arcsin(45) is some α between −π2 and π2 with sinα=45
arctan(125) is some β between −π2 and π2 with tanβ=125
We are asked to find sin(α+β). (Do you have some suspicions about why I rewrote the problem this way?)
We know that
sin(α+β)=sinαcosβ+cosαsinβ .
We know sinα=45 which is positive and with the restriction we already mentioned, we conclude that 0≤α≤π2. We want cosα.
At this point in our study of trigonometry we have already developed at least one method to find cosα (We may have three or four or more methods, but we only need one right now.)
Use your chosen method to get cosα=35
So we have:
sinα=45 and cosα=35
Similarly, given tanβ=125 and 0≤β≤π2, find sinβ and cosβ
sinβ=1213 and cosβ=513
sin(α+β)=sinαcosβ+cosαsinβ
=45513+351213
=5665.
If you want to write this without the names α and β 9Or something like that, you could write:
sin(arcsin(45)+arctan(125))
=sin(arcsin(45))cos(arctan(125))+cos(arcsin(45))sin(arctan(125))
But I, personally do not think that is more clear. (I do think it is good for students to see it this way, however.)
