Question

If sinA =3/5
and sinB =5/13
. Where A and B are acute angles. Find (cosAcosB+sinAsinB)

63/65
43/56
3/5
2/5

Answer 1

Answer on Question #46303 – Math - Trigonometry

If $\sin(A) = \frac{3}{5}$ , and $\sin(B) = \frac{5}{13}$ , where $A$ and $B$ are acute angles. Find $\cos(A)\cos(B) + \sin(A)\sin(B)$

Solution

Use the basic relationship between the sine and the cosine (the Pythagorean Identity):

$$\sin^2 x + \cos^2 x = 1$$

To calculate $\cos(A)$

$$\sin^2 (A) + \cos^2 (A) = 1$$ $$\cos^2 (A) = 1 - \sin^2 (A)$$ $$\cos (A) = \pm \sqrt{1 - \sin^2 (A)}$$

Here we have only "plus" because the A angle is acute angle.

$$\cos (A) = \sqrt{1 - \sin^2 (A)}$$

Substitute $\sin(A) = \frac{3}{5}$ and simplify

$$\cos A = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{25 - 9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$

The same for $\cos(B)$

$$\sin^2 B + \cos^2 B = 1$$ $$\cos^2 B = 1 - \sin^2 B$$ $$\cos B = \pm \sqrt{1 - \sin^2 B}$$

Take only $\cos(B) = \sqrt{1 - \sin^2 B}$ because the B angle is acute angle.

Substitute $\sin(B) = \frac{5}{13}$ and simplify

$$\cos (B) = \sqrt {1 - \left(\frac {5}{13}\right) ^ {2}} = \sqrt {1 - \frac {25}{169}} = \sqrt {\frac {169 - 25}{169}} = \sqrt {\frac {144}{169}} = \frac {12}{13}$$

So

$$\cos (A) \cos (B) + \sin (A) \sin (B) = \frac {4}{5} * \frac {12}{13} + \frac {3}{5} * \frac {5}{13} = \frac {48}{65} + \frac {15}{65} = \frac {63}{65}$$

Answer: $\frac{63}{65}$

www.AssignmentExpert.com