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

1 Answer
Shwetank Mauria
Jun 17, 2016

#sin((5pi)/2)=1#, #cos((5pi)/2)=0#, #tan((5pi)/2)=oo#, #cot((5pi)/2)=0#, #sec((5pi)/2)=oo# and #csc((5pi)/2)=1#.

Explanation:

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

As such, as #(5pi)/2=2pi+pi/2#,trigonometric ratio of #(5pi)/2# and #pi/2# are same.

Now #sin(pi/2)=1#, #cos(pi/2)=0#, #tan(pi/2)=oo#, #cot(pi/2)=0#, #sec(pi/2)=oo# and #csc(pi/2)=1#,

hence #sin((5pi)/2)=1#, #cos((5pi)/2)=0#, #tan((5pi)/2)=oo#, #cot((5pi)/2)=0#, #sec((5pi)/2)=oo# and #csc((5pi)/2)=1#.

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