Question

If `sinx+sin^2x=1` and `acos^12x+bcos^8x+c cos^6x-1=0`, then what is the value of `a^2+b^2+c^2`?

Answer 1

Question

If `sinx+sin^2x=1` and `acos^12x+bcos^8x+c cos^6x-1=0`, then what is the value of `a^2+b^2+c^2`?

Given

`sinx+sin^2x=1`

`=>sin^2x+sinx-1=0`

`=>sinx=(-1pmsqrt(1^2-4*1*(-1)))/2`

`=>sinx=(-1pmsqrt5)/2`

But we know `-1<=sinx<=1`

Hence acceptable value of

`sinx=(-1+sqrt5)/2=1/2(sqrt5-1)`

Now
`color(red)(sinx=1/2(sqrt5-1))`

`color(red)(sin^2x=1/4(5+1-2sqrt5)=1/2(3-sqrt5)`

`color(blue)(sin^3x=sinx xx sin^2x=1/2(sqrt5-1)xx1/2(3-sqrt5))`

`=>color(blue)(sin^3x=(-2+sqrt5))`

`color(magenta)(sin^4x=(1/2(3-sqrt5))^2=(7/2-3/2sqrt5)`

`color(green)(sin^6x=(-2+sqrt5)^2=(9-4sqrt5))`

Again we have

`sinx+sin^2x=1`

`=>sinx=1-sin^2x=cos^2x`

Now putting `cos^2x=sinx` in the LHS of the given expression

`asin^6x+bsin^4x+c sin^3x-1=0`

Now putting the values of `sin^6x " "sin^4x and sin^3x`

`a(9-4sqrt5)+b(7/2-3/2sqrt5)+c (-2+sqrt5)-1=0`

Now if we take rationalizing factors

`a=(9+4sqrt5)`

`b=(7/2+3/2sqrt5)`

and

`c= (-2-sqrt5)`

then above equation is satisfied.

So inserting these values we can get a possible value of

`a^2+b^2+c^2`

`=(9+4sqrt5)^2+(7/2+3/2sqrt5)^2+ (-2-sqrt5)^2`

`=(161+72sqrt5)+1/4(94+42sqrt5)+ (9+4sqrt5)`

`=(161+72sqrt5)+(23.5+10.5sqrt5)+ (9+4sqrt5)`

`=(193.5+86.5sqrt5)`

In addition here is a scratchpad note of respected Cesareo R regarding this problem .

https://socratic.org/scratchpad/87db6c91b6866e1f5e3a