Question

Find the range of `sinx(sinx+cosx)`?

Answer 1

Range of `sinx(sinx+cosx)` is `[1/2-1/sqrt2,1/2+1/sqrt2]`

Explanation:

`sinx(sinx+cosx)`

= `sinxsqrt2(sinx1/sqrt2+cosx1/sqrt2)`

= `sqrt2sinx(sinxcos45^@+cosxsin45^@)`

= `sqrt2sinxsin(x+45^@)`

= `sqrt2/2(2sinxsin(x+45^@))`

= `sqrt2/2(2sinxsin(x+45^@))`

= `1/sqrt2[cos(x-(x+45^@)-cos(x+(x+45^@)]`

= `1/sqrt2[cos(-45^@)-cos(2x+45^@)]`

= `1/sqrt2[1/sqrt2-cos(2x+45^@)]`

= `1/2-1/sqrt2cos(2x+45^@)`

As range of `cos(2x+45^@)` is `[-1,1]`

range of `sinx(sinx+cosx)` or `1/2-1/sqrt2cos(2x+45^@)` is

`[1/2-1/sqrt2,1/2+1/sqrt2]`