Asked by Trig
How do you do this..i have the answer i just need a step by step way to solve this..
sec(sin^-1 (2/5))
sec(sin^-1 (2/5))
Answers
Answered by
Reiny
sin^-1 (2/5) is the angle Ø of a right angles triangle, where the opposite side to Ø is 2 and the hypotenuse is 5 , sketch that triangle.
Using Pythagoras, the third side would then be √21
now we want sec Ø
= 1/cosØ
= 1/(√21/5) = 5/√21
check with calculator:
2
÷
5
=
2nd
sin
= ------ ( I got 23.578...° )
cos
=
2nd
x^2 ----- I have x^-1 paired with x^2
=
1.0910....
taking 5/√21 gives me the same result
Using Pythagoras, the third side would then be √21
now we want sec Ø
= 1/cosØ
= 1/(√21/5) = 5/√21
check with calculator:
2
÷
5
=
2nd
sin
= ------ ( I got 23.578...° )
cos
=
2nd
x^2 ----- I have x^-1 paired with x^2
=
1.0910....
taking 5/√21 gives me the same result
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