Question

givenn cos2A = (1/5) and the angle A is in quadrant two. Find cscA. PLEASE HELP!!!

Answer 1

Use the formula cos(2A) = 1 - 2 sin^2(A)

so 1/5 = 1 - 2sin^2(A)
2 sin^2(A) = 4/5
sin^2(A) = 2/5
sin A = sqrt (2/5) A is in quad II, so sin A is positive there.

Hence, csc A = sqrt(5/2)

Answer 2

can't the bottom not be a square root? so would the final asnwer be sqrt 10 / (5)

Answer 3

and then flip again?

Answer 4

csc A is the reciprocal of sin A.

If sin A = sqrt (2/5), then sqrt(5/2) is the reciprocal.

Answer 5

yes but since it's like sqrt of 5 divided by the sqrt of 2 wouldn't you need to get rid of the sqrt at the bottom? (the denominator)

Answer 6

You can taionalize the denominator to get rid of the square root in the denominator, but sqrt(2/5) is a number that is correct. Nowhere does it say in the problem that the final answer should have a rationalized denominator. But if you want, by all means, rationalize the denominator.

Answer 7

ok! thank you. sorry i think my teacher wanted me to. would it be 5 sqrt 10 divided by 10 then?

Answer 8

or would the final answer be the sqrt 5?

Answer 9

that's what i just got

Answer 10

sqrt(5/2) = sqrt (5)/sqrt(2).

Multiply top and bottom by sqrt(2), you sqrt(10)/2..and thats fine and has a rationalized denominator.

Answer 11

but then couldn't you simplify the sqrt 10 / 2 to 2 times the sqrt of 5 divided by 2? and the 2s would cancel out?

Answer 12

THANK YUO SO MUCH FOR YOUR HELP BY THE WAY!!

Answer 13

You're welcome.