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If A is acute and cos2A = -7/9 find without a calculator: a) cosA b) sinA
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Use a) Cos2A= [2Cos^2(A)-1]=(-7/9) b) SinA=Sqrt(1-cos^2(A))
Because A is acute, it is between 0 and 90 degrees. You can use the half angle formula, but in this case it is not really needed. You just need to see where cos2A = -7/9, then take half of that. Should be two possible answers and only one of them will be acute. Then develop your solutions from there.
\[\cos 2A=2\cos ^{2}A-1=-\frac{ 7 }{9 }\] \[2\cos ^{2}A=\frac{ -7 }{ 9 }+1=\frac{ 2 }{ 9 }\] \[\cos ^{2}A=\frac{ 1 }{ 9 }\] \[\cos A=\frac{ 1 }{ 3 } ...\left( A is accute \right)\] Again \[\cos 2A=1-2\sin ^{2}A\] You can find sin A
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