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I'm struggling with the following: We are to use the sum and difference formulas to find the exact value of the expression. The problem is simplification has been tough. As a last resort I decided to use Symbolab to find the answer and steps but the steps were not to be found. Despite lack of steps, the answer is $(\sqrt {2+\sqrt{3}})/2$

Here are my steps so far: \begin{align*} \sin(135^\circ - 30^\circ) & = (\sin 135^\circ\cos 30^\circ)-(\cos 135^\circ \sin 30^\circ)\\ & = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(-\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)\\ & = \left(\frac{\sqrt{6}}{4}\right)-\left(-\frac{\sqrt{2}}{4}\right) \end{align*} [insert final simplification step]
$(\sqrt {2+\sqrt{3}})/2$

What's the missing step here?

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  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. Also, your equations should include equals signs. $\endgroup$ – N. F. Taussig Feb 12 '16 at 12:20
  • $\begingroup$ its not an equation. Its an expression. $\endgroup$ – Diamond Louis XIV Feb 12 '16 at 18:18
  • $\begingroup$ When you evaluate the expression $\sin(135^{\circ} - 30^{\circ})$ by setting it equal to $\sin(135^{\circ})\cos(30^{\circ}) - \cos(135^\circ)\sin(30^{\circ})$, you have an equation. $\endgroup$ – N. F. Taussig Feb 12 '16 at 18:22
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\begin{align*} \frac{\sqrt{6}}{4}- \left(-\frac{\sqrt{2}}{4}\right) & = \frac{1}{4}(\sqrt6+\sqrt2)\\ & = \frac{\sqrt{2}}{4}(\sqrt{3}+1)\\ & = \frac{\sqrt{2}}{4}\sqrt{(\sqrt3+1)^2}\\ & = \frac{\sqrt{2}}{4}\sqrt{4+2\sqrt3}\\ & = \frac{\sqrt{2}}{4}\sqrt{2}\sqrt{2+\sqrt{3}}\\ & = \frac{1}{2}\sqrt{2+\sqrt{3}} \end{align*}

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  • $\begingroup$ I dont see how you get from one step to the next $\endgroup$ – Diamond Louis XIV Feb 12 '16 at 7:12
  • $\begingroup$ @DiamondLouisXIV edited to simpler form $\endgroup$ – Ilyas y. Feb 12 '16 at 7:17
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. Also, your equations should include equals signs. $\endgroup$ – N. F. Taussig Feb 12 '16 at 12:25
  • $\begingroup$ @N.F.Taussig thanks, got it $\endgroup$ – Ilyas y. Feb 12 '16 at 12:36
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$x=\frac{\sqrt6+\sqrt2}4$
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$x^2=\frac{6+2\sqrt{12}+2}{16}=\frac{8+2\sqrt{12}}{16}=\frac{8+4\sqrt3}{16}=\frac{2+\sqrt3}4$
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$x=\frac{\sqrt{2+\sqrt3}}2$

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  • $\begingroup$ The two expressions are equal because they are both positive numbers with the same square. $\endgroup$ – N. F. Taussig Feb 12 '16 at 12:38

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