Trigonometric relation problem

Question

If !!(sin\^4A)/a + (cos\^4A)/b = (1/(a+b))!!

Then !!(sin\^8A)/a^3 + (cos\^8A)/b^3 = ?!!

!!1)" "1/(a-b)^3!!

!!2)" "1/(a-b)^2!!

!!3)" "1/(a+b)^3!!

!!4)" "1/(a+b)^2!!

Answer

From the given equation we can find the value of !!(sin\^2A)!! using trigonometric relation properties.

!!(sin\^4A)/a + (cos\^4A)/b = (1/(a+b))!!

Since we know that !!sin\^2A + cos^2A = 1!!

Hence,

!!(sin\^4A)/a + (1-Sin\2A)^2/b = (1/(a+b))!!

!!(sin\^4A)/a + (1 + sin\^4A - 2sin\^2A)/b = (1/(a+b))!!

!!(sin\^4A)/a + 1/b + (sin\^4A)/b - (2sin\^2A)/b = (1/(a+b))!!

!!(sin\^4A)xx(1/a + 1/b) - (2sin\^2A)/b + 1/b - (1/(a+b))!!

!!(sin\^4A)xx(a+b)/(ab) - (2sin\^2A)/b + a/(b(a+b)) = 0!!

!!(sin\^4A)xx(a+b)/a - 2sin\^2A + a/(a+b) = 0!!........(i)

We can see that this is quadratic equation of !!sin\^2A!!.

We know that roots of quadratic equation !!(ax^2 + bx +c = 0)!! are !!x = (-b +- sqrt(b^2 - 4ac))/(2a)!!

Using the same relation in equation (i), we get

!!sin\^2A = (2 +- sqrt(4 - 4((a+b)/a)xx(a/(a+b))))/(2(a+b)/a)!!

!!=> sin\^2A = (2 +- sqrt(4 - 4))/(2(a+b)/a)!!

!!=> sin\^2A = 2/(2(a+b)/a)!!

!!=> sin\^2A = a/(a+b)!!

Then, !!sin\^8A = (sin\^2A)^4 = (a/(a+b))^4 = a^4/(a+b)^4!!

So, !!=> cos\^2A = 1 - sin\^2A = 1 - (a/(a+b)) = b/(a+b)!!

Similarly, !!cos\^8A = (cos\^2A)^4 = (b/(a+b))^4 = b^4/(a+b)^4!!

In the given question,

!!(sin\^8A)/a^3 + (cos\^8A)/b^3 = (a^4/(a+b)^4)/a^3 + (b^4/(a+b)^4)/b^3!!

!!=> (sin\^8A)/a^3 + (cos\^8A)/b^3 = (a^4/(a+b)^4)/a^3 + (b^4/(a+b)^4)/b^3!!

!!=> (sin\^8A)/a^3 + (cos\^8A)/b^3 = a/((a+b)^4) + b/((a+b)^4)!!

!!=> (sin\^8A)/a^3 + (cos\^8A)/b^3 = (a+b)/((a+b)^4)!!

!!=> (sin\^8A)/a^3 + (cos\^8A)/b^3 = 1/(a+b)^3!!

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