# Multivariable Linear Approximation – A multiplication of sin and tan functions in 2 variables – Exercise 4211

Exercise

Find an approximate value of

$$\sin 32^{\circ} \tan 40^{\circ}$$

$$\frac{85+\sqrt{3}\pi}{180}$$

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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