Show the exact value of cosine of $144^\circ$ using the law of cosines. By this law, $\cos144^\circ=1-2\sin^2 72^\circ$. And that is equivalent to $\cos144^\circ=1-2(2\sin36^\circ\cos36^\circ)^2$. What would be the next steps to solve this?
[Math] Show the exact value of cosine of 144° using law of cosines.
trigonometry
Related Solutions
Below is the Spherical Law of Cosines as it appears in UCSMP Functions, Statistics, and Trigonometry, 3rd ed., copied here because the diagram is good and helps with clarity.
If $\triangle ABC$ is a spherical triangle with arcs $a$, $b$, and $c$ (meaning the measures of the arcs, not the lengths), then $\cos c=\cos a\cos b+\sin a\sin b\cos C$.
Now, to the specific problem at hand. Let's use the diagram below, also from UCSMP Functions, Statistics, and Trigonometry, 3rd ed., for reference.
Let $A$ and $B$ be as you defined them. $N$ and $S$ are the north and south poles, respectively; $C$ and $D$ are the points on the equator that are on the same line of longitude as $A$ and $B$, respectively. Consider spherical $\triangle ABN$. $a=(90°-\text{latitude of point }B)$; $b=(90°-\text{latitude of point }A)$. $N=\text{positive difference in longitude between points }A\text{ and }B$. Use the Spherical Law of Cosines ($\cos n=\cdots$ form) to determine $n$, which is the shortest arc between the two points.
(graphics from Lesson 5-10 of UCSMP Functions, Statistics, and Trigonometry, 3rd ed., © 2010 Wright Group/McGraw Hill)
It was a good idea to do the problem a third way -- it gives more credence to the law of cosines answer than to the law of sines answer. Additionally, you seem to be somewhat uncomfortable with the law of sines method -- you mention that you consider using $25$ degrees instead of $10$ degrees, so you are at least somewhat unsure what that angle should be.
You're thus very close to the root of the issue: the angle between the parallelogram's diagonal and the bottom vector is unknown. It's not $10$ degrees (that's the angle between the x-axis and the bottom vector), and it's not $25$ degrees (that's the angle between the top vector and the bottom vector). We can't apply the law of sines because we don't have two known angles.
Best Answer
I really don't see how this can be accomplished using the Law of Cosines, but here's a way that we can go about it.
Note that $$\cos 144^\circ=\cos(2\cdot 72^\circ)=1-2\sin^2(72^\circ).$$ If we can find $\sin^2(72^\circ),$ then, we will be able to find $\cos 144^\circ.$
Now, for any angle $\theta,$ we have by sum and difference formulas, double-angle formulas, and Pythagorean identity that $$\begin{align}\sin(5\theta) &= \sin(\theta+4\theta)\\ &= \sin\theta\cos4\theta+\sin4\theta\cos\theta\\ &= \sin\theta\bigl(1-2\sin^22\theta\bigr)+2\sin2\theta\cos2\theta\cos\theta\\ &= \sin\theta-2\sin^22\theta\sin\theta+2\sin2\theta\bigl(1-2\sin^2\theta\bigr)\cos\theta\\ &= \sin\theta-2(2\sin\theta\cos\theta)^2\sin\theta+2(2\sin\theta\cos\theta)\bigl(1-2\sin^2\theta\bigr)\cos\theta\\ &= \sin\theta-8\sin^3\theta\cos^2\theta+4\sin\theta\cos^2\theta-8\sin^3\theta\cos^2\theta\\ &= \sin\theta+(4\sin\theta-16\sin^3\theta)\cos^2\theta\\ &= \sin\theta+(4\sin\theta-16\sin^3\theta)(1-\sin^2\theta)\\ &= \sin\theta+4\sin\theta-4\sin^3\theta-16\sin^3\theta+16\sin^5\theta\\ &= 5\sin\theta-20\sin^3\theta+16\sin^5\theta.\end{align}$$
In particular, for $\theta=72^\circ,$ making the substitution $s=\sin 72^\circ,$ we have $$0=\sin 360^\circ=\sin(5\cdot 72^\circ)=5s-20s^3+16s^5.$$ Observing that we can't have $\sin 72^\circ=0,$ we have $$0=5-20s^2+16s^4=16(s^2)^2-20(s^2)+5.$$ By quadratic formula, we find that $$s^2=\frac{5\pm\sqrt5}8.$$ Observing that $s\ge\sin 60^\circ=\frac{\sqrt3}2,$ we have $s^2\ge\frac34,$ so we conclude that $s^2=\frac{5+\sqrt5}8,$ and so $$\cos 144^\circ=1-2\sin^2(72^\circ)=1-2s^2=1-\frac{5+\sqrt5}4=-\frac{1+\sqrt5}4.$$