Question

Use co-function and expansion of cos[(90-a)- B] to show that: sin(a+b) = sinacosb +sinbcosa​

Answer 1

Explanation:

`\cos((90 - a) - b) = \cos(90 - a) \cos(b) + \sin(90 - a) \sin(b)`

`\sin(a) \cos(b) + \cos(a) \sin(b)`