# Quod Erat Demonstrandum

## 2015/05/19

### 某中三三角學題

Filed under: Junior Form Mathematics — johnmayhk @ 11:47 上午
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Given that $\tan \theta=\sqrt{2}$, where $\theta$ is an acute angle. Using trigonometric identities, find the value of $\sin^2\theta - \cos^2\theta$.

$\sin\theta = \frac{\sqrt{2}}{\sqrt{3}}$
$\cos\theta = \frac{1}{\sqrt{3}}$

$\tan \theta=\sqrt{2} \Rightarrow \sin \theta = \sqrt{2}\cos \theta$

$\sin^2\theta + \cos^2\theta = 1$

$(\sqrt{2}\cos \theta)^2 + \cos^2\theta = 1$

$\Rightarrow \cos^2\theta = \frac{1}{3}$

$\Rightarrow \sin^2\theta = 1-\cos^2\theta=1-\frac{1}{3}=\frac{2}{3}$

$\sin^2\theta - \cos^2\theta=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$

$\tan\phi=\frac{a+b\sin\theta}{b\cos\theta}$

$\tan\phi=\frac{a+b\sin\theta}{b\cos\theta}$

$x=a\sin(90^o-\phi)=a\cos\phi$

$\sin\alpha = \frac{x}{b} = \frac{a\cos\phi}{b}$

$\alpha = \sin^{-1}\frac{a\cos\phi}{b}$

$\theta = \phi - \alpha = \phi - \sin^{-1}\frac{a\cos\phi}{b}$

$\tan\phi=\frac{a+b\sin\theta}{b\cos\theta}$

$\frac{\sin \phi}{\cos \phi}=\frac{a+b\sin\theta}{b\cos\theta}$

$b\sin \phi \cos\theta=a\cos \phi + b\cos\phi\sin\theta$

$b(\sin \phi \cos\theta-\cos\phi\sin\theta)=a\cos \phi$

$\sin(\phi-\theta)=\frac{a\cos\phi}{b}$

$\phi-\theta=\sin^{-1}\frac{a\cos\phi}{b}$

$\theta = \phi - \sin^{-1}\frac{a\cos\phi}{b}$