Exercise
Find an approximate value of
\sin 32^{\circ} \tan 40^{\circ}
Final Answer
Solution
We will use the 2-variable linear approximation formula
f(x,y)\approx f(x_0,y_0)+f'_x(x_0,y_0)\cdot(x-x_0)+f'_y(x_0,y_0)\cdot(y-y_0)
Therefore, we need to define the following
x, y, x_0, y_0, f(x)
And place them in the formula. x and y will be the numbers that appear in the question, and the points
x_0,y_0
will be the closest values to x and y respectively, which are easy for calculations.
In our exercise, we will define
x=32^{\circ}, y=40^{\circ}
because these values appear in the question. And we set
x_0=30^{\circ}=\frac{\pi}{6}
y_0=45^{\circ}=\frac{\pi}{4}
because they are the closest values to x and y respectively that are easy for calculations.
After we have defined x and y, it is easy to find the function. Just put x instead of the number we set to be x and put y instead of the number we set to be y. In this way, we get the function
f(x,y)=\sin x\cdot\tan y
In the formula above we see the function partial derivatives. Hence, we calculate them.
f'_x(x,y)=\cos x\cdot \tan y
f'_y(x,y)=\sin x\cdot\frac{1}{\cos^2 y}
We put all the data in the formula and get
f(\frac{\pi}{6},\frac{\pi}{4})\approx f(\frac{\pi}{6},\frac{\pi}{4})+f'_x(\frac{\pi}{6},\frac{\pi}{4})\cdot(32^{\circ}-30^{\circ})+f'_y(\frac{\pi}{6},\frac{\pi}{4})\cdot(40^{\circ}-45^{\circ})=
=\sin \frac{\pi}{6}\cdot\tan\frac{\pi}{4}+\cos\frac{\pi}{6}\cdot\tan\frac{\pi}{4}\cdot 2^{\circ}+\sin\frac{\pi}{6}\cdot\frac{1}{\cos^2(\frac{\pi}{4})}\cdot(-5^{\circ})=
=\frac{1}{2}\cdot 1+\frac{\sqrt{3}}{2}\cdot 1\cdot\frac{\pi}{90}+\frac{1}{2}\cdot\frac{1}{{(\frac{\sqrt{2}}{2})}^2}\cdot(-\frac{\pi}{36})=
=\frac{1}{2}+\frac{\sqrt{3}}{2}\cdot\frac{\pi}{90}+\frac{1}{2}\cdot 2\cdot(-\frac{\pi}{36})=
=\frac{1}{2}+\frac{\sqrt{3}}{2}\cdot\frac{\pi}{90}-\frac{\pi}{36}=
=\frac{1}{2}+\frac{\sqrt{3}\pi}{180}-\frac{\pi}{36}=
=\frac{90+\sqrt{3}\pi-5}{180}=
=\frac{85+\sqrt{3}\pi}{180}
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