Question Video: Selecting a Trigonometric Expression Equal to a given One Using the Cofunction and the Double Angle Identities | Nagwa Question Video: Selecting a Trigonometric Expression Equal to a given One Using the Cofunction and the Double Angle Identities | Nagwa

Question Video: Selecting a Trigonometric Expression Equal to a given One Using the Cofunction and the Double Angle Identities Mathematics • Second Year of Secondary School

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Which of the following is equal to β(1 β sin 2π₯)? [A] |cos π₯ β sin π₯| [B] cos π₯ β sin π₯ [C] |cos π₯ + sin π₯| [D] cos π₯ + sin π₯ [E] sin π₯ β cos π₯

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Video Transcript

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

Okay, so here weβre evaluating this expression. And we want to see which of these five answer options it equals. The first thing we can notice is that weβre taking the sin of two times some angle π₯. We can think of this then as the sine of a double angle, where π₯ is that angle. Recalling the double-angle identity for the sine function, we know that the sin of two π₯ equals two times the sin of π₯ times the cos of π₯. Making this substitution into our square root gives us this result.

And now letβs consider this factor of one. By the Pythagorean identity, this unassuming number one is equal to the sine squared of an angle plus the cosine squared of that same angle. Making that substitution gives us this expression. So far, it seems as though weβre complicating rather than simplifying the expression under our square root. But now that we have these three terms β sin squared π₯, cos squared π₯, and two sin π₯ cos π₯ β say that we can write them as the quantity cos π₯ minus sin π₯ squared. We see then that our squaring operation and then taking the square root will effectively invert one another. So we might expect our final result to be the cos of π₯ minus the sin of π₯.

Here though, we have to be careful to make sure this result is truly equal to our original expression. When we think about the sine function, we know it has a maximum of one and a minimum of negative one. This means if we think about it for a moment, that one minus the sin of two π₯ will never be negative. Since this is true for our original expression, it must also be true for our final one. However, itβs certainly possible for the cos of π₯ minus the sin of π₯ to be negative. To correct this and make our expression truly equal to the square root of one minus the sin of two π₯, weβll put absolute value bars around it. And so this is our final answer. Itβs the absolute value of the cos of π₯ minus the sin of π₯ that equals the square root of one minus the sin of two π₯.

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