trigonometry
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mmm4444bot
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r267747 said:
Why did you post this equation?
BigGlenntheHeavy
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\(\displaystyle Prove: \ sin(10^{0})+sin(20^{0})+sin(40^{0})+sin(50^{0}) \ = \ sin(70^{0})+sin(80^{0})\)
\(\displaystyle [sin(10^{0})+sin(20^{0})]+[sin(40^{0})+sin(50^{0})] \ = \ [sin(70^{0})+sin(80^{0})]\)
\(\displaystyle 2sin(15^{0})cos(5^{0})+2sin(45^{0})cos(5^{0}) \ = \ 2sin(75^{0})cos(5^{0}), \ Sum-to-Product \ Formulas\)
\(\displaystyle sin(15^{0})+sin(45^{0}) \ = \ sin(75^{0})\)
\(\displaystyle sin(60^{0}-45^{0})+sin(45^{0}) \ = \ sin(30^{0}+45^{0})\)
\(\displaystyle sin(60^{0})cos(45^{0})-cos(60^{0})sin(45^{0})+sin(45^{0}) \ = \ sin(30^{0})cos(45^{0})+cos(30^{0})sin(45^{0})\)
\(\displaystyle \bigg(\frac{\sqrt3}{2}\bigg)\bigg(\frac{1}{\sqrt2}\bigg)-\bigg(\frac{1}{2}\bigg)\bigg(\frac{1}{\sqrt2}\bigg)+\frac{1}{\sqrt2} \ = \ \bigg(\frac{1}{2}\bigg)\bigg(\frac{1}{\sqrt2}\bigg)+\bigg(\frac{\sqrt3}{2}\bigg)\bigg(\frac{1}{\sqrt2}\bigg)\)
\(\displaystyle = \ \bigg(\frac{\sqrt3}{2\sqrt2}\bigg)-\bigg(\frac{1}{2\sqrt2}\bigg)+\frac{1}{\sqrt2} \ = \ \bigg(\frac{1}{2\sqrt2}\bigg)+\bigg(\frac{\sqrt3}{2\sqrt2}\bigg)\)
\(\displaystyle Hence, \ \frac{\sqrt3}{2\sqrt2}+\frac{1}{2\sqrt2} \ = \ \frac{\sqrt3}{2\sqrt2}+\frac{1}{2\sqrt2}, \ QED\)
\(\displaystyle [sin(10^{0})+sin(20^{0})]+[sin(40^{0})+sin(50^{0})] \ = \ [sin(70^{0})+sin(80^{0})]\)
\(\displaystyle 2sin(15^{0})cos(5^{0})+2sin(45^{0})cos(5^{0}) \ = \ 2sin(75^{0})cos(5^{0}), \ Sum-to-Product \ Formulas\)
\(\displaystyle sin(15^{0})+sin(45^{0}) \ = \ sin(75^{0})\)
\(\displaystyle sin(60^{0}-45^{0})+sin(45^{0}) \ = \ sin(30^{0}+45^{0})\)
\(\displaystyle sin(60^{0})cos(45^{0})-cos(60^{0})sin(45^{0})+sin(45^{0}) \ = \ sin(30^{0})cos(45^{0})+cos(30^{0})sin(45^{0})\)
\(\displaystyle \bigg(\frac{\sqrt3}{2}\bigg)\bigg(\frac{1}{\sqrt2}\bigg)-\bigg(\frac{1}{2}\bigg)\bigg(\frac{1}{\sqrt2}\bigg)+\frac{1}{\sqrt2} \ = \ \bigg(\frac{1}{2}\bigg)\bigg(\frac{1}{\sqrt2}\bigg)+\bigg(\frac{\sqrt3}{2}\bigg)\bigg(\frac{1}{\sqrt2}\bigg)\)
\(\displaystyle = \ \bigg(\frac{\sqrt3}{2\sqrt2}\bigg)-\bigg(\frac{1}{2\sqrt2}\bigg)+\frac{1}{\sqrt2} \ = \ \bigg(\frac{1}{2\sqrt2}\bigg)+\bigg(\frac{\sqrt3}{2\sqrt2}\bigg)\)
\(\displaystyle Hence, \ \frac{\sqrt3}{2\sqrt2}+\frac{1}{2\sqrt2} \ = \ \frac{\sqrt3}{2\sqrt2}+\frac{1}{2\sqrt2}, \ QED\)