Question

use the following figure below, which shows the graphs of ( y=sin x ) and ( y=cos x ) on the closed interval ( 0,2 pi ), to answer each question. (a) (2 points) when the value of ( sin x ) is a maximum, what is the corresponding value of ( cos x )? (b) (2 points) when the value of ( cos x ) is a minimum, what is the corresponding value of ( sin x )? (c) (2 points) for which ( x )-values in the interval ( 0,2 pi ) is ( sin x=cos x )?

Explanation:

Step1: Analyze the maximum of \(y = \sin x\)

The function \(y=\sin x\) has a maximum value of \(1\) on the interval \([0,2\pi]\) when \(x=\frac{\pi}{2}\).

Step2: Calculate \(\cos x\) at \(x = \frac{\pi}{2}\)

Substitute \(x=\frac{\pi}{2}\) into \(y = \cos x\), we get \(\cos(\frac{\pi}{2})=0\).

Step3: Analyze the minimum of \(y=\cos x\)

The function \(y = \cos x\) has a minimum value of \(- 1\) on the interval \([0,2\pi]\) when \(x=\pi\).

Step4: Calculate \(\sin x\) at \(x=\pi\)

Substitute \(x = \pi\) into \(y=\sin x\), we get \(\sin(\pi)=0\).

Step5: Solve \(\sin x=\cos x\)

We know that \(\sin x=\cos x\) implies \(\tan x = 1\) (since \(\tan x=\frac{\sin x}{\cos x}\), \(\cos x
eq0\) when \(\tan x\) is defined). The general solution of \(\tan x=1\) is \(x = n\pi+\frac{\pi}{4},n\in\mathbb{Z}\).
For \(n = 0\), \(x=\frac{\pi}{4}\); for \(n = 1\), \(x=\pi+\frac{\pi}{4}=\frac{5\pi}{4}\) in the interval \([0,2\pi]\).

Answer:

(a) When \(\sin x\) is maximum (\(x = \frac{\pi}{2}\)), \(\cos x=0\).
(b) When \(\cos x\) is minimum (\(x=\pi\)), \(\sin x = 0\).
(c) The \(x -\)values for which \(\sin x=\cos x\) in \([0,2\pi]\) are \(x=\frac{\pi}{4}\) and \(x=\frac{5\pi}{4}\).