Let P= [0,4 ]: ^2 1 and S=a Z:2( ^8 - ^8 ) 2 =a^2, P.

Question

Let $P={	heta \in [0,4\pi ]:{	an }^{2}	heta 
e 1}$ and $S={a\in Z:2({\cos }^{8}	heta -{\sin }^{8}	heta )\sec 2	heta ={a}^{2},	heta \in P}$. Then $n(S)$ is:

Accepted Answer

$$2(\cos^8\theta-\sin^8\theta)\sec2\theta = 2(\cos^4\theta+\sin^4\theta)(\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta)\sec2\theta$$

$$= 2(\cos^4\theta+\sin^4\theta)\cos2\theta\sec2\theta = 2(\cos^4\theta+\sin^4\theta) = 2\left(1-\tfrac12\sin^22\theta\right) = a^2$$

So $a^2 = 2-\sin^22\theta \in [1,2]$. Since $a$ is an integer, $a^2$ must be a perfect square lying in $[1,2]$, forcing $a^2=1$, which requires $\sin^22\theta = 1$, i.e. $2\theta = (2n+1)\dfrac\pi2$, so $\theta = (2n+1)\dfrac\pi4$.

But at such $\theta$, $\tan\theta = \pm1$, i.e. $\tan^2\theta = 1$ --- which is excluded from $P$. Hence no $\theta \in P$ produces an integer value of $a$, so $S = \varnothing$ and $n(S) = 0$.

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