Evaluate 4[sin4 30+cos4 60]-2/3[sin2 60-cos2 45 ] + 1/2 tan2 60

**butler12** · July 15th 2011, 09:37 PM

4[sin4 30+cos4 60]-2/3[sin2 60-cos2 45 ] + 1/2 tan2 60

**icedtrees** · July 15th 2011, 10:19 PM

To do this question exact values must be known, that is:

$sin(30) = cos(60) = \frac{1}{2}$
$tan(60) = \sqrt(3)$
$sin(45) = cos(45) = \frac{1}{\sqrt(2)}$

If you did not already know:

$tan(30) = \frac{1}{\sqrt(3)}$
$sin(60) = cos(30) = \frac{\sqrt(3)}{2}$
$tan(45) = 1$

Now, all that is needed is a substitution:

$4(\frac{1}{2})^4 + 4(\frac{1}{2})^4 - \frac{2}{3}[(\frac{1}{2})^2 - (\frac{1}{\sqrt(2)})^2] + \frac{1}{2}[\sqrt(3)]^2$

and then it should be easy to solve from there

**butler12** · July 15th 2011, 11:03 PM

hi thanks for this solutions, and do you have any ideas you can give me for solving my questions. because this is just a try i gave to this place for solutions. This question i had got from my institute pioneermathematics and they have solutions too.

**icedtrees** · July 15th 2011, 11:36 PM

I can't give you ideas for solving other questions if I don't know what the questions are. Are you asking for the final answer to the question? I evaluated it to be 11/6, but I'm sure you could have done it as well

**mr fantastic** · July 16th 2011, 02:31 PM

Originally Posted by butler12

hi thanks for this solutions, and do you have any ideas you can give me for solving my questions. because this is just a try i gave to this place for solutions. This question i had got from my institute pioneermathematics and they have solutions too.

You were told exactly how to do the question!

It is insulting to post questions in order to 'test' us.

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