Suggested: (2cosa+1)(cos2a-1)(2cos2a-1)=2cos4a+1 - (2cosa+1)(cos2a-1)=2cos2a+1 - if a+b+c=π prove that cos2a+cos2b+cos2c=1-2cosacosbcosc - (1+sin2a+cos2a)^2=4cos^2a(1+sin2a) - cos^6a-sin^6a=cos2a(1-1/4 sin^2 2a) - cos2a/2+cos2b/2+cos2c/2 - cos2a-cos2b+cos2c=1-4sina cosb sinc - using the formula cosa=√1+cos2a/2 - sin2a/cos2a+cos2a/sin2a=1/sin2a cos2a-2 - show that cos2a/2+cos2b/2+cos2c/2=2+r/2 r - cos2a=cos^2a-sin^2a - (cos2a-cos2b)^2+(sin2a+sin2b)^2=4sin^2(a+b) - cos2a+cos2b+cos2c=1-2cosacosbcosc - cos2a=1-2sin2a - 2cos2a Browse related:
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