Dayaan M. answered • 3d
Earned A’s Twice in Precalculus | 5 Years of Tutoring Experience
I want to be upfront with you: as this is written, it cannot be proven, because it is not actually an identity. Let me show you why, because the work of checking is the same work you would do to prove one, and it will also show you what the problem probably meant to say.
Start with the left side. It is a difference of squares, since (a + 1)(a - 1) = a^2 - 1:
(sin x + 1)(sin x - 1) = sin^2(x) - 1
The Pythagorean identity says sin^2(x) + cos^2(x) = 1, so sin^2(x) - 1 is just -cos^2(x):
left side = -cos^2(x)
Now the right side. The thing to use here is that cot x is cos x over sin x:
cot(x) sin(x) - 2cos(x) = (cos(x)/sin(x)) sin(x) - 2cos(x)
The sin x on top and the sin x on the bottom cancel, leaving:
cos(x) - 2cos(x) = -cos(x)
right side = -cos(x)
So the left side simplifies to -cos^2(x) and the right side to -cos(x), and those are not the same function. If you notice, they DO happen to agree at a few places, like x = 90 degrees where both are 0, which is probably why the problem slipped through. But try x = 45 degrees. The left side gives -1/2, while the right side gives about -0.707. Different values, so it is not an identity.
If the right side had been cot(x) sin(x) cos(x) - 2cos^2(x), then it would work out, since that simplifies to cos^2(x) - 2cos^2(x) = -cos^2(x), which matches the left side exactly. My guess is that some exponents got lost when the problem was typed up.
So, our final answer is that the statement as written is false, and the proof only goes through if the right side is cot(x) sin(x) cos(x) - 2cos^2(x). Worth checking your original sheet to see which one it is.
