# Thread: maximum value of sine

1. ## maximum value of sine

show that sinx (1+cosx) has a maximum value when x=Pie/3

2. Originally Posted by Shivanand
show that sinx (1+cosx) has a maximum value when x=Pie/3
$f(x) = sin(x)(1+cos(x)) = sin(x) + sin(x)cos(x)$

Differentiate!

$f'(x) = cos(x)( \, 1+cos(x) \, ) + sin(x)(\, -sin(x) \, )$

Maximum occurs when $f'(x) = 0$

$f'(x) = cos(x)(1+cos(x) + sin(x)(-sin(x)) = 0$

$cos(x) + cos^2(x) - sin^2(x) = 0$ (Remember that $cos^2(x) - sin^2(x) = cos(2x)$)

$cos(x) + cos(2x) = 0$

$cos(x) + 2cos^2(x) -1 = 0$ Remember that $cos(2x) =2cos^2(x) - 1$

$cos(x)(1+2cos(x)) = 1$

Take it from here?

3. Originally Posted by Mush
$f(x) = sin(x)(1+cos(x)) {\color{red}= sin(x) + sin(x)cos(x)}$ Mr F says: The red stuff in this line is unnecessary (since you don't use it in your solution) and in fact unhelpful and potentially confusing to the OP.

Differentiate!

$f'(x) = cos(x)( \, 1+cos(x) \, ) + sin(x)(\, -sin(x) \, )$

Maximum occurs when $f'(x) = 0$

$f'(x) = cos(x)(1+cos(x) + sin(x)(-sin(x)) = 0$

$cos(x) + cos^2(x) - sin^2(x) = 0$ (Remember that $cos^2(x) - sin^2(x) = cos(2x)$) Mr F says: Alternatively and more efficiently, remember the Pythagorean Identity ${\color{red}\cos^2 x + \sin^2 x = 1 \Rightarrow \sin^2 x = 1 - \cos^2 x}$. I'll pick up from here in my main reply below.

$cos(x) + cos(2x) = 0$

$cos(x) + 2cos^2(x) -1 = 0$ Remember that $cos(2x) =2cos^2(x) - 1$

$cos(x)(1+2cos(x)) = 1$ Mr F says: This line does not lead anywhere helpful and may well reinforce a common misconception regarding the null factor law and its misuse.
Take it from here?
$\Rightarrow \cos x + 2 \cos^2 x -1 = 0$

$\Rightarrow 2 \cos^2 x + \cos x - 1 = 0$

$\Rightarrow (2 \cos x - 1)(\cos x + 1) = 0$

and you should be able to see what happens now.

4. Hello,

$\sin(x)(1+\cos(x))=\sin(x)+\sin(x)\cos(x)=\sin(x)+ \frac 12 \sin(2x)$

Differentiate :
$\cos(x)+\cos(2x)$

Set it equal to 0 :
$\cos(x)+\cos(2x)=0$

$\cos(2x)=\cos(\pi \pm x)$

And we know that $\cos(a)=\cos(b) \Leftrightarrow a=b \text{ or } a=-b \quad +2k \pi$