The question is:
tan(arcsinx)=
I know that inverse of sin is 1/sqrt(1-x^2) however this question still confuses me. The outcome is x/sqrt(1-x^2), how did the tan create the additional x ?
2 answers
oops double posted
make a sketch of a right-angled triangle, where the base angle is Ø
consider arcsin x
that means, in our triangle state the angle Ø so that sin Ø = x or sin Ø = x/1
so in your triangle label the adjacent side x and the hypotenuse 1
by Pythagoras, y^2 + x^2 = 1^2
y^2 = 1 - x^2
y = √(1 - x^2) , you had that
now tan(arcsin (x) )
= tan Ø
= y/x
= √(1-x^2)/x
consider arcsin x
that means, in our triangle state the angle Ø so that sin Ø = x or sin Ø = x/1
so in your triangle label the adjacent side x and the hypotenuse 1
by Pythagoras, y^2 + x^2 = 1^2
y^2 = 1 - x^2
y = √(1 - x^2) , you had that
now tan(arcsin (x) )
= tan Ø
= y/x
= √(1-x^2)/x
2 answers