So, I have this equation:sin⁡x=4sin⁡10∘sin⁡40∘sin⁡(70∘−x)

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bailaretzy33 · Accepted Answer

First off all it is easy to prove this identity: sin⁡(3x)=4sin⁡(x)⋅sin⁡(60∘−x)⋅sin⁡(60∘+x)Now,sin⁡x=4sin⁡10∘sin⁡40∘sin⁡(70∘−x)sin⁡x=4sin⁡10∘sin⁡50∘sin⁡(70∘)sin⁡(50)sin⁡(70)⋅sin⁡(40)⋅sin⁡(70−x)=sin⁡(30)sin⁡(40)sin⁡(70−x)sin⁡(50)sin⁡(70)sin⁡(30)sin⁡(40)sin⁡(70−x)sin⁡(50)sin⁡(70)=sin⁡(30)sin⁡(40)sin⁡(70−x)sin⁡(50)cos⁡(20)=sin⁡(20)sin⁡(70−x)sin⁡(50)So we are left with, sin⁡(x)sin⁡(50)=sin⁡(20)sin⁡(70−x) . Its clear that x=20. One can prove it is only solution, by expanding...sin⁡(x)sin⁡(50)=sin⁡(20)sin⁡(70)cos⁡(x)−sin⁡(20)sin⁡(x)cos⁡(70)sin⁡(70)cos⁡(20)−sin⁡(20)cos⁡(70)=sin⁡(70−20)=sin⁡(20)sin⁡(70)cot⁡(x)−sin⁡(20)sin⁡(70)Thus tan⁡(x)=tan⁡(20)

Camryn Diaz · Accepted Answer

Using Werner Formulas,2sin⁡10sin⁡40=cos⁡30−cos⁡50Again,4sin⁡10sin⁡40sin⁡(70−x)=2(cos⁡30−cos⁡50)sin⁡(70−x)=sin⁡(100−x)+sin⁡(40−x)−[sin⁡(120−x)+sin⁡(20−x)]Now,sin⁡x=4sin⁡10sin⁡40sin⁡(70−x)⇒sin⁡x=sin⁡(100−x)+sin⁡(40−x)−[sin⁡(120−x)+sin⁡(20−x)]⇒sin⁡x−sin⁡(40−x)=[sin⁡(100−x)−sin⁡(120−x)]+sin⁡(x−20)⇒2sin⁡(x−20)cos⁡20=−2sin⁡10cos⁡(110−x)+sin⁡(x−20)As cos⁡(110−x)=cos⁡[90−(x−20)], this becomessin⁡(x−20)(2cos⁡20−1)=−2sin⁡10sin⁡(x−20)⇒sin⁡(x−20)(2cos⁡20−1+2sin⁡10)=0But cos⁡20+sin⁡10=cos⁡20+cos⁡80=2cos⁡50cos⁡30=3cos⁡50≠1⇒sin⁡(x−20)=0⇒x−20=180∘n+20 where n is any integer

So, I have this equation: \sin x=4\sin10^\circ\sin40^\circ\sin(70^\circ-x)

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