Question

Write the expression sec (sin^-1 x) as an algebraic expression in x (without trig or inverse trig functions).

Answer 1

the key is to convert "sec" into expression with "sine"
\[\sec x = \frac{1}{\cos x} = \frac{1}{\sqrt{1 - \sin^{2} x}}\]
since
\[\sin (\sin^{-1} x) = x\]
\[\sec(\sin^{-1} ) = \frac{1}{\sqrt{1-x^2}}\]

Answer 2

another approach is to say
\[x = \sin \theta\]
then find sec(theta) using triangle

Answer 3

ahh makes sense. thank you. i have a similar problem

sin(sin^-1 x + cos^-1 x) rewrite as an expression in x

Answer 4

if you could help me on this that would be great

Answer 5

first use angle sum identity
\[\sin(a +b) = \sin a \cos b + \sin b \cos a\]
\[\sin(a) = \cos(b) = x\]
\[\sin b = \sin(\cos^{-1} x) = \sqrt{1-x^2}\]
\[\cos a= \cos(\sin^{-1} x) = \sqrt{1-x^2}\]

*I hope the a,b notation didnt confuse you :)

Answer 6

that helps. thanks