cos(2theta)=2cos^2(theta)-1; sin(2theta)=2sin(theta)cos(theta); tan(2theta)=sin(2theta)/cos(2theta). These are trigonometric identities, true for all values of theta.
In Q4, draw a triangle ABC with the following coordinates for the vertices: B(0,0), A(3,0), C(0,-4). In this triangle, AB=3, BC=-4, AC=sqrt(3^2+4^2)=5 (hypotenuse) using Pythagoras' theorem. Angle BAC=theta. So cos(theta)=AB/AC=3/5; sin(theta)=BC/AC=-4/5, and sin(2theta)=2sin(theta)cos(theta)=2*3/5*(-4/5)=-24/25, cos(2theta)=2(3/5)^2-1=18/25-1=(18-25)/25=-7/25; tan(2theta)=-24/25÷(-7/25)=24/7. Draw a triangle DEF with E(0,0), D(-7,0), F(0,-24) in Q3. Angle EDF=2theta, sin(2theta)=EF/DF=-24/25, cos(theta)=DE/DF=-7/25, tan(2theta)=EF/DE=-24/-7=24/7.