Updated greek laters @ latex compiling

astrosander edited
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revision #10120 parent #9673 GitHub dbb31ba ← older newer →
en/2.5.17.md +3 −3
@@ -77,11 +77,11 @@
Let's regroup this into a quadratic equation
<div class="scroll-wrapper">$$p_1^2\left(1+\frac{m_2}{m_1}\right)-2p_0p_1\cos\alpha+p_0^2\left(1-\frac{m_2}{m_1}\right)=0$$</div>
Solving relative to $p_1$
−<div class="scroll-wrapper">$$p_1=\frac{2p_0\cos\alpha\pm \sqrt{4p_0^2\cos^2\alpha-4p_0^2\left(1-\frac{m_2^2}{m_1^2}\right)}}{2\left(1+\frac{m_2}{m_1}\right)}$$</div>
+<div class="scroll-wrapper">$$p_1=\frac{2p_0\cos\alpha\pm \sqrt{4p_0^2\cos^2 \alpha-4p_0^2\left(1-\frac{m_2^2}{m_1^2}\right)}}{2\left(1+\frac{m_2}{m_1}\right)}$$</div>
Since $\Delta = 0$
−$$\cos^2\alpha=1-\frac{m_2^2}{m_1^2}$$
+$$\cos^2 \alpha=1-\frac{m_2^2}{m_1^2}$$
From where we find the required ratio
−$$\frac{m_2^2}{m_1^2}=1-\cos^2\alpha=\sin^2\alpha$$
+$$\frac{m_2^2}{m_1^2}=1-\cos^2 \alpha=\sin^2 \alpha$$
$$\boxed{\sin\alpha=\frac{m_2}{m_1}}$$
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