Jason D.

asked • 03/25/17

Trigonometry

if sec(θ)=9/10
and tan(θ) < 0
find sin(θ)
 that is my problem i believe the answer  is -√19/10
but im not sure why the square root of 19 turned out negative.
i know the unit circle is involved but im confused on why it turns out that way i guess.
 

Mark M.

tutor
Since -1 ≤ cosθ ≤ 1, one of three things can be said about secθ, the reciprocal of cosθ:
 
   If cosθ > 0, then secθ ≥ 1.
 
   If cosθ < 0, then secθ ≤ -1
 
   If cosθ = 0, then secθ is undefined 
 
So, there is no value of θ for which secθ = 9/10.
 
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03/25/17

1 Expert Answer

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Michael J. answered • 03/25/17

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Let  x=θ
 
1/cos(x) = 9/10
 
cos(x) = 10/9
 
tan(x) is negative in the 2nd and 4th quadrants.
 
 
Then,
 
sin(x) = √(1 - cos2(x))
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Arthur D.

tutor
Michael, cosine can't be 10/9, it's too big. You end up with the minus sign under the radical sign. Am I missing something ?
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03/25/17

Kemal G.

Hi Jason,
 
Are you sure that sec(theta) = 9/10. I know the range of the secant function to be sec x ≥ 1 and sec x ≤ -1. If this is true, then you probably made a mistake writing sec (θ).
 
If  sec(θ) = 10/9 then we have a right triangle with hyp = 10 and adj = 9. Then, the opposite side equals 10^2 - 9^2, which is
100 - 81 = 19
opp = √19
 
Sin(θ) = √19/10
 
Since sec(θ)>0 and tan(θ)<0, the angle is in the 4th quadrant where sin θ <0 so
the answer is -√19/10.
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03/25/17

Arthur D.

tutor
If sec=9/10 then sec=1/cos and cos=10/9 which makes adjacent=10 and hypotenuse=9 which makes opposite=√-19.
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03/25/17

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