How do you do simplify #cos [sec ^-1 (-5)]#?
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#cos[sec^-1(-5)]#
Finding the value of the Composition:
Lets being to solve the equation by working on the inside. Immediately we can see that #sec^-1(5/(-1))# isn't a special angle. If you noticed how I attached the negative sign to "1" is because sec = #r/x#. The radius can not be negative. and "1" serves as the x-value in this equation.
Note: The inverse of cosine (secant) is restricted between #0 < x < pi#. (This would matter if it were asking for sin instead of cos because the y-value would become negative and give you a wrong answer).
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Now cosine is #x/r#. So simply put the values where they belong.
#cos((-1)/5)# = #-1/5#
This is not a special angle so leave the answer as a improper fraction or decimal. Whatever your instructor prefers. :)
