1. ## Maximum value

Show that $sinx(1+cosx)$ has a maximum value when $x = \frac{\pi}{3}$

2. Hey!

You need so show that

$f(x) := sin(x) [1 + cos(x)]$

$f'(x) = 2(cos(x))^2 + cos(x) - 1$

$f'(x) = 0 <=> x = -\frac{\pi}{3}, x = \frac{+\pi}{3} , x = ...$

$f''(x) = - 4 sin(x) cos(x) - sin(x)$

$f''(\frac{\pi}{3}) = -\frac{3*\sqrt{3}}{2} < 0$

=> Maximum

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Is there any problem for you?

Regards, Rapha

3. Let $f(x) = \sin{x}(1 + \cos{x})$

Differentiating,

$f'(x) = -\sin^2{x} + \cos{x}(1 + \cos{x})$

$= \cos^2{x} - \sin^2{x} + \cos{x}$

$= \cos^2{x} - (1 - \cos^2{x}) + \cos{x}$

$= 2\cos^2{x} + \cos{x} - 1$

$= (2\cos{x} - 1)(\cos{x} + 1)$

Equate to zero and solve. For the x values you obtain, substitute back into f(x) and see which x value gives a higher y-value. You can now prove what is being asked.

4. ## Trigonometry

Hello varunnayudu
Originally Posted by varunnayudu
Show that $sinx(1+cosx)$ has a maximum value when $x = \frac{\pi}{3}$
You might find it easier to use the formula $\sin 2x = 2\sin x \cos x$ before differentiating. This means you don't get any powers of 2. Instead you get:

$y = \sin x(1+\cos x)$

$=\sin x + \frac{1}{2}\sin 2x$

$\implies \frac{dy}{dx}=\cos x + \cos 2x$

$=0$ when

$\cos x = -\cos 2x$

Then use the fact that

$\cos A = -\cos B \implies A=\pi - B$ or $\pi + B$ or $3\pi \pm B$, etc.

Taking the first one of these possibilities gives

$x = \pi - 2x$

$\implies x=\frac{\pi}{3}$

Now differentiate $\cos x + \cos 2x$ and show that this value of $x$ gives a maximum value of $y$.