Question

If Sin A = 6/10 and Tan B = 5/12 where A and B are acute angles, find Sin (A+B), and cos 2B without using a calculator. All calculations must be carried out as fractions.

Answer 1

Answer on Question #84714 – Math – Trigonometry

Question

If $\sin A = 6/10$ and $\tan B = 5/12$ where $A$ and $B$ are acute angles, find $\sin(A + B)$, and $\cos(2B)$ without using a calculator. All calculations must be carried out as fractions.

Solution

A and B are acute angles => 0 < cos A < 1, 0 < cos B < 1

Pythagorean identity

$$\sin^2 A + \cos^2 A = 1$$ $$\sin^2 B + \cos^2 B = 1$$

Then

$$\left\{ \begin{array}{c} \cos^2 A = 1 - \sin^2 A \\ 0 < \cos A < 1 \end{array} \right. => \cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{6}{10}\right)^2} = \frac{8}{10},$$ $$\left\{ \begin{array}{c} \sin^2 B \\ \cos^2 B + \cos^2 B \\ 0 < \cos B < 1 \end{array} \right. => \left\{ \begin{array}{c} \tan^2 B + 1 = \frac{1}{\cos^2 B} \\ 0 < \cos B < 1 \end{array} \right. =>$$ $$\cos B = \sqrt{\frac{1}{\tan^2 B + 1}} = \sqrt{\frac{1}{\left(\frac{5}{12}\right)^2 + 1}} = \frac{12}{13},$$ $$\sin B = \tan B \cos B = \frac{5}{12} \left(\frac{12}{13}\right) = \frac{5}{13},$$ $$\sin (A + B) = \sin A \cos B + \cos A \sin B = \frac{6}{10} \left(\frac{12}{13}\right) + \frac{8}{10} \left(\frac{5}{13}\right) = \frac{112}{130} = \frac{56}{65},$$ $$\cos (2B) = 2 \cos^2 B - 1 = 2 \left(\frac{12}{13}\right)^2 - 1 = \frac{119}{169}.$$

Answer:

$$\sin (A + B) = \frac{56}{65},$$ $$\cos (2B) = \frac{119}{169}.$$

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