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cos theta =(squareroot70)/4 and 0 degrees
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find \[\sin \theta ,then \sin 2\theta=2\sin \theta \cos \theta \]
@ganeshie8
use below to find value of \(\sin \theta\) : \(\sin ^2 \theta + \cos^2 \theta = 1\)
plugin the give \(\cos \theta \) above and solve \(\sin \theta \)
\[1-\sqrt{70}/4\]
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\[\frac{ \sqrt{70} }{ 4 }>1, which~ is ~\not` correct ~as \left| \cos \theta \right|\le 1\]
\[1-(\sqrt{70}/4)^2\]
check your statement.
im lost^^^
thanks
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your question has a typo : cos theta =(squareroot7\(\color{Red}{0}\))/4 and 0 degrees<theta<90 degrees
check the original question again, that \(\color{Red}{0}\) doesnt belong there ^
im done yea i see that now thanks for pointing out
np :)
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