Question

If sina + sinb = m and cosa + cosb = n then prove that cos(a-b) = 1/2(m^2 + n^2 -2)​

Answer 1

Given

• sin a + sin b = m,
• cos a + cos b = n.

Prove that

• cos (a - b) = 1/2(m² + n² -2)​

Solution

• cos (a - b) =
• cos a cos b + sin a sin b=
• 1/2( 2cos a cos b + 2sin a sin b) =
• 1/2( cos²a + 2cos a cos b + cos²b + sin²a + 2sin a sin b + sin²b - (cos²a + sin²a + cos²b + sin²b)) =
• 1/2( (cos a + cos b)² + (sin a + sin b)² - (1 + 1)) =
• 1/2(n² + m² - 2)

Proved

Used identities

• cos (a - b) = cos a cos b + sin a sin b,
• sin²a + cos²a = 1,
• (a + b)² = a² + 2ab + b²