JA 5A2If cos A =-then the value of sinsiisvalue4ㅡㅡㅡ1S2(a) 1/32(c

1.	JA 5A2If cos A =-then the value of sinsiisvalue4ㅡㅡㅡ1S2(a) 1/32(c) 11/32(b) 11/8(d) 11/16
Answer» the formula for sin(A/2)sin(5A/2) is 1/2(cos(A/2-5A/2)-cos(A/2+5A/2)) so sin(A/2)sin(5A/2) = 1/2(Cos(-2A)-cos(3A)) given cosA = 3/4 , so cos2A = 2cos²A-1 = 2(3/4)²-1 = 1/8and cos3A = 4cos³A -3cosA = 4(3/4)³-3(3/4) = 27/16-9/4= -9/16 putting these values.. in the equation we get = 1/2(1/8-(-9/16)) = 1/2*(11/16) = 11/32. option C

JA 5A2If cos A =-then the value of sinsiisvalue4ㅡㅡㅡ1S2(a) 1/32(c) 11/32(b) 11/8(d) 11/16

Answer»

the formula for sin(A/2)sin(5A/2) is 1/2*(cos(A/2-5A/2)-cos(A/2+5A/2))

so sin(A/2)sin(5A/2) = 1/2*(Cos(-2A)-cos(3A))

given cosA = 3/4 , so cos2A = 2cos²A-1 = 2*(3/4)²-1 = 1/8and cos3A = 4cos³A -3cosA = 4*(3/4)³-3*(3/4) = 27/16-9/4= -9/16

putting these values.. in the equation we get = 1/2*(1/8-(-9/16)) = 1/2*(11/16) = 11/32.

option C