Complete the identity


Answer:

sin⁡(α+β)/sin⁡(α-β) ==(tan⁡ α+tan ⁡β)/(tan ⁡α-tan ⁡β )

Step-by-step explanation:

We have to complete

sin⁡(α+β)/sin⁡(α-β) = ?

The identities that will be used:

sin⁡(α+β)=sin⁡ α cos ⁡β+cos ⁡α sin ⁡β

and

sin⁡(α-β)=sin⁡ α cos⁡ β-cos⁡ α sin⁡ β  

Now:

=   sin⁡(α+β)/sin⁡(α-β)  

=(sin⁡ α cos⁡ β+cos ⁡α sin⁡ β)/(sin ⁡α cos ⁡β-cos ⁡α sin ⁡β)

In order to bring the equation in compact form we wil divide both numerator and denominator with  cos⁡ α cos⁡ β  

=  (((sin ⁡α cos ⁡β+cos ⁡α sin ⁡β))/(cos ⁡α cos ⁡β ))/(((sin α  cos ⁡β-cos ⁡α sin ⁡β))/(cos ⁡α  cos ⁡β))

=((sin⁡ α cos⁡β)/(cos ⁡α cos ⁡β )+(cos ⁡α sin ⁡β)/(cos ⁡α cos ⁡β ))/((sin ⁡α cos ⁡β)/(cos ⁡α  cos ⁡β )-(cos ⁡α sin ⁡β)/(cos ⁡α cos ⁡β))  

=(sin⁡ α/cos ⁡α + sin ⁡β/cos ⁡β )/(sin ⁡α/cos ⁡β - sin ⁡β/cos ⁡β)

=(tan ⁡α+tan ⁡β)/(tan ⁡α-tan ⁡β )

So,

sin⁡(α+β)/sin⁡(α-β) ==(tan⁡ α+tan ⁡β)/(tan ⁡α-tan ⁡β)


Rate answer
Wrong answer?

If your question is not fully disclosed, then try using the search on the site and find other answers on the subject Mathematics.

Find another answers

Load image