graph the function.\ny = sec ( x - \\frac { \\pi } { 2 } )

Explanation:

Step1: Use the identity

We know that (\sec(A - B)=\frac{1}{\cos(A - B)}), and (\cos\left(x-\frac{\pi}{2}\right)=\sin x) (using the co - function identity (\cos(A - B)=\cos A\cos B+\sin A\sin B), here (A = x), (B=\frac{\pi}{2}), (\cos x\cos\frac{\pi}{2}+\sin x\sin\frac{\pi}{2}=\sin x)). So (y = \sec\left(x-\frac{\pi}{2}\right)=\frac{1}{\sin x}=\csc x)

Step2: Analyze the key features of (y = \csc x)

• Period: The period of (y=\csc x=\frac{1}{\sin x}) is (2\pi) since the period of (y = \sin x) is (2\pi) and (\csc x) is the reciprocal of (\sin x).
• Vertical asymptotes: The vertical asymptotes of (y=\csc x) occur where (\sin x = 0). So (x = n\pi), (n\in\mathbb{Z})
• Local maxima and minima: The local maxima of (y=\csc x) occur at (x=\frac{\pi}{2}+ 2n\pi), (y = 1) and local minima occur at (x=\frac{3\pi}{2}+2n\pi), (y=-1)

Step3: Plot the graph

• Mark the vertical asymptotes at (x = 0,\pm\pi,\pm2\pi,\cdots)
• Plot the points (\left(\frac{\pi}{2},1\right)), (\left(\frac{3\pi}{2},-1\right)), (\left(\frac{5\pi}{2},1\right)), (\left(-\frac{\pi}{2},-1\right)) etc.
• Sketch the two - part curves (one above the (x) - axis between ((2n\pi,(2n + 1)\pi)) and one below the (x) - axis between (((2n+1)\pi,(2n + 2)\pi))) for (n\in\mathbb{Z})

The graph of (y=\sec\left(x-\frac{\pi}{2}\right)) is the same as the graph of (y = \csc x) with vertical asymptotes at (x=n\pi), (n\in\mathbb{Z}), period (2\pi), local maxima at (x=\frac{\pi}{2}+2n\pi), (y = 1) and local minima at (x=\frac{3\pi}{2}+2n\pi), (y=-1)