maximum value of sine

**Shivanand** · December 17th 2008, 01:10 AM

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

**Mush** · December 17th 2008, 01:29 AM

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?

**mr fantastic** · December 17th 2008, 03:21 AM

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.

**Moo** · December 17th 2008, 06:17 AM

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

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