Question

if sin⁴x/a + cos⁴x/b = 1/(a+b) then value of sin ⁸x/a³+cos⁸x/b³=? ​

Answer 1

Step-by-step explanation:

Sin⁴x/a + cos⁴x/b = 1/(a + b)

we know, sin²x + cos²x = 1 ⇒cos²x = 1 - sin²x

cos⁴x = (1 - sin²x)² = 1 + sin⁴x - 2sin²x , use it above

sin⁴x/a + (1 + sin⁴x - 2sin²x)/b = 1/(a + b)

sin⁴x/a + 1/b + sin⁴x/b - 2sin²x/b = 1/(a + b)

sin⁴x(1/a + 1/b) -2sin²x/b = 1/(a + b) - 1/b

sin⁴x(a + b)/ab - 2asin²x/ab = (b - a - b )/(a + b)b

(a + b)²(sin²x)² - 2a(a + b) sin²x = -a²

{(a + b)sin²x}² -2.a.(a + b)sin²x + a² = 0 [ this is like (a - b)² = a² - 2ab + b² ]

{(a + b)sin²x - a}² = 0

sin²x = a/(a + b)

⇒1 - sin²x = cos²x = 1 - a/(a + b) = b/(a + b)

hence, sin²x = a/(a + b) and cos²x = b/(a + b)

so,

sin⁸x = a⁴/(a + b)⁴ and cos⁸x = b⁴/(a + b)⁴

now, sin⁸x/a³ = a/(a + b)⁴

cos⁸x/b³ = b/(a + b)⁴

So, sin⁸x/a³ + cos⁸x/b³ = a/(a + b)⁴ + b/(a + b)⁴ = (a + b)/(a + b)⁴ = 1/(a + b)³