(sin x/2+cos x/2)^2 - 1/tg x/2-sin x/2cos x/2=2ctg^2 x/2  

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

The request of the problem is vague, hence, supposing that you need to prove that the given expression holds, you should expand the square such that:

`sin^2 (x/2) + 2sin(x/2)cos(x/2) + cos^2(x/2) - 1/(sin(x/2))/(cos(x/2))) - sin(x/2)cos(x/2) = 2(cos^2(x/2))/(sin^2(x/2))`

Notice that `(sin(x/2))/(cos(x/2))`  substitutes `tan(x/2)`  and `(cos^2(x/2))/(sin^2(x/2))`  substitutes `cot^2(x/2).`

You need to use the fundamental formula of trigonometry such that:

`sin^2 (x/2)+ cos^2(x/2) = 1`

`1 + sin(x/2)cos(x/2) - (cos(x/2))/(sin(x/2)) =2(cos^2(x/2))/(sin^2(x/2))`

You should bring the terms to a common denominator such that:

`1 + sin^2(x/2)cos(x/2) - cos(x/2)sin(x/2) = 2cos^2(x/2)`

`sin^2(x/2)cos(x/2) - cos(x/2)sin(x/2) = 2cos^2(x/2) - 1`

`sin^2(x/2)cos(x/2) - cos(x/2)sin(x/2) = cos x`

`(1+cos x)/2*cos(x/2) - cos(x/2)sin(x/2) = cos x`

`(1+cos x)*cos(x/2) - 2cos(x/2)sin(x/2) = 2cos x`

`(1+cos x)*cos(x/2) = 2cos x + sin x`

Hence, evaluating the given expression yields that it does not represent an identity.

Sources:

We’ve answered 318,957 questions. We can answer yours, too.

Ask a question