Inverse Trigonometric Functions

[Stein]

Please help me prove this!!! I'm stucked. I already know the inverse identities yet i can't prove this... any help would much be appreciated. thanks in advance!

Prove each identity:
1.) Arcsin (1/square root of 10) + Arccot 2 = Pi/2
2.) Arcsin (11/10) = 3 Arccos (square root of 15/4)

 

[stapel]

Prove each identity:
1.) Arcsin (1/square root of 10) + Arccot 2 = Pi/2
2.) Arcsin (11/10) = 3 Arccos (square root of 15/4)

Have these been copied correctly? For instance, the left-hand side of the second equation means the same thing as "sin(@) = 11/10 = 1.1", but the value of sine cannot be greater than 1. :shock:

 

[Ishuda]

Please help me prove this!!! I'm stucked. I already know the inverse identities yet i can't prove this... any help would much be appreciated. thanks in advance!

Prove each identity:
1.) Arcsin (1/square root of 10) + Arccot 2 = Pi/2
2.) Arcsin (11/10) = 3 Arccos (square root of 15/4)

I think 1.) may have been copied incorrectly also. In any case, the way I would go about this is to note that given the angle with a particular Arcsin [I assume the capital A means principle value] or Arccot or ..., you know all the trig values of that angle. For example suppose
$\displaystyle \theta=Arcsin(\frac{1}{\sqrt10})$
then
$\displaystyle sin(\theta)=\frac{1}{\sqrt10}$
$\displaystyle cos(\theta)=\frac{3}{\sqrt10}$
$\displaystyle tan(\theta)=\frac{1}{3}$
$\displaystyle cot(\theta)=3$
etc.
Now let
$\displaystyle \phi=Arccot(2)$
and expand
$\displaystyle sin(\theta + \phi)$

Edit: BTW, you need to justify why that would be sufficient and that has to do with the capital A.

 

[Stein]

Thanks for the help though. I will let you know asap.