Demonstrate that arcsin(1/`sqrt5` )+arccos(3/`sqrt10` )=`pi` /4?

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should re-write the expression using the angles `alpha` and `theta` for `arcsin(1/sqrt5)` and `arccos(3/sqrt10)` , such that:

`alpha + theta = pi/4`

You should take the sine function both sides, such that:

`sin(alpha + theta) = sin(pi/4)`

Expanding the sine of sum of angles, yields:

`sin alpha*cos theta + sin theta*cos alpha = sqrt2/2`

Replacing back `arcsin(1/sqrt5)` for `alpha` and `arccos(3/sqrt10)` for `theta` , yields:

`sin (arcsin(1/sqrt5))*cos (arccos(3/sqrt10)) + sin (arccos(3/sqrt10))*cos (arcsin(1/sqrt5)) = sqrt2/2`

You need to use the following identities, such that:

`sin (arcsin x) = cos(arccos x) = x`

`sin(arccos x) = cos(arcsin x) = sqrt(1 - x^2)`

Reasoning by analogy, yields:

`(1/sqrt5)*(3/sqrt10) + sqrt(1 - 9/10)*sqrt(1 - 1/5) = sqrt2/2`

`3/(sqrt(5*10)) + sqrt((10-9)/10)*sqrt((5-1)/5) = sqrt2/2`

`3/(5sqrt2) + 1/sqrt10*2/sqrt5 = sqrt2/2`

`(3 + 2)/(5sqrt2) = sqrt2/2`

`5/5sqrt2 = sqrt2/2`

`1/sqrt2 = sqrt2/2`

You need to rationalize to the left side, such that:

`sqrt 2/2 = sqrt2/2`

Hence, the last relation proves that the given identity holds.

arccos(3/sqrt10)

Sources:

We’ve answered 318,991 questions. We can answer yours, too.

Ask a question