Video: Selecting a Trigonometric Expression Equal to a Given One Using Cofunction and Double Angle Identities

Which of the following is equal to √(1 βˆ’ cos 2π‘₯)? [A] |sin π‘₯| [B] 2|cos π‘₯| [C] √2|cos π‘₯| [D] 2|sin π‘₯| [E] √2|sin π‘₯|

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Video Transcript

Which of the following is equal to the square root of one minus the cos of two π‘₯? (A) The absolute value of the sin of π‘₯. (B) Two times the absolute value of the cos of π‘₯. (C) The square root of two times the absolute value of the cos of π‘₯. (D) Two times the absolute value of the sin of π‘₯. (E) The square root of two times the absolute value of the sin of π‘₯.

Okay, we want to see about converting this given expression to one of the five of our answer options. Thinking along those lines, the first thing we can notice is that we’re taking the cos of two times some angle π‘₯. This suggests we make use of the double-angle identity of the cosine function. And in fact, there are three different forms that this identity takes. We can choose any of them. But notice that if we choose this third one, then upon making that substitution for cos of two π‘₯ under our square root, we would have negative one being added to positive one adding up to zero. This would simplify the expression under the square root. So let’s indeed choose this third form of the double-angle identity.

When we make this substitution, indeed we find out that this negative one added to a positive one gives us zero. And multiplying all the sines through, we get the square root of two times the sin squared of π‘₯. This equals the square root of two times the square root of the sin squared of π‘₯. And here we have to be careful because we might be tempted to say that the square root of the sin squared of π‘₯ equals simply the sin of π‘₯. Note though that while the square root of the sin squared of π‘₯ would never be negative, sin π‘₯ by itself could be. As we simplify this expression then, we’ll want to include absolute value bars around the sin of π‘₯. This ensures that no matter what the value of π‘₯, we’ll never get a negative overall result.

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