Given Sina + SINB = 1 / 3, find the range of square 2b of sina cos! Given sin (pie-a) = 1 / 2, calculate cos (3 pie + a)

Given Sina + SINB = 1 / 3, find the range of square 2b of sina cos! Given sin (pie-a) = 1 / 2, calculate cos (3 pie + a)

What is the value range of the square 2b of sina cos?
Sin (pie-a) = 1 / 2 to calculate cos (3 pie + a)
Cos (3 faction + a) = cos [4 faction - (faction-a)] = cos 4 faction cos (faction-a) = ± (radical 3) / 2

cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC
ABC is a triangle, which is proved by (COSA) square - (CoSb) square = sin (a + b) * sin (B-A)
Cosa square - CoSb square + sinc square = 2cosa * SINB * sinc

Left = sin (a + b) sin (B-A) + Sin & # 178; C = sin (180-c) sin (B-A) + Sin & # 178; C = sincsin (B-A) + Sin & # 178; C = sinc [sin (B-A) + sinc] = sinc [sin (B-A) + sin (B + a)] = sinc (sinbcosa cosbsina + sinbcosa + cosbsina) = sinc * 2sinbcosa = right proposition is proved

If Tan (π / 4-A) = - 5 / 12 A belongs to (π / 4,3 π / 4) cos (3 π / 2 + 2b) = 7 / 25 B belongs to (0, π / 4), find the value of sin (a + b)

∵ a belongs to (π / 4,3 π / 4), ∵ Tan (π / 4-A) = - 5 / 12, ∵ sin (π / 4-A) = - 5 / 13, cos (π / 4-A) = 12 / 13 (1) ∵ cos (3 π / 2 + 2b) = 7 / 25, and COS (3 π / 2 + 2b) = cos [2 π - (3 π / 2 + 2b)] = cos (π / 2-2b) ∵ cos (π / 2-2b) = 7 / 25

Simplification of sin (π - a) cos (2 / π - a) / 1-cos ^ 2A

sin(π-a)cos(2/π-a)/1-cos^2a
=sina*sina/sin^2a
=1