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半角公式05
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2022-10-14 07:48
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<p><strong>半角公式</strong></p> <p><strong><span class="math-tex">\( \sin \dfrac{\theta}{2}=\pm \sqrt{\dfrac{1-\cos \theta}{2}} \) </span></strong></p> <p><strong><span class="math-tex">\( \cos \dfrac{\theta}{2}=\pm \sqrt{\dfrac{1+\cos \theta}{2}} \) </span></strong></p> <p><strong><span class="math-tex">\( \tan \dfrac{\theta}{2} =\csc \theta-\cot \theta =\pm \sqrt{\dfrac{1-\cos \theta}{1+\cos \theta}} =\dfrac{\sin \theta}{1+\cos \theta}=\dfrac{1-\cos \theta}{\sin \theta} \) </span></strong></p> <p><strong><span class="math-tex">\( \cot \dfrac{\theta}{2} =\csc \theta+\cot \theta =\pm \sqrt{\dfrac{1+\cos \theta}{1-\cos \theta}} \) </span></strong></p> <p><strong><span class="math-tex">\( \sec \dfrac{\theta}{2}=\pm \sqrt{\dfrac{2 \sec \theta}{\sec \theta+1}} \) </span></strong></p> <p><strong><span class="math-tex">\( \csc \dfrac{\theta}{2}=\pm \sqrt{\dfrac{2 \sec \theta}{\sec \theta-1}} \) </span></strong></p> <p> </p> <p>典型例题</p> <p>已知<span class="math-tex">\( \sin \theta= -\dfrac{4}{5} \) </span>,求<span class="math-tex">\( \sin \dfrac{\theta}{2} \) </span>,<span class="math-tex">\( \theta \) </span> 为第四象限的角。</p> <p>解:<span class="math-tex">\( \sin \theta= -\dfrac{4}{5} \) </span>, 所以 <span class="math-tex">\( \cos \theta= \dfrac{3}{5} \) </span>, 且<span class="math-tex">\( \dfrac{\theta}{2} \) </span>为第二或者第四象限的角</p> <p>所以 <span class="math-tex">\( \sin \dfrac{\theta}{2}=\pm \sqrt{\dfrac{1-\cos \theta}{2}} \) </span>=<span class="math-tex">\( \pm \dfrac{\sqrt{5}}{5} \) </span></p> <p>1</p> <p>证明:<span class="math-tex">\( \sin 2 \theta =2 \sin \theta \cos \theta \) </span></p> <p>证:根据 <a href="../kbase/lists.aspx?kid=42" target="_blank" rel="noopener">两角和与两角差</a>恒等式,有</p> <p><span class="math-tex">\( \sin (\alpha +\beta)=\sin \alpha \cos \beta +\cos \alpha \sin \beta \) </span></p> <p>令<span class="math-tex">\( \alpha=\beta \) </span>, 则有 <span class="math-tex">\( \sin 2 \theta =2 \sin \theta \cos \theta \) </span></p> <p> </p> <p>下图解释了正弦2背公式的几何意义。在左图中,三角形的面积为</p> <p><span class="math-tex">\( S_1=\dfrac{1}{2} (sin \theta + sin \theta) * \cos \theta =\sin \theta cos \theta \) </span></p> <p>将此三角形旋转180度,则有三角形面积为</p> <p><span class="math-tex">\( S_2=\dfrac{1}{2} *1 * \sin {2\theta} \) </span>=<span class="math-tex">\( \dfrac{1}{2} \sin {2\theta} \) </span></p> <p>因为 <span class="math-tex">\( S_1=S_2 \) </span></p> <p>所以 <span class="math-tex">\( \sin \theta cos \theta = \dfrac{1}{2} \sin {2\theta} \) </span> 即</p> <p><span class="math-tex">\( \sin 2 \theta =2 \sin \theta \cos \theta \) </span></p> <p><img src="../uploads/2022-10/fc0bdd.jpg" width="350px"></p> <p> </p> <p><strong>典型例题</strong></p> <p>已知 <span class="math-tex">\( \tan 2 x=-2 \sqrt{2} \quad \pi<2 x<2 \pi \) </span>,求 <span class="math-tex">\( \dfrac{2 \cos ^2 \dfrac{x}{2}-\sin x-1}{\sqrt{2} \sin \left(x+\dfrac{\pi}{4}\right)} \) </span></p> <p>解:<span class="math-tex">\( \tan 2x=\dfrac{2 \tan x}{1-\tan^2x} =-\sqrt{2} \) </span></p> <p><span class="math-tex">\( \therefore \sqrt{2} {\tan^2 x- 2\tan x}- \sqrt{2}=0 \) </span></p> <p><span class="math-tex">\( \therefore \tan x=\sqrt{2} \) </span> 或 <span class="math-tex">\( \tan x=-\dfrac{\sqrt{2}}{2} \) </span></p> <p><span class="math-tex">\( \because \dfrac{\pi}{2}< x < \pi \) </span> </p> <p><span class="math-tex">\( \therefore \tan x <0 \) </span></p> <p><span class="math-tex">\( \therefore \quad \tan x=-\frac{\sqrt{2}}{2} \) </span></p> <p>原式<span class="math-tex">\( =\dfrac{\cos x-\sin x}{\sin x+\cos x} \) </span>,分子分母同除以<span class="math-tex">\( \cos x \) </span></p> <p><span class="math-tex">\( =\dfrac{1-\tan x}{1+\tan x}=\dfrac{1+\dfrac{\sqrt{2}}{2}}{1-\dfrac{\sqrt{2}}{2}}=3+2 \sqrt{2} \) </span></p> <p>1</p> <p> </p>
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