Good morning! I'm having trouble with this problem… It's just taking me forever and I'm worn out and I'm lost on how to use a double angle identity for $72=2⋅36$
The problem reads as follows
An exact value for $\cos36°$ can be found using the following procedure. Begin by considering $\sin108°$. Note that $108=72+36$ and use the sine sum identity. Also note that $72=2⋅36$ and use double angle identities. If there are any common factors in each term, factor them out and cancel them if they are not equal to zero. You should eventually obtain a quadratic equation containing $\cos36°$. Use the quadratic formula to obtain the exact value for $\cos36°$. (Note that the quadratic formula should give two solutions. One can be disregarded – why?
Thank you in advance to anyone who can help.
Best Answer
\begin{align} \sin 108^\circ&=\sin (72^\circ+36^\circ)\\ \sin(180^\circ-72^\circ)&=\sin 72^\circ\cos36^\circ+\cos 72^\circ\sin36^\circ\\ \sin72^\circ&=\sin (2\cdot36^\circ)\cos36^\circ+\cos (2\cdot36^\circ)\sin36^\circ\\ \sin (2\cdot36^\circ)&=2\sin36^\circ\cos36^\circ\cos36^\circ+(2\cos^236^\circ-1)\sin36^\circ\\ 2\sin36^\circ\cos36^\circ&=\sin36^\circ(2\cos^236^\circ+2\cos^236^\circ-1)\\ 2\cos36^\circ&=4\cos^236^\circ-1\\ 4\cos^236^\circ-2\cos36^\circ-1&=0\tag1 \end{align} Let $y=\cos36^\circ$, then $(1)$ becomes $$ 4y^2-2y-1=0\tag2 $$ Use complete square method or quadratic formula to obtain the roots of $(2)$ and take the positive value only because $\cos36^\circ>0$ since it lies on the first quadrant.