I simplified this to:
$\sqrt {150+30\sqrt 5} + 14 + 9\sqrt{6-2\sqrt5}= 10\sqrt5+\sqrt{30+6\sqrt5} + 2\sqrt{30-6\sqrt5}+\sqrt{30-10 \sqrt5}$
any thoughts
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We need to prove that $$\cos108^{\circ}+\cos60^{\circ}=1+\cos144^{\circ}$$ or $$\cos108^{\circ}-\cos144^{\circ}=\frac{1}{2}$$ or $$2\sin18^{\circ}\sin126^{\circ}=\frac{1}{4}$$ or $$\sin18^{\circ}\cos36^{\circ}=\frac{1}{4}$$ or $$\frac{\sqrt5-1}{4}\cdot\frac{\sqrt5+1}{4}=\frac{1}{4},$$ which is obvious.
Id est, your statement is true.
Done!
About your identity.
We need to prove that $$\sqrt {150+30\sqrt 5} + 14 + 9(\sqrt5-1)= 10\sqrt5+\sqrt{30+6\sqrt5} + 2\sqrt{30-6\sqrt5}+\sqrt5(\sqrt5-1)$$ or $$\sqrt {150+30\sqrt 5}=\sqrt{30+6\sqrt5} + 2\sqrt{30-6\sqrt5}$$ or $$150+30\sqrt5=30+6\sqrt5+4\sqrt{(30+6\sqrt5)(30-6\sqrt5)}+4(30-6\sqrt5)$$ or $$48\sqrt5=4\sqrt{900-180}.$$ Done!
Michael Rozenberg
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OK cute but what about the above equation – Howard Jul 29 '17 at 21:31
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@Howard I think there is a mistake in your computations. $LS-RS=-2.288...$ – Michael Rozenberg Jul 29 '17 at 21:35
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39.85831599 for both – Howard Jul 29 '17 at 21:40
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@Howard See here: http://www.wolframalpha.com/input/?i=%5Csqrt+%7B150%2B30%5Csqrt+5%7D+%2B+14+%2B+9%5Csqrt%7B6-2%5Csqrt5%7D-(10%5Csqrt5%2B%5Csqrt%7B30%2B6%5Csqrt5%7D+%2B+2%5Csqrt%7B30-6%5Csqrt5%7D%2B%5Csqrt%7B30-2%5Csqrt5%7D) – Michael Rozenberg Jul 29 '17 at 21:46
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found and corrected my typo – Howard Jul 29 '17 at 21:59
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$$\cos108^\circ-\cos144^\circ=\cos(180-144)^\circ-\cos(180-108)^\circ=?$$ https://math.stackexchange.com/questions/827540/proving-trigonometric-equation-cos36-circ-cos72-circ-1-2 – lab bhattacharjee Jul 30 '17 at 04:00
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@lab bhattacharjee I think In math all problems are duplicate because they are math problems. What do you think? – Michael Rozenberg Jul 30 '17 at 04:35
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Hint: use the sum to product formula for the left hand side...
$$\cos \alpha \cos \beta = \frac {cos (\alpha - \beta) + \cos (\alpha + \beta)}{2}$$
Then use the power reduction formula for cosine on the right hand side...
$$\cos^2 \alpha = \frac {1 + \cos 2 \alpha}{2}$$
bjcolby15
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Yes I got that very elegant but I am interested with working with the radicals perhaps I need a different question – Howard Jul 29 '17 at 22:06
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