## amy0799 one year ago Let f(x) = sinxcosx + 0 for

1. amy0799

$0 \le x \le \frac { 3\pi }{ 2}$ find all the values for which f'(x)=1

2. campbell_st

the question really contains + 0....?

3. amy0799

oops its suppose to be sinxcosx+x

4. campbell_st

ok.... so you need the product rule to do the sin(x)cos(x) or you can use a double angle substitution... do you know sin(2x) = 2 sin(x)cos(x)..?

5. amy0799

ill do the product rule f'(x)=cosxcosx+cosx(-sinx)+1

6. campbell_st

well not quite u = sin(x) v = cos(x) u' = cos(x) v' = -sin(x) so the derivative is $\frac{dy}{dx} = \sin(x) \times - \sin(x) + \cos(x) \times \cos(x) + 1$ or $\frac{dy}{dx} = \cos^2(x) - \sin^2(x) + 1$ does that make sense

7. amy0799

ooh I see what I did wrong

8. campbell_st

so the derivative is equal to 1 then $1 = \cos^2(x) - \sin^2(x) + 1~~~or~~~0 = \cos^2(x) - \sin^2(x)$ I'd now make a substitution of $\sin^2(x) = 1 - \cos^2(x)$ so you have $0 = \cos^2(x) -(1 - \cos^2(x)$ or $0 = 2\cos^2(x) - 1$ now you should be able to solve for x

9. amy0799

what do I put in x? 3pi/2? @campbell_st

10. campbell_st

no you solve so $1 = 2\cos^2(x)$ then $\frac{1}{2} = \cos^2(x)$ so $\pm \frac{1}{\sqrt{2}} = \cos(x)$ this is an exact value.... that is equal to $x = \frac{\pi}{4}$ in the 1st quadrant so there are angles in the 2nd quadrant $\pi - \frac{\pi}{4}$ and 3rd quadrant $\pi + \frac{\pi}{4}$ just simplify the angles in the 2nd and 3rd quadrants by writing them as improper fractions.

11. amy0799

how did you get pi/4? @campbell_st

12. campbell_st

because I know $\frac{1}{\sqrt{2}} = \frac{\pi}{4}$ its called an exact value, you may know it with a rationalized denominator as $\frac{\sqrt{2}}{2} = \frac{\pi}{4}$

13. campbell_st

but if you want to check, use a calculator and just enter $\cos^{-1}(\frac{1}{\sqrt{2}})$

14. amy0799

ooh ok so what do you mena by simplify the angles in the 2nd and 3rd quadrants by writing them as improper fractions?

15. campbell_st

well the angles are normally written using fractions so 3rd quadrant $\pi + \frac{\pi}{4} = \frac{5\pi}{4}$ you need to work out the 2nd quadrant angle