Question

prove that :sin (A+B) + sin (A - B)/
cos (A+B) + cos (A - B)
= tan A​

Answer 1

Answer:

Step-by-step explanation:

[sin(A + B) + sin(A - B)] / [cos(A + B) + cos(A - B)] = tan(A)  

Recall the following two facts:  

sin(A)cos(B) = (1/2)[sin(A + B) + sin(A - B)]  

cos(A)cos(B) = (1/2)[cos(A + B) + cos(A - B)]  

Sub those into the left side:  

[2 sin(A)cos(B)] / [2cos(A)cos(B)]  

tan(A)  

Thus LS = RS and QED. Alternatively, you could expand everything out in the left side with the sine and cosine addition and subtraction formulas.  

*********  

sin(α) + sin[α +(2π)/(3)] + sin [α + (4π)/(3)] = 0  

Apply sin(a + b) = sin(a)cos(b) + sin(b)cos(a) in the left side:  

sin(α) + sin(α)cos(2π / 3) + sin(2π / 3)cos(α) + sin(α)cos(4π / 3) + sin(4π / 3)cos(α)  

Use the following facts (from the unit circle or otherwise):  

sin(2π / 3) = √(3) / 2  

sin(4π / 3) = -√(3) / 2  

cos(2π / 3) = -1 / 2  

cos(4π / 3) = -1 / 2  

sin(α) + (-1/2)sin(α) + (√(3) / 2)cos(α) - (1/2)sin(α) + (-√(3) / 2)cos(α)  

0  

Thus LS = RS and QED.  

Done!