Gelfand trigonometry question

Question

If we start with a lemma that states that when $ a^2+b^2=1$ there exists an angle $ \theta $ such that $ a=\cos\theta $ and $ b=\sin\theta$

Suppose that $\alpha$ is some angle if $a=4\cos^3\alpha-3\cos\alpha $ and $ b=3\sin\alpha -4\sin^3\alpha $ show that there is an angle $\theta $ such that $ a=\cos\theta $ and $ b=\sin\theta$

Thanks. I think I have to expand and then reduce. But it isn't coming together. Any suggestions?

Answer 1

If you are familiar with the triple angle formulas

$$a = 4\cos^3\alpha - 3\cos\alpha = \cos(3\alpha)$$

and

$$b = 3\sin\alpha - 4\sin^3\alpha = \sin(3\alpha)$$

then you can take $\theta = 3\alpha$.

If not, you can use the Pythagorean Identity $\sin^2\varphi + \cos^2\varphi = 1$ to simplify the problem.

\begin{align*} a & = 4\cos^3\alpha - 3\cos\alpha\\ & = 4\cos\alpha(1 - \sin^2\alpha) - 3\cos\alpha\\ & = 4\cos\alpha - 4\cos\alpha\sin^2\alpha - 3\cos\alpha\\ & = \cos\alpha - 4\cos\alpha\sin^2\alpha \end{align*}

\begin{align*} b & = 3\sin\alpha - 4\sin^3\alpha\\ & = 3\sin\alpha - 4\sin\alpha(1 - \cos^2\alpha)\\ & = 3\sin\alpha - 4\sin\alpha + 4\sin\alpha\cos^2\alpha\\ & = -\sin\alpha + 4\sin\alpha\cos^2\alpha \end{align*}

Therefore,

\begin{align*} a^2 + b^2 & = (4\cos^3\alpha - 3\cos\alpha)^2 + (3\sin\alpha - 4\sin\alpha)^2\\ & = (\cos\alpha - 4\cos\alpha\sin^2\alpha)^2 + (-\sin\alpha + 4\sin\alpha\cos^2\alpha)^2\\ & = \cos^2\alpha - 8\cos^2\sin^2\alpha + 16\cos^2\sin^4\alpha + \sin^2\alpha - 8\sin^2\alpha\cos^2\alpha + 16\cos^4\alpha\sin^2\alpha\\ & = 1 - 16\cos^2\alpha\sin^2\alpha + 16\cos^2\alpha\sin^4\alpha + 16\sin^2\alpha\cos^4\alpha\\ & = 1 + 16\cos^2\alpha\sin^2\alpha(-1 + \sin^2\alpha + \cos^2\alpha)\\ & = 1 + 16\cos^2\alpha\sin^2\alpha(-1 + 1)\\ & = 1 + 0\\ & = 1 \end{align*}

Hence, by the theorem in the text by Gelfand and Saul, there exists $\theta$ such that $a = \cos\theta$ and $b = \sin\theta$.

Answer 2

should be easy (edit: not really :-) $$ (4c^3-3c)^2+(3s-4s^3)= 16c^6-24c^4+9c^2+9s^2-24s^4+16s^6=? $$ and collect powers $$ 16c^6+16s^6-24c^4-24s^4+9c^2+9s^2=16(c^6+s^6)-24(c^4+s^4)+9(c^2+s^2) $$ Now $9(c^2+s^2)=1$ and the other terms give you $-8$ $$ 24=24\cdot 1=24 (c^2+s^2)^2=24(c^4+2s^2c^2+s^4)=24(c^4+s^4)+48s^2c^2 $$ and $$ 16=16\cdot 1=16 (c^2+s^2)^3=16(c^6+3c^4s^2+3c^2s^4+s^6)=16(c^6+s^6)+48c^4s^2+48c^2s^4 =16(c^6+s^6)+48c^2s^2(c^2+s^2) $$ which is $$ =16(c^6+s^6)+48c^2s^2 $$ Now subtract!