Question

Find the value of the expression `3[sin^4 ((3pi)/2 - alpha) + sin^4 (3pi + alpha)] - 2[sin^6 (pi/2 + alpha) + sin^6 (5pi - alpha)]`

Answer 1

Let, y = `3[sin^4 ((3pi)/2 - alpha) + sin^4 (3pi + alpha)] - 2[sin^6 (pi/2 + alpha) + sin^6 (5pi - alpha)]`

We know that,

`sin((3pi)/2  –  alpha)` = –cos α

sin(3π + α) = –sin α

`sin(pi/2 + alpha)` = cos α

sin(5π – α) = sin α

Therefore,

y = 3[(–cos α)⁴ + (–sin α)⁴] – 2[cos⁶ α + sin⁶ α]

⇒ y = 3 [cos⁴α + sin⁴α] – 2[sin⁶α + cos⁶α]

⇒ y = 3[(sin²α + cos²α)² – 2sin²α cos²α] – 2[(sin²α)³ + (cos²α)³]

Since, we know that,

sin²α + cos²α = 1

Also, we know that,

a³ + b³ = (a + b)(a² – ab + b²)

⇒ y = 3[1 – 2sin²α cos²α] – 2[(sin²α + cos²α)( cos⁴α + sin⁴α – sin²α cos²α)]

⇒ y = 3[1 – 2sin²α cos²α] – 2[cos⁴α + sin⁴α – sin²α cos²α]

⇒ y = 3[1 – 2sin²α cos²α] – 2[(sin²α + cos²α)² – 2sin²α cos²α – sin²α cos²α]

⇒ y = 3[1 – 2sin²α cos²α] – 2[1 – 3sin²α cos²α]

⇒ y = 3 – 6sin²α cos²α – 2 + 6 sin²α cos²α

⇒ y = 1