If sin A = \(\frac{3}{5}\), cos B = \(\frac{4}{5}\) find sin(A + B) a

1.	If sin A = \(\frac{3}{5}\), cos B = \(\frac{4}{5}\) find sin(A + B) and cos(A – B). Where A and B are acute angles.
Answer» Given sin A = \(\frac{3}{5}\), cos B = \(\frac{4}{5}\) ⇒ cos A = \(\frac{4}{5}\) & sin B = \(\frac{3}{5}\) ∵ sin²A + cos²A = 1 (i) sin(A + B) = sinA cosB + cosA sinB = \(\frac{3}{5}\). \(\frac{4}{5}\) + \(\frac{4}{5}\).\(\frac{3}{5}\) = \(\frac{24}{25}\) (ii) cos(A – B) = cosAcosB + sinA sinB = \(\frac{4}{5}\). \(\frac{4}{5}\) + \(\frac{3}{5}\) .\(\frac{3}{5}\) = \(\frac{16}{25}\) + \(\frac{9}{25}\) = \(\frac{25}{25}\) = 1

If sin A = \(\frac{3}{5}\), cos B = \(\frac{4}{5}\) find sin(A + B) and cos(A – B). Where A and B are acute angles.

Answer»

Given sin A = \(\frac{3}{5}\), cos B = \(\frac{4}{5}\)

⇒ cos A = \(\frac{4}{5}\) & sin B = \(\frac{3}{5}\)

∵ sin²A + cos²A = 1

(i) sin(A + B) = sinA cosB + cosA sinB

= \(\frac{3}{5}\). \(\frac{4}{5}\) + \(\frac{4}{5}\).\(\frac{3}{5}\) = \(\frac{24}{25}\)

(ii) cos(A – B) = cosAcosB + sinA sinB

= \(\frac{4}{5}\). \(\frac{4}{5}\) + \(\frac{3}{5}\) .\(\frac{3}{5}\) = \(\frac{16}{25}\) + \(\frac{9}{25}\) = \(\frac{25}{25}\) = 1