Question

sin^4x/a+cos^4/a=1/a+b then find sin^8/a^3+cos^8/b^3=?

Answer 1

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

hence, $\boxed{\boxed{\bold{\frac{sin^8x}{a^3}+\frac{cos^8x}{b^3}=\frac{1}{(a+b)^3}}}}$

Answer 2

Given : Sin⁴x/a + cos⁴x/b = 1/a+b

To prove :
sin⁸x/a³ + cos⁸x/b³ =?

Solution:

Sin⁴x/a + cos⁴x/b = 1/a+b

As we know that Sin²x +Cos²x=1
∴Cos²x=1-sin²x
Cos⁴x=(Cos²x)²=(1-sin²x)²=1+sin⁴x-2sin²x

⇒Sin⁴x/a  +[1+sin⁴x-2sin²x]/b =1/a+b

⇒Sin⁴x/a+1/b + sin⁴x/b -2sin²x/b=(1/a+b)

⇒Sin⁴x/a+ sin⁴x/b -2sin²x/b=(1/a+b)-1/b

⇒Sin⁴x(1/a +1/b)-2sin²xa/ab=[b-a-b]/b(a+b)

⇒Sin⁴x(1/a +1/b)-2asin²x/ab=-a/b(a+b)

⇒Sin⁴x(a+b)/ab-2asin²x/ab=-a/b(a+b)


⇒(a+b)²(Sin²x)² - 2a(a+b)sin²x= - a²

⇒(a+b)²(Sin²x)² - 2a(a+b)sin²x +  a² =0
This equation is similar to a²-2ab+b²=0 ⇒(a-b)²
[(a+b) Sin²x-a]²=0
sin²x=a/a+b

Cos²x=1-sin²x= 1- a/(a+b)= b/a+b
∴sin²x=a/a+b and Cos²x= b/a+b

Now :
sin⁸x=a⁴/(a+b)⁴  and Cos⁸x= b⁴/(a+b)⁴

sin⁸x/a³= a/(a+b)⁴   and Cos⁸x/b³=b/(a+b)⁴
Now 

sin⁸x/a³ + Cos⁸x/b³ =   a/(a+b)⁴ + b/(a+b)⁴
=a+b/(a+b)⁴
⇒sin⁸x/a³ + Cos⁸x/b³=1/(a+b)³