Question

Multiple Choice Which of the following is an equation of the tangent line to y=\sin x+\cos x\( at \)x=\pi ? (A) \(y=-x+\pi-1\) (B)\(y=-x+\pi+1\) (C) \(y=-x-\pi+1\) (D) \(y=-x-\pi-1\) (E) \(y=x-\pi+1\)

Short answer

The correct answer is (A) \(y=-x+\pi-1\).

Step-by-step solution

Calculate Derivative

The derivative of \(y=\sin x+\cos x\) is found by taking the derivative separately of both \(\sin x\) and \(\cos x\) and adding them together. The derivative of \(\sin x\) is \(\cos x\) and the derivative of \(\cos x\) is \(-\sin x\). Therefore, the derivative of \(y=\sin x+\cos x\) is \(\cos x - \sin x\).

Evaluate the Derivative at Given Point

Substitute \(\pi\) into the derivative, \(y'=\cos x - \sin x\), resulting in \(y'=\cos\pi - \sin\pi\). Evaluating the cosine of \(\pi\) gives -1 and the sine of \(\pi\) gives 0, therefore, \(y'=-1-0=-1\). This means, the slope of the tangent line to the function at \(x=\pi\) is -1.

Find y-intercept for Tangent Line

The y-intercept can be found by substituting the x-value, \(\pi\), and slope, -1, into the point-slope equation \(y-y_1 = m(x-x_1)\), where \(y_1\) is the y-value of the function at \(x=\pi\) (The y-value can be found by substituting \(\pi\) into the original function \(y=\sin x+\cos x\), giving \( y_1 = \sin\pi + \cos\pi = 0 + (-1) = -1\). Hence, we get \(y+1=-1(x-\pi)\). Simplifying this equation gives the tangent line equation \(y=-x+\pi-1\).