Asked by Jordan
So how do find the arcsec(-3)?
secy = x
secx = -3
1/cosx = -3
1 = -3cosx
-1/3 = cosx
Now what angle is this?
secy = x
secx = -3
1/cosx = -3
1 = -3cosx
-1/3 = cosx
Now what angle is this?
Answers
Answered by
Reiny
by arcsec(-3) you are looking for an angle Ø so that
secØ = -3
or
cosØ = -1/3
we know that the cosine is negative in quads II and III
so Ø = 180° - 70.53° OR Ø = 180° + 70.53°
Ø = appr 109.5° or appr. 250.5°
I had my calculator set to degrees, if you need radians, set it to that, then
find 2ndF cos (/13) to get 1.91..
and it to and subtract it from π for the two answers in radians.
secØ = -3
or
cosØ = -1/3
we know that the cosine is negative in quads II and III
so Ø = 180° - 70.53° OR Ø = 180° + 70.53°
Ø = appr 109.5° or appr. 250.5°
I had my calculator set to degrees, if you need radians, set it to that, then
find 2ndF cos (/13) to get 1.91..
and it to and subtract it from π for the two answers in radians.
Answered by
Reiny
small typo:
2ndF cos (/13) should have been 2ndF cos (1/3)
2ndF cos (/13) should have been 2ndF cos (1/3)
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