Question

a+b=60°then sinA+sinB-sinAsinB​

Answer 1

Answer:

Given :-

(A - B) = 60° .

To Find :-

sin²A + sin²B - sinA*sinB = ?

SOLUTION :-

→ sin²A + sin²B - sinA*sinB

Adding & subtracting sinA*sinB ,

→ sin²A + sin²B - sinA*sinB - sinA*sinB + sinA*sinB

→ sin²A + sin²B - 2sinA*sinB + sinA*sinB

Comparing it with a² + b² - 2ab = (a - b)² ,

→ (sinA - sinB)² + sinA*sinB

using sinC - sinD = 2sin(C-D/2)*cos(C+D/2)

→ [ 2sin(A - B/2)*cos(A+B/2) ]² + sinA*sinB

Putting (A - B)= 60° ,

→ [ 2 * sin30° * cos(A+B/2) ]² + sinA*sinB

→ (2 * 1/2 * cos(A+B/2) )² + sinA*sinB

→ cos²(A+B/2) + sinA*sinB

Now,

→ cos²(A+B/2) + (2*sinA*sinB)/2

using 2*sinA*sinB = cos(A-B) - cos(A+B)

→ cos²(A+B/2) + (1/2) [ cos(A-B) - cos(A+B) ]

→ cos²(A+B/2) + (1/2) [ cos60° - cos(A+B) ]

→ cos²(A+B/2) + (1/2) [ (1/2) - cos(A+B) ]

→ cos²(A+B/2) - cos(A+B)/2 + (1/4)

Now, using cosA = 2cos²(A/2) - 1 , we get,

→ cos²(A+B/2) - (1/2) {2cos²(A+B/2) - 1} + (1/4)

→ cos²(A+B/2) - cos²(A+B/2) - (1/2) + (1/4)

→ (1/4) - (1/2)

→ (1 - 2)/4

→ (-1)/4 (Ans).

(Nice Question).