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So we have this even/odd properties in trigonometry and we had this problem:

$\sec (-(\frac{\pi}6))$

And ok i was like first off i can change the - because of even odd properties so it becomes:

$-(\sec(\frac{\pi}6))$

So next I substitute $\sec$ to $\frac1{\cos}$ ....

$-(\frac1{\cos (\frac{\pi}6)}))$

There I get $-2$ ... but the answer is $\frac{2\sqrt{3}}{3}$
I really don't understand what I did wrong here... And I figured out that she was right.. I searched up the question online and it did give the answer? What am I doing wrong here?

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2 Answers 2

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You have the even/odd property of $\sec$ wrong. $$\sec(-x)=\sec(x)$$

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  • $\begingroup$ @Dsafds np, just a minor issue. You also evaluated $\cos\left(\frac{\pi}6\right)$ incorrectly. $\endgroup$
    – Simply Beautiful Art
    Mar 7, 2016 at 22:44
  • $\begingroup$ Hmm i will take note of that too! Thanks! Now i dont know which anwser to accept.. I had such an obvious anwser both of you got it... $\endgroup$
    – amanuel2
    Mar 7, 2016 at 22:45
  • $\begingroup$ Isnt Cos pie/6 = 1/2?? $\endgroup$
    – amanuel2
    Mar 7, 2016 at 22:47
  • $\begingroup$ @Dsafds No, it should be $\frac{\sqrt{3}}2$, as Kf-Sansoo noted. $\endgroup$
    – Simply Beautiful Art
    Mar 7, 2016 at 22:49
  • $\begingroup$ Sorry made a mistake.... I was thinking about pie/3 .. $\frac{\pi}3$ $\endgroup$
    – amanuel2
    Mar 7, 2016 at 22:50
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$\sec\left(-\dfrac{\pi}{6}\right)= \dfrac{1}{\cos\left(-\dfrac{\pi}{6}\right)}= \dfrac{1}{\cos\left(\dfrac{\pi}{6}\right)}= \dfrac{1}{\dfrac{\sqrt{3}}{2}}= \dfrac{2}{\sqrt{3}}= \dfrac{2\sqrt{3}}{3}$

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  • $\begingroup$ Like Simple Art said i had the sec even property wrong! Man im stupid..... Sorry for irritance! $\endgroup$
    – amanuel2
    Mar 7, 2016 at 22:43

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