Without using the value for $\sin 18^°$ or $\cos 36^°$, prove that $4\sin 36^°\cos 18^°=\sqrt 5$
I tried to square it use some trigonometric formulas but could solve it.
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2See this PDF: https://math.la.asu.edu/~surgent/mat170/Exact_Trig_Values.pdf – Batman Jul 02 '14 at 07:45
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this is the same thing as putting the values.!! we dont have to plug in values.!! – maths lover Jul 02 '14 at 07:49
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What? You can calculate $\sin 36^\circ$ as in that pdf, and then use the half angle formula in the inane form $\cos 18^\circ = \sqrt{\frac{1}{2} ( 1 + \cos 36^\circ ) } = \sqrt{\frac{1}{2} ( 1 + \sqrt{1 - (\sin 36^\circ)^2 })} $ if you want to avoid those two values... – Batman Jul 02 '14 at 07:54
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so basically we are using the value of sin36 – maths lover Jul 02 '14 at 08:02
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Sure, which you didn't preclude (you only precluded $\sin 18^\circ$ and $\cos 36^\circ$) - the question is inane beyond the point presented at that link. – Batman Jul 02 '14 at 08:04
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ya basically we dont have to use values which are not common..!! – maths lover Jul 02 '14 at 08:04
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1It's hard to imagine an easier way to solve this than how the pdf suggests, with a 36-72-72 triangle. – DanielV Jul 02 '14 at 08:42
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ohk.!!!!!!!!!!!!!!!!!!!!! – maths lover Jul 02 '14 at 09:02