Question

on the following unit circle, θ is in radians. which two of the following expressions are equivalent to cos(θ)? choose 2 answers: a cos(-θ) b cos(π - θ) c cos(π + θ) d cos(2π + θ)

Explanation:

Step1: Recall cosine even - odd identity

The cosine function is an even function, which means that for any angle \(\theta\), \(\cos(-\theta)=\cos(\theta)\). This is a fundamental property of the cosine function, as the cosine of an angle and the cosine of its negative (which is a reflection over the \(x\) - axis in the unit circle) give the same \(x\) - coordinate on the unit circle.

Step2: Recall cosine periodicity

The cosine function has a period of \(2\pi\). That is, for any angle \(\theta\), \(\cos(\theta + 2k\pi)=\cos(\theta)\) where \(k\in\mathbb{Z}\). When \(k = 1\), we have \(\cos(2\pi+\theta)=\cos(\theta)\) because adding \(2\pi\) to an angle \(\theta\) in the unit circle brings us back to the same point on the unit circle, so the \(x\) - coordinate (which is \(\cos\) of the angle) remains the same.

Step3: Analyze \(\cos(\pi-\theta)\)

Using the cosine difference formula \(\cos(A - B)=\cos A\cos B+\sin A\sin B\), for \(A=\pi\) and \(B = \theta\), we have \(\cos(\pi-\theta)=\cos\pi\cos\theta+\sin\pi\sin\theta\). Since \(\cos\pi=- 1\) and \(\sin\pi = 0\), \(\cos(\pi-\theta)=-\cos\theta\).

Step4: Analyze \(\cos(\pi+\theta)\)

Using the cosine sum formula \(\cos(A + B)=\cos A\cos B-\sin A\sin B\), for \(A=\pi\) and \(B=\theta\), we have \(\cos(\pi + \theta)=\cos\pi\cos\theta-\sin\pi\sin\theta\). Since \(\cos\pi=-1\) and \(\sin\pi = 0\), \(\cos(\pi+\theta)=-\cos\theta\).

Answer:

A. \(\cos(-\theta)\), D. \(\cos(2\pi+\theta)\)