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Trigonometry
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Find exact value: sin(2 cos^-1(-3/11))
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there are infinite
how so?
cos(x) = -3/11 has infinite solutions
forget what I said, they just want the one
cos^(-1)(-3/11) = 105.8 now sin(105.8) = .9622
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4sqrt(7)/11
thank you!
\[put \cos^{-1} \left( \frac{ -3 }{ 11 } \right)=x\] \[\cos x=\frac{ -3 }{ 11 }\] hence x lies in second or third quadrant. \[\sin x=\sqrt{1-\left( \frac{ -3 }{11 } \right)^{2}}=\sqrt{1-\frac{ 9 }{121 }}=\frac{ \sqrt{112} }{11 }\] sin x can be positive or negative according as x lies in second or third quadrant. plug in sin 2x= 2 sin x cos x
wow thank you so much!
yw
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woops I didn't notice the 2
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