Determining the exact value of each expression

**melaniemich87** · October 23rd 2008, 02:06 PM

Hi guys, I'm having trouble with obtaining the correct results for this expression.

The question is: Determine the exact value of each expression.

[cos(pi/4)tan(5pi/6)] / [sin(3pi/4)]

Somehow, I obtained the answer square root of three... but the answer is supposed to be square root of 6 divided by 6.

Thank you in advance for all of your help.

**Henderson** · October 23rd 2008, 02:13 PM

Are you sure either answer is right, or that you copied the problem correctly?

I get an answer of $-\frac{\sqrt3}{3}$ .

$\frac{(\frac{\sqrt2}{2})\cdot(-\frac{\sqrt3}{3})}{(\frac{\sqrt2}{2})}$

**11rdc11** · October 23rd 2008, 02:50 PM

Originally Posted by Henderson

Are you sure either answer is right, or that you copied the problem correctly?

I get an answer of $-\frac{\sqrt3}{3}$ .

$\frac{(\frac{\sqrt2}{2})\cdot(-\frac{\sqrt3}{3})}{(\frac{\sqrt2}{2})}$

Ditto what Henderson said

**melaniemich87** · October 23rd 2008, 02:55 PM

Oh no! I didn't copy it right! It's supposed to be cos(4pi/3) instead of cos(pi/4)

**skeeter** · October 23rd 2008, 03:43 PM

well, you should know that $\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}$

sub it in for the $\frac{\sqrt{2}}{2}$ in the numerator of the expressions given and re-evaluate.

**melaniemich87** · October 23rd 2008, 03:57 PM

how did you get square root of 2 over 2 as the denominator tho?

-cos(pi/3) = -1/2
-tan(pi/6) = - 1/ square root of 3
sin(pi/4) = 1/ square root of 2

**skeeter** · October 23rd 2008, 04:28 PM

$\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

**melaniemich87** · October 23rd 2008, 06:35 PM

i mean.. how did you get your answer? the answer that i got wat 2/(2*square root of 3)

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