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Part 1: Find the tangent line approximation to cos x at x=π/4.


Part 2: To one decimal place, estimate the maximum value of the error on the interval 0≤x≤π/2.

Verified Answer :

sdmatera

Part 1: To find the tangent line, we need a point and a slope because we are using point slope form: y-y1=m(x-x1)
To get the y-value for our point, plug pi/4 into the original as x
y=cos(x)
y=cos(pi/4)
y=√2/2
Then to find a slope, we need to get the derivative:
y=cos(x)
y'=-sin(x) (and plug in pi/4 as x to find the slope at this point)
=-sin(pi/4)
=-√2/2

So the tangent line is y-√2/2=-√2/2(x-pi/4)

Part 2: I'm not positive on this but I think you're supposed to use this tangent line to approximate the value at x=pi/2 and find the degree of error. 
y-√2/2=-√2/2(pi/2-pi/4)
y=-pi√2/8+√2/2
y=1.3 rounded to the nearest tenth

So since we know that cos(pi/2) is actually 0, the maximum value of error would be 1.3 (1.3-0=1.3)

Hope this helps


yoodyannapolis

Equation of the tangent line is \({2}(y+x)= \frac{\pi}{4}+1\) and maximum value of the error is 1.3

Part 1: Equation of the tangent line is \(y-y_{1} =m(x-x_{1} )\)

Let  y=cos(x) then  

\(y(\frac{\pi}{4}) =cos(\frac{\pi}{4}) \\y=\frac{1}{\sqrt{2} }\)

Now, find a slope  

\(y=cos(x)\\y'=-sin(x)\\y'(\frac{\pi}{4} ) =-\frac{1}{\sqrt{2} }\)


Equation of the tangent line is:

\(y-\frac{1}\sqrt{2} } =-\frac{1}{\sqrt{2} } (x-\frac{\pi}{4} )\\{2}(y+x)= \frac{\pi}{4}+1\)


Part 2: Tangent line to approximate the value at x=pi/2  

\(y=\frac{-\sqrt{2} \pi}{8}+\frac{1}{\sqrt{2} } \\y=1.3\)


The maximum value of error is 1.3

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