Question

Let f(x) = sinxcosx + 0 for

Answer 1

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

Answer 2

the question really contains + 0....?

Answer 3

oops its suppose to be sinxcosx+x

Answer 4

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)..?

Answer 5

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

Answer 6

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

Answer 7

ooh I see what I did wrong

Answer 8

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

Answer 9

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

Answer 10

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.

Answer 11

how did you get pi/4?
@campbell_st

Answer 12

because I know

\[\frac{1}{\sqrt{2}} = \frac{\pi}{4}\]

its called an exact value, you may know it with a rationalized denominator as

\[\frac{\sqrt{2}}{2} = \frac{\pi}{4}\]

Answer 13

but if you want to check, use a calculator and just enter

\[\cos^{-1}(\frac{1}{\sqrt{2}})\]

Answer 14

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

Answer 15

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