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Trigonometry: Double-Angle and Half-Angle Formulas

Sol A $\cos^4\theta-\sin^4\theta=(\cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^2\theta)$ $=(\cos 2\theta)(1)=\cos 2\theta$.

Sol 1 $\cos 2\theta=2\cos^2 \theta-1=2(4/7)^2-1=2(16/49)-1=32/49-1=-17/49$.

Sol 2 $\cos^2 \theta=1-\sin^2 \theta=1-(5/13)^2=1-25/169=144/169$, so $\cos\theta=-12/13$ since $\tan\theta=\frac{\sin\theta}{\cos\theta}$ with $\sin\theta>0$ and $\tan\theta<0 \Rightarrow\cos\theta<0$.

Therefore $\sin 2\theta=2\sin\theta \cos\theta=2(5/13)(-12/13)=-120/169$.

Sol 3

\begin{displaymath}\cos\pi/8=+\sqrt{\frac{1+\cos(\pi/4)}{2}}\end{displaymath}


\begin{displaymath}=\sqrt{\frac{1+\sqrt{2}/2}{2}}=\sqrt{\frac{2+\sqrt{2}}{4}}\end{displaymath}


\begin{displaymath}=\frac{\sqrt{2+\sqrt{2}}}{2}\end{displaymath}

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Sol 4 $\sin\frac{\theta}{2}+\cos\frac{\theta}{2})^2$

\begin{displaymath}=\sin^2 \theta/2+2\sin\theta/2 \cos\theta/2+\cos^2 \theta/2\end{displaymath}


\begin{displaymath}=(\sin^2 \theta/2+\cos^2 \theta/2)+2\sin\theta/2 \cos\theta/2=1+\sin\theta\end{displaymath}

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Sol 5 $\sin^2\theta \cos^2\theta=(\sin\theta \cos\theta)^2$ $=(\frac{1}{2}(2\sin\theta\cos\theta))^2=(\frac{1}{2}\sin 2\theta)^2$

$=\frac{1}{4}\sin^2 2\theta=\frac{1}{4}(\frac{1}{2}(1-\cos 4\theta))$ $=\frac{1}{8}(1-\cos 4\theta)$.

Sol 6 $\cos^4 \theta=(\cos^2 \theta)^2=(1/2(1+\cos 2\theta))^2$ $=1/4(1+2\cos 2\theta+\cos^2 2\theta)$

$=1/4(1+2\cos 2\theta+1/2(1+\cos 4\theta))$ $=1/4(1+2\cos 2\theta+1/2+1/2\cos 4\theta)$

$=1/4(3/2+2\cos 2\theta+1/2\cos 4\theta)$.

Sol 7 $\cos^2 \theta=1-\sin^2 \theta=1-(4/5)^2=1-16/25=9/25$, so

$\cos\theta=-3/5$ since $\cos\theta<0$ for $\pi/2<\theta<\pi$. Then

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

$=4(4/5)(-3/5)(9/25-16/25)=4(4/5)(-3/5)(-7/25)=336/625$.

Sol 8 $\cos^2\theta=1-\sin^2\theta=1-(-5/8)^2=1-25/64=39/64$, so

$\cos\theta=\frac{\sqrt{39}}{8}$ since $\cos\theta>0$ for $270^\circ<\theta<360^\circ$.

Since $135^\circ<\frac{\theta}{2}<180^\circ$, $\cos\frac{\theta}{2}<0$, so


\begin{displaymath}\cos\frac{\theta}{2}=-\sqrt{\frac{1+\cos\theta}{2}}\end{displaymath}


\begin{displaymath}=-\sqrt{\frac{1+\sqrt{39}/8}{2}}=-\sqrt{\frac{8+\sqrt{39}}{16}}\end{displaymath}


\begin{displaymath}=-\frac{\sqrt{8+\sqrt{39}}}{4}\end{displaymath}

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Lawrence Marx 2002-07-11