving Trigonometric Functions

Evaluate |10 cos π‘₯, βˆ’2 sin π‘₯ and 10 sin π‘₯, 2 cos π‘₯|.

02:40

Video Transcript

Evaluate the determinant of the matrix with elements 10 cos π‘₯, negative two sin π‘₯, 10 sin π‘₯, two cos π‘₯.

Firstly, let’s just clarify the notation used in the question. This pair of vertical lines on either side of the matrix mean that we’re looking to calculate the determinant of that matrix. Let’s recall what the determinant means. The determinant of the two-by-two matrix with elements π‘Ž, 𝑏, 𝑐, and 𝑑 is equal to the scalar value π‘Žπ‘‘ minus 𝑏𝑐. That’s the product of the elements on the leading diagonal, π‘Ž and 𝑑, minus the product of the elements on the other diagonal, 𝑏 and 𝑐.

So to calculate the determinant of the matrix we’ve been given then, we take the elements on the leading diagonal, which are 10 cos π‘₯ and then two cos π‘₯. And we take the product of these two elements. We then subtract the product of the elements on the other diagonal, which are negative two sin π‘₯ and 10 sin π‘₯.

So we have 10 cos π‘₯ multiplied by two cos π‘₯ minus negative two sin π‘₯ multiplied by 10 sin π‘₯. 10 multiplied by two is 20. And cos π‘₯ multiplied by cos π‘₯ is cos squared π‘₯. So the first term simplifies to 20 cos squared π‘₯.

In the second, negative two multiplied by 10 is negative 20. And sin π‘₯ multiplied by sin π‘₯ is sin squared π‘₯. So we have 20 cos squared π‘₯ minus negative 20 sin squared π‘₯. The two negatives together form a positive. If we are subtracting negative 20 sin squared π‘₯, then this is the same as adding 20 sin squared π‘₯. So we have 20 cos squared π‘₯ plus 20 sin squared π‘₯.

At this point, we need to recall one of our trigonometric identities, which is that, for any angle π‘₯, cos squared of π‘₯ plus sin squared of π‘₯ is always equal to one. We can see this using the unit circle because cos π‘₯ and sin π‘₯ give the horizontal and vertical sides in a right-angle triangle. And one is the length of the hypotenuse. So by applying the Pythagorean theorem, we prove this result.

So we can factorize 20 from our expression for the determinant of this matrix, giving 20 multiplied by cos squared π‘₯ plus sin squared π‘₯. As cos squared π‘₯ plus sin squared π‘₯ is equal to one, we have 20 multiplied by one, which is equal to 20.

So we found that the determinant of the two-by-two matrix with elements 10 cos π‘₯, negative two sin π‘₯, 10 sin π‘₯, two cos π‘₯ is 20.

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