How would you find the exact value of the six trigonometric function of 5pi/2?

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How would you find the exact value of the six trigonometric function of 5pi/2?

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Answer 1 · 2016-06-17T03:30:59

$sin (\frac{5 π}{2}) = 1$ $sin((5pi)/2)=1$ , $cos (\frac{5 π}{2}) = 0$ $cos((5pi)/2)=0$ , $tan (\frac{5 π}{2}) = \infty$ $tan((5pi)/2)=oo$ , $cot (\frac{5 π}{2}) = 0$ $cot((5pi)/2)=0$ , $sec (\frac{5 π}{2}) = \infty$ $sec((5pi)/2)=oo$ and $csc (\frac{5 π}{2}) = 1$ $csc((5pi)/2)=1$ .

Explanation:

Trigonometric ratio of an angle $θ$ $theta$ is same as that of $2 n π + θ$ $2npi+theta$ , where $n$ $n$ is an integer. In other words trigonometric ratio of an angle does not change if $2 π$ $2pi$ or its multiple is added to the angle or is subtracted from the angle.

As such, as $\frac{5 π}{2} = 2 π + \frac{π}{2}$ $(5pi)/2=2pi+pi/2$ ,trigonometric ratio of $\frac{5 π}{2}$ $(5pi)/2$ and $\frac{π}{2}$ $pi/2$ are same.

Now $sin (\frac{π}{2}) = 1$ $sin(pi/2)=1$ , $cos (\frac{π}{2}) = 0$ $cos(pi/2)=0$ , $tan (\frac{π}{2}) = \infty$ $tan(pi/2)=oo$ , $cot (\frac{π}{2}) = 0$ $cot(pi/2)=0$ , $sec (\frac{π}{2}) = \infty$ $sec(pi/2)=oo$ and $csc (\frac{π}{2}) = 1$ $csc(pi/2)=1$ ,

hence $sin (\frac{5 π}{2}) = 1$ $sin((5pi)/2)=1$ , $cos (\frac{5 π}{2}) = 0$ $cos((5pi)/2)=0$ , $tan (\frac{5 π}{2}) = \infty$ $tan((5pi)/2)=oo$ , $cot (\frac{5 π}{2}) = 0$ $cot((5pi)/2)=0$ , $sec (\frac{5 π}{2}) = \infty$ $sec((5pi)/2)=oo$ and $csc (\frac{5 π}{2}) = 1$ $csc((5pi)/2)=1$ .

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How would you find the exact value of the six trigonometric function of 5pi/2?

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