Chapter 7 - Section 7.1 - Trigonometric Identities - 7.1 Exercises - Page 544: 104

Verify the following identity: $sin(b)^6 + cos(b)^6 = 1 - 3 sin(b)^2 cos(b)^2$

Verify the following identity: $sin(b)^6 + cos(b)^6 = 1 - 3 sin(b)^2 cos(b)^2$ $cos(b)^6 = (cos(b)^2)^3 = 1/8 (1 + cos(2 b))^3$: $((1 + cos(2 b))^3)/(8) + sin(b)^6 = ^?1 - 3 cos(b)^2 sin(b)^2$ $(1 + cos(2 b))^3 = 1 + 3 cos(2 b) + 3 cos(2 b)^2 + cos(2 b)^3$: $1/81 + 3 cos(2 b) + 3 cos(2 b)^2 + cos(2 b)^3 + sin(b)^6 = ^?1 - 3 cos(b)^2 sin(b)^2$ $cos(2 b)^2 = 1/2 (1 + cos(4 b))$: $(1 + 3 cos(2 b) + 3×(1 + cos(4 b))/(2) + cos(2 b)^3)/(8) + sin(b)^6 = ^?1 - 3 cos(b)^2 sin(b)^2$ $(1 + cos(4 b))/2 = 1/2 + 1/2 cos(4 b)$: $(1 + 3 cos(2 b) + 3 1/2 + (cos(4 b))/(2) + cos(2 b)^3)/(8) + sin(b)^6 = ^?1 - 3 cos(b)^2 sin(b)^2$ $3 (1/2 + (cos(4 b))/2) = 3/2 + 3/2 cos(4 b)$: $(1 + 3 cos(2 b) + 3/2 + (3 cos(4 b))/(2) + cos(2 b)^3)/(8) + sin(b)^6 = ^?1 - 3 cos(b)^2 sin(b)^2$ $cos(2 b)^3 = 1/4 (3 cos(2 b) + cos(6 b))$: $(1 + 3 cos(2 b) + 3/2 + (3 cos(4 b))/(2) + (3 cos(2 b) + cos(6 b))/(4))/(8) + sin(b)^6 = ^?1 - 3 cos(b)^2 sin(b)^2$ $(3 cos(2 b) + cos(6 b))/4 = 3/4 cos(2 b) + 1/4 cos(6 b)$: $(1 + 3 cos(2 b) + 3/2 + (3 cos(4 b))/(2) + (3 cos(2 b))/(4) + (cos(6 b))/(4))/(8) + sin(b)^6 = ^?1 - 3 cos(b)^2 sin(b)^2$ $1 + 3 cos(2 b) + 3/2 + (3 cos(4 b))/(2) + (3 cos(2 b))/(4) + (cos(6 b))/4$ $= 5/2 + 15/4 cos(2 b) + 3/2 cos(4 b) + 1/4 cos(6 b)$: $1/85/2 + 15/4 cos(2 b) + 3/2 cos(4 b) + 1/4 cos(6 b) + sin(b)^6 = ^?1 - 3 cos(b)^2 sin(b)^2$ $(5/2 + (15 cos(2 b))/(4) + (3 cos(4 b))/(2) + (cos(6 b))/4)/(8)$ $= 5/16 + 15/32 cos(2 b) + 3/16 cos(4 b) + 1/32 cos(6 b)$: $5/16 + (15 cos(2 b))/(32) + (3 cos(4 b))/(16) + (cos(6 b))/(32) + sin(b)^6 = ^?1 - 3 cos(b)^2 sin(b)^2$ $sin(b)^6 = (sin(b)^2)^3 = 1/8 (1 - cos(2 b))^3$: $5/16 + (15 cos(2 b))/(32) + (3 cos(4 b))/(16) + (cos(6 b))/32 + ((1 - cos(2 b))^3)/(8) = ^?1 - 3 cos(b)^2 sin(b)^2$ $(1 - cos(2 b))^3 = 1 - 3 cos(2 b) + 3 cos(2 b)^2 - cos(2 b)^3$: $5/16 + (15 cos(2 b))/(32) + (3 cos(4 b))/(16) + (cos(6 b))/32 + 1/81 - 3 cos(2 b) + 3 cos(2 b)^2 - cos(2 b)^3 = ^?1 - 3 cos(b)^2 sin(b)^2$ $cos(2 b)^2 = 1/2 (1 + cos(4 b))$: $5/16 + (15 cos(2 b))/(32) + (3 cos(4 b))/(16) + (cos(6 b))/32 + (1 - 3 cos(2 b) + 3×(1 + cos(4 b))/(2) - cos(2 b)^3)/(8) = ^?1 - 3 cos(b)^2 sin(b)^2$ $(1 + cos(4 b))/2 = 1/2 + 1/2 cos(4 b)$: $5/16 + (15 cos(2 b))/(32) + (3 cos(4 b))/(16) + (cos(6 b))/32 + (1 - 3 cos(2 b) + 3 1/2 + (cos(4 b))/(2) - cos(2 b)^3)/(8) = ^?1 - 3 cos(b)^2 sin(b)^2$ $cos(2 b)^3 = 1/4 (3 cos(2 b) + cos(6 b))$: $5/16 + (15 cos(2 b))/(32) + (3 cos(4 b))/(16) + (cos(6 b))/32 + (1 - 3 cos(2 b) + 3 (1/2 + (cos(4 b))/2) - (3 cos(2 b) + cos(6 b))/(4))/(8) = ^?1 - 3 cos(b)^2 sin(b)^2$ (3 cos(2 b) + cos(6 b))/4 = 3/4 cos(2 b) + 1/4 cos(6 b): 5/16 + (15 cos(2 b))/(32) + (3 cos(4 b))/(16) + (cos(6 b))/32 + (1 - 3 cos(2 b) + 3 (1/2 + (cos(4 b))/2) - (3 cos(2 b))/(4) + (cos(6 b))/(4))/(8) = ^?1 - 3 cos(b)^2 sin(b)^2 $3 (1/2 + (cos(4 b))/2) = 3/2 + 3/2 cos(4 b)$: $5/16 + (15 cos(2 b))/(32) + (3 cos(4 b))/(16) + (cos(6 b))/32 + (1 - 3 cos(2 b) + 3/2 + (3 cos(4 b))/(2) - ((3 cos(2 b))/(4) + (cos(6 b))/4))/(8) = ^?1 - 3 cos(b)^2 sin(b)^2$ $-((3 cos(2 b))/(4) + (cos(6 b))/4) = -3/4 cos(2 b) - 1/4 cos(6 b)$: $5/16 + (15 cos(2 b))/(32) + (3 cos(4 b))/(16) + (cos(6 b))/32 + (1 - 3 cos(2 b) + 3/2 + (3 cos(4 b))/(2) + -3/4 cos(2 b) - 1/4 cos(6 b))/(8) = ^?1 - 3 cos(b)^2 sin(b)^2$ $1 - 3 cos(2 b) + 3/2 + (3 cos(4 b))/(2) - (3 cos(2 b))/(4) - (cos(6 b))/(4) = 5/2 - 15/4 cos(2 b) + 3/2 cos(4 b) - 1/4 cos(6 b)$: $5/16 + (15 cos(2 b))/(32) + (3 cos(4 b))/(16) + (cos(6 b))/32 + 1/85/2 - 15/4 cos(2 b) + 3/2 cos(4 b) - 1/4 cos(6 b) = ^?1 - 3 cos(b)^2 sin(b)^2$ $(5/2 - (15 cos(2 b))/(4) + (3 cos(4 b))/(2) - (cos(6 b))/(4))/(8) = 5/16 - 15/32 cos(2 b) + 3/16 cos(4 b) - 1/32 cos(6 b)$: $5/16 + (15 cos(2 b))/(32) + (3 cos(4 b))/(16) + (cos(6 b))/32 + 5/16 - (15 cos(2 b))/(32) + (3 cos(4 b))/(16) - (cos(6 b))/(32) = ^?1 - 3 cos(b)^2 sin(b)^2$ $5/16 + (15 cos(2 b))/(32) + (3 cos(4 b))/(16) + (cos(6 b))/32 + 5/16 - (15 cos(2 b))/(32) + (3 cos(4 b))/(16) - (cos(6 b))/(32)$ $= 5/8 + 3/8 cos(4 b)$: $5/8 + 3/8 cos(4 b) = ^?1 - 3 cos(b)^2 sin(b)^2$ $cos(b)^2 = 1/2 (1 + cos(2 b))$: $5/8 + (3 cos(4 b))/(8) = ^?1 - 3(1 + cos(2 b))/(2) sin(b)^2$ $(1 + cos(2 b))/2 = 1/2 + 1/2 cos(2 b)$: $5/8 + (3 cos(4 b))/(8) = ^?1 - 31/2 + (cos(2 b))/(2) sin(b)^2$ $sin(b)^2 = 1/2 (1 - cos(2 b))$: $5/8 + (3 cos(4 b))/(8) = ^?1 - 3(1 - cos(2 b))/(2) (1/2 + (cos(2 b))/2)$ $(1 - cos(2 b))/2 = 1/2 - 1/2 cos(2 b)$: $5/8 + (3 cos(4 b))/(8) = ^?1 - 31/2 - (cos(2 b))/(2) (1/2 + (cos(2 b))/2)$ $-3 (1/2 + (cos(2 b))/2) (1/2 - (cos(2 b))/(2)) = 3/4 cos(2 b)^2 - 3/4$: $5/8 + (3 cos(4 b))/(8) = ^?1 + (3 cos(2 b)^2)/(4) - (3)/4$ $cos(2 b)^2 = 1/2 (1 + cos(4 b))$: $5/8 + (3 cos(4 b))/(8) = ^?1 - 3/4 + (3×(1 + cos(4 b))/(2))/(4)$ $(1 + cos(4 b))/2 = 1/2 + 1/2 cos(4 b)$: $5/8 + (3 cos(4 b))/(8) = ^?1 - 3/4 + (3 1/2 + (cos(4 b))/(2))/(4)$ $(3 (1/2 + (cos(4 b))/2))/(4) = 3/8 + 3/8 cos(4 b)$: $5/8 + (3 cos(4 b))/(8) = ^?1 - 3/4 + 3/8 + (3 cos(4 b))/(8)$ $1 - 3/4 + 3/8 + (3 cos(4 b))/(8) = 5/8 + 3/8 cos(4 b)$: $5/8 + (3 cos(4 b))/(8) = ^?5/8 + 3/8 cos(4 b)$ The left hand side and right hand side are identical, thus verifying the identity.