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6s^(2)1+(cos^(2)2+cos^(2)3+...+cos^(2)90...

6s^(2)1+(cos^(2)2+cos^(2)3+...+cos^(2)90^(@)=0

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cos^(2)1^(@)+ cos^(2)2^(@)+cos^(2)3^(@)+....+cos^(2)90^(@)=

cos^(2)5^(@)+cos^(2)10^(@)+cos^(2)15^(@)+.....+cos^(2) 90^(@)=

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If A+B+C=2S then cos^(2)S+cos^(2)(S-A)+cos^(2)(S-B)+cos^(2)(S-C)=

If A+B+C=2S , " then " cos^(2)S + cos^(2)(S-A)+cos^(2)(S-B)+cos^(2)(S-C)=

The value of (1)/(2) sin ^(2) 90^(@) sin^(2) 30^(@) cos ^(2) 45^(@) + 4 tan ^(2) 30^(@) + (1)/(2) sin^(2) 90^(@) - 2 cos ^(2) 90^(@) is :

Prove that cos A cos 2A cos 2^(2) A cos 2^(3)A....cos 2^(n-1) A=(sin 2^(n) A)/(2^(n) sin A)

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Prove that : cos^(2)A+cos^(2)(A+pi/3)+cos^(2)(A-pi/3) =1/2[2cos^(2)A+2cos^(2)(A+60^(@))+2cos^(2)(A-60^(@))] =1/2[1+cos2A+1+cos(2A+120^(@))+1+cos(2A-120^(@))] 1/2[3+cos2A+cos(2A+120^(@))+cos(2A-120^(@))] 1/2[3+cos2A+2cos(2A+120^(@)+2A-120^(@))/(2).cos(2A+120^(@)-2A+120^(@))/(2)] =1/2[3+cos2A+2cos2Acos120^(@)] =1/2[3cos2A+2cos2Acos(90^(@)+30^(@))] =1/2[3+cos2A-2.cos2A.(1/2)] =1/2(3+cos2A-cos2A) =3/2 = RHS Hence Proved.