Question

If \(\angle \mathrm{A}\) and \(\angle \mathrm{B}\) are both in quadrant I, and \(\sin \mathrm{A}=\frac{4}{5}\) and \(\cos \mathrm{B}=\frac{12}{13},\) evaluate each of the following, a) \(\cos (A-B)\) b) \(\sin (A+B)\) c) cos \(2 \mathrm{A}\) d) sin \(2 \mathrm{A}\)

Short answer

a) \frac{56}{65}, b) \frac{63}{65}, c) -\frac{7}{25}, d) \frac{24}{25}

Step-by-step solution

Calculate \(\cos \mathrm{A}\)

Given \(\sin \mathrm{A} = \frac{4}{5}\), use the Pythagorean identity \(\sin^2\mathrm{A} + \cos^2\mathrm{A} = 1\).Since \(\sin \mathrm{A} = \frac{4}{5}\), we have \(\sin^2 \mathrm{A} = \left(\frac{4}{5}\right)^2 = \frac{16}{25}\).Thus, \(\cos^2 \mathrm{A} = 1 - \frac{16}{25} = \frac{9}{25}\).As \mathrm{A} is in quadrant I, \(\cos \mathrm{A} = \sqrt{\frac{9}{25}} = \frac{3}{5}\).

Calculate \(\sin \mathrm{B}\)

Given \(\cos \mathrm{B} = \frac{12}{13}\), use the Pythagorean identity \(\sin^2 \mathrm{B} + \cos^2 \mathrm{B} = 1\).Since \(\cos \mathrm{B} = \frac{12}{13}\), we have \(\cos^2 \mathrm{B} = \left(\frac{12}{13}\right)^2 = \frac{144}{169}\).Thus, \(\sin^2 \mathrm{B} = 1 - \frac{144}{169} = \frac{25}{169}\).As \mathrm{B} is in quadrant I, \(\sin \mathrm{B} = \sqrt{\frac{25}{169}} = \frac{5}{13}\).

Calculate \(\cos (\mathrm{A} - \mathrm{B})\)

Use the formula for the cosine of a difference: \(\cos (\mathrm{A} - \mathrm{B}) = \cos \mathrm{A} \cos \mathrm{B} + \sin \mathrm{A} \sin \mathrm{B}\).Substitute the known values: \(\cos (\mathrm{A} - \mathrm{B}) = \left(\frac{3}{5}\right) \left(\frac{12}{13}\right) + \left(\frac{4}{5}\right) \left(\frac{5}{13}\right)\).Simplify: \(\cos (\mathrm{A} - \mathrm{B}) = \frac{36}{65} + \frac{20}{65} = \frac{56}{65}\).

Calculate \(\sin (\mathrm{A} + \mathrm{B})\)

Use the formula for the sine of a sum: \(\sin (\mathrm{A} + \mathrm{B}) = \sin \mathrm{A} \cos \mathrm{B} + \cos \mathrm{A} \sin \mathrm{B}\).Substitute the known values: \(\sin (\mathrm{A} + \mathrm{B}) = \left(\frac{4}{5}\right) \left(\frac{12}{13}\right) + \left(\frac{3}{5}\right) \left(\frac{5}{13}\right)\).Simplify: \(\sin (\mathrm{A} + \mathrm{B}) = \frac{48}{65} + \frac{15}{65} = \frac{63}{65}\).

Calculate \(\cos 2\mathrm{A}\)

Use the double angle formula for cosine: \(\cos 2\mathrm{A} = 2 \cos^2 \mathrm{A} - 1\).Substitute the known value of \(\cos \mathrm{A}\): \(\cos 2\mathrm{A} = 2 \left(\frac{3}{5}\right)^2 - 1\).Simplify: \(\cos 2\mathrm{A} = 2 \left(\frac{9}{25}\right) - 1 = \frac{18}{25} - 1 = \frac{18}{25} - \frac{25}{25} = -\frac{7}{25}\).

Calculate \(\sin 2\mathrm{A}\)

Use the double angle formula for sine: \(\sin 2\mathrm{A} = 2 \sin \mathrm{A} \cos \mathrm{A}\).Substitute the known values: \(\sin 2\mathrm{A} = 2 \left(\frac{4}{5}\right) \left(\frac{3}{5}\right)\).Simplify: \(\sin 2\mathrm{A} = 2 \left(\frac{12}{25}\right) = \frac{24}{25}\).

Key concepts

Pythagorean identity

The Pythagorean identity is a fundamental trigonometric concept that states: \ \[ \sin^2\theta + \cos^2\theta = 1 \] \
This identity helps relate the sine and cosine of an angle. If you know one, you can find the other. For example, in the problem, \sin \mathrm{A} = \frac{4}{5}, \ so \[ \sin^2 \mathrm{A} = \left(\frac{4}{5}\right)^2 = \frac{16}{25} \] \ Then, you find \cos^2\mathrm{A} \ by rearranging the identity: \[ \cos^2\mathrm{A} = 1 - \sin^2\mathrm{A} = 1 - \frac{16}{25} = \frac{9}{25} \] \ Since \theta \rm{A} is in the first quadrant where cosine is positive, \[ \cos \mathrm{A} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

cosine of a difference formula

The cosine of a difference formula is used to find the cosine of the difference of two angles. The formula is given by: \ \[ \cos (\theta - \beta) = \cos \theta \cos \beta + \sin \theta \sin \beta \] \
This is very useful when you know the individual sines and cosines of the angles but not their difference. In the problem, to find \cos (\mathrm{A} - \rm{B}), we use known values: \[ \cos (\mathrm{A} - \rm{B}) = \frac{3}{5} \cdot \frac{12}{13} + \frac{4}{5} \cdot \frac{5}{13} \] \ Simplifying, we get: \[ \cos (\mathrm{A} - \rm{B}) = \frac{36}{65} + \frac{20}{65} = \frac{56}{65} \]

sine of a sum formula

The sine of a sum formula helps to find the sine of the sum of two angles. The formula is: \ \[ \sin (\theta + \beta) = \sin \theta \cos \beta + \cos \theta \sin \beta \] \
When both individual angles' sines and cosines are known, this formula simplifies finding their sum. For the given problem, to find \sin (\mathrm{A} + \rm{B}), we use: \[ \sin (\mathrm{A} + \rm{B}) = \frac{4}{5} \cdot \frac{12}{13} + \frac{3}{5} \cdot \frac{5}{13} \] \ Simplifying, the result is: \[ \sin (\mathrm{A} + \rm{B}) = \frac{48}{65} + \frac{15}{65} = \frac{63}{65} \]

double angle formulas

Double angle formulas relate the trigonometric functions of double an angle to the functions of the angle itself. There are separate formulas for sine and cosine. \
For cosine, the formula is: \[ \cos 2\theta = 2 \cos^2 \theta - 1 \] \ For sine, it's: \[ \sin 2\theta = 2 \sin \theta \cos \theta \] \
These formulas are used to find the double angles when you know the original angles. In the given exercise, \cos 2\mathrm{A} is calculated by: \[ \cos 2\mathrm{A} = 2 \frac{3}{5}^2 - 1 = 2 \frac{9}{25} - 1 = \frac{18}{25} - \frac{25}{25} = -\frac{7}{25} \] \ To find \sin 2\mathrm{A}, use the provided values: \[ \sin 2\mathrm{A} = 2 \frac{4}{5} \cdot \frac{3}{5} = 2 \frac{12}{25} = \frac{24}{25} \]