Trig Identities to Construct a Triangle

Question

Using twice the formula $\cos (2q) = 2 \cos^2 (q)-1$ calculate $\cos 36°$ and $\cos 72°$.

Answer 1

We know that $\cos 72^\circ = \sin 18^\circ $. Also check that $\cos (2\times 72^\circ) = \cos (3\times 72^\circ) $. Thus, $$\Rightarrow 2\cos^2 (72^\circ) -1 =4\cos^3 (72^\circ) -3 \cos (72^\circ) $$ $$\Rightarrow \cos 72^\circ = \frac {\sqrt {5}-1}{4} $$

Now using the above formula, we have $$\cos 36^\circ = \frac {1}{\sqrt {2}} \sqrt {\cos 72^\circ +1} = \frac {\sqrt {5}+1}{4} $$

For the second part, what do you mean by construct? If you want to construct a triangle using instruments, then you can draw a base of required length and from one end, draw a line making an angle of $72^\circ $ from that vertex and then cut a point of required length and join it with the other end of the base.

Hope it helps.

Answer 2

Continuing from @Rohan ‘s work. Since we have that formula for $\cos 36^0$. The starting point is to draw a line whose length is $\sqrt 5$.

1) Create the right $\triangle OAB$ such that $OA = 1$ and $OB = 2$. Then, $AB = \sqrt 5$.

2) Extend BA to C such that $AC = 1$. Then, $BC = \sqrt 5 + 1$.

3) Construct //gm ACDE such that $AE = 4$. Then $CD = AE = 4$.

4) Construct BF such that $BF \bot BC$.

5) Draw the red circle with center at C and radius $= CD = 4$ cutting BF at F.

Hope you can finish the rest.

Answer 3

Rigorously interpreting the question we can start with: $\cos 36°=x$ so that: $$ \cos 72°= \cos(2\cdot 36°)=2x^2-1 \qquad \cos 144°=\cos (2 \cdot 72°)=2(2x^2-1)^2-1 $$ so we have: $$ \cos(144°)=\cos(180°-36°) \quad \iff \quad 2(2x^2-1)^2-1=-x $$ that ''use twice'' the formula in OP. This equation factorize as $(x+1)(2x-1)(4x^2-2x-1)=0$ so the acceptable solution is $$ x=\cos 36°=\frac{1+\sqrt{5}}{4} $$

For the triangle note that $36°$ is the central angle of a regular decagon, so ....see the figure

(for the construction of the decagon you can see here)