# Expressions for `sin(t),cos(t),tan(t)` through `tan(t/2)`

Using identities `sin(t)=2sin(t/2)cos(t/2)` and `cos^2(t/2)+sin^2(t/2)=1` we can write that

`sin(t)=(sin(t))/1=(2sin(t/2)cos(t/2))/(cos^2(t/2)+sin^2(t/2))`.

Dividing both numerator and denominator by `cos^2(t/2)` will give

`color(blue)(sin(t)=(2tan(t/2))/(1+tan^2(t/2)))`.

Now, using identities `cos(t)=cos^2(t/2)-sin^2(t/2)` and `cos^2(t/2)+sin^2(t/2)=1` we can write that

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

Dividing both numerator and denominator by `cos^2(t/2)` will give

`color(green)(cos(t)=(1-tan^2(t/2))/(1+tan^2(t/2)))`.

Now, `tan(t)=sin(t)/cos(t)=((2tan(t/2))/(1+tan^2(t/2)))/((1-tan^2(t/2))/(1+tan^2(t/2)))`.

This can be rewritten as `color(red)(tan(t)=(2tan(t/2))/(1-tan^2(t/2)))`.

These three formulas hold when `t!=(2n+1)pi,n in ZZ`.

**Example**. Find `(2+cos(t))/(4-5sin(t))` if `tan(t/2)=-2/3`.

We have that `sin(t)=(2tan(t/2))/(1+tan^2(t/2))=(2*(-2/3))/(1+(-2/3)^2)=-12/13`.

Also `cos(t)=(1-tan^2(t/2))/(1+tan^2(t/2))=(1-(-2/3)^2)/(1+(-2/3)^2)=5/13`.

Finally, `(2+cos(t))/(4-5sin(t))=(2+3*5/13)/(4-5*(-12/13))=41/112`.