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Trig q (1 Viewer)
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using the second equation
+2\cos^2(\beta)=4\\\\\frac{3}{2}(1+\cos(2\alpha))+(1+\cos(2\beta))=4\\\\3+3\cos(2\alpha)+2+2\cos(2\beta)=8\\\\3\cos(2\alpha)=3-2\cos(2\beta)\\\\\cos(2\alpha)=\frac{3-2\cos(2\beta)}{3})
using Pythagoras we can say that=\frac{\sqrt{3^2-(3-2\cos(2\beta))^2}}{3})
=\frac{\sqrt{9-(9-12\cos(2\beta)+4\cos^2(2\beta))}}{3}\\\\\sin(2\alpha)=\frac{\sqrt{12\cos(2\beta)-4\cos^2(2\beta)}}{3})
now we sub this expression into the first equation
-6\cos^2(2\beta)}}{3}\right)-2\sin(2\beta)=0\\\\\\\sqrt{12\cos(2\beta)-4\cos^2(2\beta)}=2\sin(2\beta)\\\\12\cos(2\beta)-4\cos^2(2\beta)=4\sin^2(2\beta)\\\\12\cos(2\beta)-4\cos^2(2\beta)=4-4\cos^2(2\beta)\\\\\cos(2\beta)=\frac{1}{3}\\\\\beta=\frac{1}{2}\cos^{-1}\left(\frac{1}{3}\right))
now that we have found beta, just sub this value into=\frac{3-2\cos(2\beta)}{3})
or the first equation to find alpha which should be )
or some other equivalent value.
using Pythagoras we can say that
now we sub this expression into the first equation
now that we have found beta, just sub this value into
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compound angle/double angle
practice
yanujw
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Try to derive it yourself. Use the relationship between
In a sense the thought process is similar to the way you get the expression
Spoiler if you couldn't find it out
Here is what I basically skipped in those two lines
\\\\=\frac{1}{2}(2\cos^2(\alpha))\\\\=\frac{1}{2}(\cos^2(\alpha)+\sin^2(\alpha)+\cos^2(\alpha)-\sin^2(\alpha))\\\\=\frac{1}{2}(\cos(\alpha-\alpha)+\cos(2\alpha))\\\\=\frac{1}{2}(1+\cos(2\alpha)))
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