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`        sin4A/a + cos4A/b = 1/(a+b)what is the value sin8A/a3 + cos8A/b3`
2 years ago

```							cos^4x=(cos^2x)^2=(1-sin^2x)^2=1+ sin^4x-2sin^2x =>sin^4x / a +cos^4x / b=1/a+b =>sin^4x/a + (1+sin^4x-2sin^2x)/b = 1/(a+b) [b*sin^4x + a(sin^4x-2sin^2x+1)] /ab = 1/(a+b) =>[(a+b)sin^4x-2a sin^2x+a]/ab = 1/a+b =>(a+b)^2 sin^4x - 2a(a+b)sin^2x + a(a+b) =ab =>(a+b)^2 sin^4x - 2a(a+b)sin^2x + a^2 => [(a+b)sin^2x-a]^2 = 0 =>(a+b)sin^2x - a = 0 sin^2x=a/(a+b).........(1) (take fourth power of both side) =>sin^8x=a^4/(a+b)^4 (divide by a^3 both side) =>sin^8x/a^3=a/(a+b)^4. ........(2) =>cos^2x=1 - sin^2x=1-a/(a+b)=b/(a+b). (from eq 1 substituting value of sin^2x) =>cos^2x=b/(a+b)...........(3) (take fourth power of both side) =>cos^8x=b^4/(a+b)^4 (divide by b^3 both side) =>cos^8x/b^3 =b/(a+b)^4......(4) (adding eq 2@4) =>sin^8x/a^3 + cos^8x/b^3=a/(a+b)^4 + b/(a+b)^4 =(a+b)/(a+b)^4=1/(a+b)^3
```
2 years ago
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