Question

What does `sin(arc cot(5))+5sin(arc tan(2))` equal?

Answer 1

`sin(cot^-1(5))+5sin(tan^-1(2)) = (sqrt(26)+52sqrt(5))/26`

Explanation:

The trick is to evaluate `sin(cot^-1(5))+5sin(tan^-1(2))` is
breaking down the problem and solving each part separately.

Let us start with evaluating `cot^-1(5)`

Let `theta=cot^-1(5)`

Then `cot(theta) = 5`

`cot(theta) = "adjancent"/"opposite"`

If I have to show that in a right triangle.

The hypotenuse would be `sqrt(1^2+5^2)`
Hypotenuse `= sqrt(26)`

`sin(theta) = "Opposite"/"hypotenuse"`
`sin(theta) = 1/sqrt(26)`

Rationalizing the denominator we get
`sin(theta) = (sqrt(26))/26`

Remember `theta` was `cot^-1(5))`

`sin(cot^-1(5)) = (sqrt(26))/26`

Now let us work on the second part.

`5sin(tan^-1(2))`
`Theta=tan^-1(2)`
`tan(Theta) = 2`
`tan(Theta) = "Opposite"/"Adjacent"`

Let us put that in a right triangle.

Now to find the hypotenuse

Hypotenuse `=sqrt(2^2+1^2)`
Hypotenuse `=sqrt(5)`

`sin(Theta) = "Opposite"/"hypotenuse"`

`sin(Theta) =2/sqrt(5)`

Rationalizing the denominator

`sin(Theta)=(2sqrt(5))/5`

Remember `Theta=tan^-1(2)`

`sin(tan^-1(2))=(2sqrt(5))/5`

`5sin(tan^-1(2))=5**(2sqrt(5))/5`

`5sin(tan^-1(2))=2sqrt(5)`

Now the final steps

`sin(cot^-1(5))+5sin(tan^-1(2)) = sqrt(26)/26+2sqrt(5)`

`sin(cot^-1(5))+5sin(tan^-1(2)) = (sqrt(26)+52sqrt(5))/26`