### Instruction: Compute $\cos 36^\circ.$.### Response: This is not one of the standard angles that I have memorized, so I will need to use some trigonometry or geometry to find it.
I recall that there is a formula for $\cos 3\theta$ in terms of $\cos \theta,$ which might be useful if I can relate $36^\circ$ to some multiple of $3^\circ.$
In fact, $36^\circ = 3 \times 12^\circ,$ so I can try to use the formula $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta.$
Substituting $\theta = 12^\circ,$ I get $\cos 36^\circ = 4\cos^3 12^\circ - 3\cos 12^\circ.$
Now I need to find $\cos 12^\circ,$ which is still not a standard angle, but I can use another trick to simplify it.
I notice that $12^\circ = 15^\circ - 3^\circ,$ and I know that $\cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}$ from the half-angle formula for $\cos 30^\circ.$
So I can use the difference formula for cosine to get $\cos 12^\circ = \cos 15^\circ \cos 3^\circ + \sin 15^\circ \sin 3^\circ.$
I also know that $\cos 3^\circ \approx 0.9986$ and $\sin 3^\circ \approx 0.0523$ from using a calculator or a Taylor series approximation.
And I can use the Pythagorean identity to get $\sin 15^\circ = \sqrt{1 - \cos^2 15^\circ}.$
Plugging in these values, I get $\cos 12^\circ \approx \frac{\sqrt{6} + \sqrt{2}}{4} \times 0.9986 + \sqrt{1 - \left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2} \times 0.0523.$
Simplifying this expression, I get $\cos 12^\circ \approx 0.9781.$
Finally, I can plug this back into the expression for $\cos 36^\circ = 4\cos^3 12^\circ - 3\cos 12^\circ.$
This gives me $\cos 36^\circ \approx 4 \times 0.9781^3 - 3 \times 0.9781.$
This simplifies to $\cos 36^\circ \approx 0.8090.$