SOLUTION: Can someone please explain this equation to me? and i also believe that the answer would be b.
a=sin^-1 -x then a=
a. cos^-1 1/sqrt 1-x^2
b. tan^-1 -x/sqrt 1-x^2
c. pi/2-co
Algebra ->
Trigonometry-basics
-> SOLUTION: Can someone please explain this equation to me? and i also believe that the answer would be b.
a=sin^-1 -x then a=
a. cos^-1 1/sqrt 1-x^2
b. tan^-1 -x/sqrt 1-x^2
c. pi/2-co
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Question 756834: Can someone please explain this equation to me? and i also believe that the answer would be b.
a=sin^-1 -x then a=
a. cos^-1 1/sqrt 1-x^2
b. tan^-1 -x/sqrt 1-x^2
c. pi/2-cos 1/-x
d. tan^-1 sqrt 1+x^2/-x Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! a=sin^-1 -x then a=
a. cos^-1 1/sqrt 1-x^2
b. tan^-1 -x/sqrt 1-x^2
c. pi/2-cos 1/-x
d. tan^-1 sqrt 1+x^2/-x
..
a=sin^-1 -x
This expression reads:
an angle (a) whose sin=-x
In other words,
sin(a)=-x
cos(a)=sqrt(1-sin^2(a)=sqrt(1-x^2)
tan(a)=sin(a)/cos(a)=-x/sqrt(1-x^2)
this reads:
angle(a) whose tangent=-x/sqrt(1-x^2) (ans b.)
Remember: the inverse is always an angle
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