Vraag 6

Formules

Goniometrie

\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\\ & \end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \\ & \end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \\ & \end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & (t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \\ & \end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \\ & \end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \\ & \end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\,\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\,\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\end{aligned}\begin{aligned} & \sin(t+u)=\sin(t)\cos(u)+\cos(t)\sin(u)\\ & \sin(t-u)=\sin(t)\cos(u)-\cos(t)\sin(u)\\ & \cos(t+u)=\cos(t)\cos(u)-\sin(t)\sin(u)\\ & \cos(t-u)=\cos(t)\cos(u)+\sin(t)\sin(u)\\ & \sin(2t)=2\sin(t)\cos(t)\\ & \cos(2t)=\cos^2(t)-\sin^2(t)=2\cos^2(t)-1=1-2\sin^2(t)\end{aligned}\begin{aligned} & \sin (t+u)=\sin (t) \cos (u)+\cos (t) \sin (u) \\ & \sin (t-u)=\sin (t) \cos (u)-\cos (t) \sin (u) \\ & \cos (t+u)=\cos (t) \cos (u)-\sin (t) \sin (u) \\ & \cos (t-u)=\cos (t) \cos (u)+\sin (t) \sin (u) \\ & \sin (2 t)=2 \sin (t) \cos (t) \\ & \cos (2 t)=\cos ^{2}(t)-\sin ^{2}(t)=2 \cos ^{2}(t)-1=1-2 \sin ^{2}(t) \end{aligned}\begin{aligned} & \sin (t+u)=\sin (t) \cos (u)+\cos (t) \sin (u) \\ & \sin (t-u)=\sin (t) \cos (u)-\cos (t) \sin (u) \\ & \cos (t+u)=\cos (t) \cos (u)-\sin (t) \sin (u) \\ & \cos (t-u)=\cos (t) \cos (u)+\sin (t) \sin (u) \\ & \sin (2 t)=2 \sin (t) \cos (t) \\ & \cos (2 t)=\cos ^{2}(t)-\sin ^{2}(t)=2 \cos ^{2}(t)-1=1-2 \sin ^{2}(t) \end{aligned}\begin{aligned} & \sin (t+u)=\sin (t) \cos (u)+\cos (t) \sin (u) \\ & \sin (t-u)=\sin (t) \cos (u)-\cos (t) \sin (u) \\ & \cos (t+u)=\cos (t) \cos (u)-\sin (t) \sin (u) \\ & \cos (t-u)=\cos (t) \cos (u)+\sin (t) \sin (u) \\ & \sin (2 t)=2 \sin (t) \cos (t) \\ & \cos (2 t)=\cos ^{2}(t)-\sin ^{2}(t)=2 \cos ^{2}(t)-1=1-2 \sin ^{2}(t) \end{aligned}\begin{aligned} & \sin (t+u)=\sin (t) \cos (u)+\cos (t) \sin (u) \\ & \sin (t-u)=\sin (t) \cos (u)-\cos (t) \sin (u) \\ & \cos (t+u)=\cos (t) \cos (u)-\sin (t) \sin (u) \\ & \cos (t-u)=\cos (t) \cos (u)+\sin (t) \sin (u) \\ & \sin (2 t)=2 \sin (t) \cos (t) \\ & \cos (2 t)=\cos ^{2}(t)-\sin ^{2}(t)=2 \cos ^{2}(t)-1=1-2 \sin ^{2}(t) \end{aligned}