Inverse trig functions

**>_<SHY_GUY>_<** · April 15th 2008, 06:30 PM

one thing to just clear out...im kinda blank today...why is cosine of 0 = 1 and sine of 0 = 0?

how is arctan square root of 3 divided by 3 = pi/6?

how is (arctan x = pi/4) equal to 1?

how is (arctan x = -pi/6) equal to the negative square root of 3 divided by 3?

and how is square root 3 /3 part of the 30-60-90 triangle? i though it was sqaureroot 3/2?

thank you

**Mathstud28** · April 15th 2008, 06:36 PM

Originally Posted by >_<SHY_GUY>_<

one thing to just clear out...im kinda blank today...why is cosine of 0 = 1 and sine of 0 = 0?

how is arctan square root of 3 divided by 3 = pi/6?

how is (arctan x = pi/4) equal to 1?

how is (arctan x = -pi/6) equal to the negative square root of 3 divided by 3?

and how is square root 3 /3 part of the 30-60-90 triangle? i though it was sqaureroot 3/2?

thank you

first part $cos(x)=\frac{e^{ix}+e^{-ix}}{2}\Rightarrow{cos(0)=\frac{e^{0i}+e^{-0i}}{2}}\Rightarrow{\frac{1+1}{2}=1}$ and for the other $sin(x)=\frac{e^{ix}-e^{-ix}}{2}\Rightarrow{ \frac{1-1}{2}=0}$ we jumped the steps using the same logic as above...for the second thing because $tan\bigg(\frac{\pi}{4}\bigg)=1$ ...for the next one follow same logic and for the next one...for the last one it is because $tan(30)=\frac{\sqrt{3}}{3}$

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