Chapter A5.3, Problem 28E

To determine

To find: The sine and cosine of an angle from the give equation.

Answer to Problem 28E

The sine and cosine of an angle is,

  $\sin \alpha =\frac{1}{\sqrt{5}}$

  $\cos \alpha =\frac{2}{\sqrt{5}}$

Explanation of Solution

Given:

  $14x^2+16xy+2y^2-10x+26,370y-17=0.$

Calculation:

Comparing with the standard equation for a quadratic curve,

  $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$

Thus,

  $A=14,B=16,C=2$

For the cross-product term to vanish, the coordinate axes must be rotated by an angle $\alpha$ , such that ,

  $\begin{array}{l}\cot 2\alpha =\frac{A-C}{B}\\ =\frac{14-2}{16}\\ =\frac{3}{4}\end{array}$

Therefore,

  $\begin{array}{l}\cos 2\alpha =\frac{3}{5}\\ \left(0\le 2\alpha <\frac{\pi }{2}\right)\end{array}$

Now,

  $\sin \alpha =\sqrt{\frac{1-\cos 2\alpha }{2}}$

  $=\sqrt{\frac{1-\frac{3}{5}}{2}}$

  $\sin \alpha =\frac{1}{\sqrt{5}}$

  $\begin{array}{l}\cos \alpha =\sqrt{\frac{1+\cos 2\alpha }{2}}\\ =\sqrt{\frac{1+\frac{3}{5}}{2}}\\ =\frac{2}{\sqrt{5}}\end{array}$

  $\cos \alpha =\frac{2}{\sqrt{5}}$

Conclusion:

Hence, the sine and cosine of an angle is ,

  $\sin \alpha =\frac{1}{\sqrt{5}},$$\cos \alpha =\frac{2}{\sqrt{5}}$