# 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.