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Simpify \(\dfrac{\sin 25^\circ \sin 35^\circ \sin 85^\circ}{\sin 75^\circ}\)

 

I don't which identities will help here.

 Jul 19, 2020

Best Answer 

 #3
avatar+25469 
+2

Simpify \(\dfrac{\sin 25^\circ \sin 35^\circ \sin 85^\circ}{\sin 75^\circ}\)

 

Formula:

\(\begin{array}{|rcll|} \hline \sin(x)\sin(y) &=& \dfrac{1}{2}\Big( \cos(x-y)-\cos(x+y) \Big) \\\\ \sin(x)\cos(y) &=& \dfrac{1}{2}\Big( \sin(x-y)+\sin(x+y) \Big) \\\\ \cos(60^\circ) &=& \dfrac12 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline &&\mathbf{ \dfrac{\sin(25^\circ) \sin(35^\circ) \sin(85^\circ)}{\sin(75^\circ)} } \\\\ && \qquad \boxed{ \sin(x)\sin(y) = \dfrac{1}{2}\Big( \cos(x-y)-\cos(x+y) \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\Big( \cos(35^\circ-25^\circ)-\cos(35^\circ+25^\circ) \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\Big( \cos(10^\circ)-\cos(60^\circ) \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\Big( \cos(10^\circ)-\dfrac{1}{2} \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\cos(10^\circ)-\dfrac{1}{4} } \\\\ &=& \dfrac{\Big(\dfrac{1}{2}\cos(10^\circ)-\dfrac{1}{4}\Big) \sin(85^\circ)}{\sin(75^\circ)} \\\\ &=& \dfrac{\dfrac{1}{2}*\cos(10^\circ)\sin(85^\circ)-\dfrac{1}{4}*\sin(85^\circ)}{\sin(75^\circ)} \\\\ && \qquad \boxed{ \sin(x)\cos(y) = \dfrac{1}{2}\Big( \sin(x-y)+\sin(x+y) \Big) \\ \sin(85^\circ)\cos(10^\circ) = \dfrac{1}{2}\Big( \sin(85^\circ-10^\circ)+\sin(85^\circ+10^\circ) \Big) \\ \sin(85^\circ)\cos(10^\circ) = \dfrac{1}{2}\Big( \sin(75^\circ)+\sin(95^\circ) \Big) \quad | \quad \sin(95^\circ)=\sin(180^\circ-95^\circ)=\sin(85^\circ) \\ \sin(85^\circ)\cos(10^\circ) = \dfrac{1}{2}\Big( \sin(75^\circ)+\sin(85^\circ) \Big) } \\\\ &=& \dfrac{\dfrac{1}{2}*\dfrac{1}{2}\Big( \sin(75^\circ)+\sin(85^\circ) \Big)-\dfrac{1}{4}*\sin(85^\circ)}{\sin(75^\circ)} \\\\ &=& \dfrac{\dfrac{1}{4}*\sin(75^\circ)+\dfrac{1}{4}*\sin(85^\circ)-\dfrac{1}{4}*\sin(85^\circ)}{\sin(75^\circ)} \\\\ &=& \dfrac{\dfrac{1}{4}*\sin(75^\circ)}{\sin(75^\circ)} \\\\ &=& \mathbf{\dfrac{1}{4}} \\ \hline \end{array}\)

 

 

laugh

 Jul 20, 2020
 #1
avatar+110089 
-1

I have not got this out but this is what I have tried to use

 

 

75=45+30

25=30-5

35=30+5

sin85=cos5

 

You still end up with ratios of 5 degrees so I am missing something.

 Jul 20, 2020
 #2
avatar+110089 
-1

I know the answer is 0.25  but I do not know how to get it.

 Jul 20, 2020
 #3
avatar+25469 
+2
Best Answer

Simpify \(\dfrac{\sin 25^\circ \sin 35^\circ \sin 85^\circ}{\sin 75^\circ}\)

 

Formula:

\(\begin{array}{|rcll|} \hline \sin(x)\sin(y) &=& \dfrac{1}{2}\Big( \cos(x-y)-\cos(x+y) \Big) \\\\ \sin(x)\cos(y) &=& \dfrac{1}{2}\Big( \sin(x-y)+\sin(x+y) \Big) \\\\ \cos(60^\circ) &=& \dfrac12 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline &&\mathbf{ \dfrac{\sin(25^\circ) \sin(35^\circ) \sin(85^\circ)}{\sin(75^\circ)} } \\\\ && \qquad \boxed{ \sin(x)\sin(y) = \dfrac{1}{2}\Big( \cos(x-y)-\cos(x+y) \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\Big( \cos(35^\circ-25^\circ)-\cos(35^\circ+25^\circ) \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\Big( \cos(10^\circ)-\cos(60^\circ) \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\Big( \cos(10^\circ)-\dfrac{1}{2} \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\cos(10^\circ)-\dfrac{1}{4} } \\\\ &=& \dfrac{\Big(\dfrac{1}{2}\cos(10^\circ)-\dfrac{1}{4}\Big) \sin(85^\circ)}{\sin(75^\circ)} \\\\ &=& \dfrac{\dfrac{1}{2}*\cos(10^\circ)\sin(85^\circ)-\dfrac{1}{4}*\sin(85^\circ)}{\sin(75^\circ)} \\\\ && \qquad \boxed{ \sin(x)\cos(y) = \dfrac{1}{2}\Big( \sin(x-y)+\sin(x+y) \Big) \\ \sin(85^\circ)\cos(10^\circ) = \dfrac{1}{2}\Big( \sin(85^\circ-10^\circ)+\sin(85^\circ+10^\circ) \Big) \\ \sin(85^\circ)\cos(10^\circ) = \dfrac{1}{2}\Big( \sin(75^\circ)+\sin(95^\circ) \Big) \quad | \quad \sin(95^\circ)=\sin(180^\circ-95^\circ)=\sin(85^\circ) \\ \sin(85^\circ)\cos(10^\circ) = \dfrac{1}{2}\Big( \sin(75^\circ)+\sin(85^\circ) \Big) } \\\\ &=& \dfrac{\dfrac{1}{2}*\dfrac{1}{2}\Big( \sin(75^\circ)+\sin(85^\circ) \Big)-\dfrac{1}{4}*\sin(85^\circ)}{\sin(75^\circ)} \\\\ &=& \dfrac{\dfrac{1}{4}*\sin(75^\circ)+\dfrac{1}{4}*\sin(85^\circ)-\dfrac{1}{4}*\sin(85^\circ)}{\sin(75^\circ)} \\\\ &=& \dfrac{\dfrac{1}{4}*\sin(75^\circ)}{\sin(75^\circ)} \\\\ &=& \mathbf{\dfrac{1}{4}} \\ \hline \end{array}\)

 

 

laugh

heureka Jul 20, 2020
 #4
avatar+110089 
0

Thanks very much Heureka :)

Melody  Jul 20, 2020
 #5
avatar+25469 
+2

Thank you, Melody!

 

laugh

heureka  Jul 20, 2020

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