If (sin^4 X)/a + (cos^4 X)/b = 1Prove that 1/(a+b)^3 = (sin^8 X)/a^3 + (cos^8 Y)/b^3

giorgiana1976 | Student

Since `(sin^4 X)/a + (cos^4 Y)/b = 1`  and `1 = ((a+b)^3*(sin^8 X))/(a^3) + ((a+b)^3*(cos^8 Y))/(b^3), ` we'll equate them and we'll check if the expression represents an identity.

`(sin^4 X)/a + (cos^4 Y)/b = ((a+b)^3*(sin^8 X))/(a^3) + ((a+b)^3*(cos^8 Y))/(b^3)`

We'll multiply the 1st fraction from the left side by `a^2*b^3`  and the second fraction by `a^3*b^2` .

`[a^2*b^3*(sin^4 X) + a^3*b^2*(cos^4 Y)]/(a^3*b^3) = [b^3*(a+b)^3*(sin^8 X) + a^3*(a+b)^3*(cos^8 Y)]/(a^3*b^3)`

Since the denominators are equal, we'll have to prove that the numerators are the same:

`[a^2*b^3*(sin^4 X) + a^3*b^2*(cos^4 Y)] = [b^3*(a+b)^3*(sin^8 X) + a^3*(a+b)^3*(cos^8 Y)]`

`` `[a^2*b^3*(sin^4 X) + a^3*b^2*(cos^4 Y)]/(a+b)^3= b^3*(sin^8 X) + a^3*(cos^8 Y)` `{a^2*b^2*[b*(sin^4 X) + a*(cos^4 Y)]}/(a+b)^3= b^3*(sin^8 X) + a^3*(cos^8 Y)`

Since we cannot use the Pythagorean identity `sin^2 a + cos^2 a = 1` , because the angles x and y are not the same, the expression above does not represent an identity. 

qwertymak | Student

Sorry its (sin^4 X)/a + (cos^4 Y)/b = 1

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