Using Inverse Trigonometric Function

[nycmathguy]

Section 4.7


Can you do 39 as a guide for me to do the rest?

 

[MathLover1]

39

tan(theta)=x/4
then theta=tan^-1(x/4 )

 

[nycmathguy]

39

tan(theta)=x/4
then theta=tan^-1(x/4 )

tan(theta) = opp/adj

tan(theta) = x/4

arctan(tan (theta)) = arctan(x/4)

theta = arctan(x/4)

Is this what you did? If so, please don't skip any parts.

 

[MathLover1]

arctan(tan (theta)) = arctan(x/4) ..........this is unnecessary step,

when you have tan(theta) = x/4, you know that theta = arctan(x/4)

 

[nycmathguy]

arctan(tan (theta)) = arctan(x/4) ..........this is unnecessary step,

when you have tan(theta) = x/4, you know that theta = arctan(x/4)

Ok. Thanks. I will do a few more later or tomorrow morning.

 

[nycmathguy]

arctan(tan (theta)) = arctan(x/4) ..........this is unnecessary step,

when you have tan(theta) = x/4, you know that theta = arctan(x/4)

Questions 40, 42 and 44.

(40) cos (theta) = 4/x

Then theta = arccos(4/x)

(42) tan (theta) = (x + 1)/10

Then theta = arctan[(x + 1)/10]

(44) sin (theta) = (x - 1)/(x^2 - 1)

Then theta = arcsin[(x - 1)/(x - 1)(x + 1)]

theta = arcsin[1/(x + 1)]

Note: For (44), I simplified the algebraic fraction in the argument for arcsin. Is this the right thing to do?

 

[MathLover1]

correct

 

[nycmathguy]

correct

I could not have done it without you. I thank you.