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What is the exact value of sin(195°)? A. (√2 + √6)/4 B. (√6 - √2)/4 C. (√2 - √6)/4 D. (√6 - √3)/4
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I got (√6 - √2)/4 which is B.)
\[\sin195=\sin(180+15)=\sin180\cos15+\cos180\sin15=-\sin15\] \[\cos30=1-2\sin^{2}15\] \[2\sin^{2}15=1-\cos30=1-\frac{ \sqrt{3} }{ 2 }\] \[\sin195=-\sqrt{\frac{ 1 }{2 }+\frac{ \sqrt{3} }{ 4 }}\]
\[\sin 195=\sin(150+45)=\sin150\cos 45+\cos 150\sin 45\] \[=\sin (180-30)\cos 45+\cos (180-30)\sin 45\] \[=\sin 30\cos 45-\cos 30\sin 45\] \[=\frac{ 1 }{ 2 }(\frac{ \sqrt{2} }{ 2 })-\frac{ \sqrt{3} }{ 2 }(\frac{ \sqrt{2} }{ 2})\] \[=\frac{ \sqrt{2} }{ 4}-\frac{ \sqrt{6} }{ 4}=\frac{ \sqrt{2}-\sqrt{6} }{ 4}\] so C
Hmmmmmm interesting thanks
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