If A=B=60deg,then prove that :cos(A-B)=cos.A cos.B+SinA.Sin B
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A = B = 60 degrees
cos(A - B) = cosA . cosB + sinA . sinB
cos(60-60) = cos 60 . cos 60 + sin60 . sin 60
cos 0 = ( 1/2 × 1/2 ) + (✓3/2 × ✓3/2)
1 = 1/4 + 3/4
1 = 1
therefore, cos( A - B) = cos A . cosB + sinA . sinB
cos(A - B) = cosA . cosB + sinA . sinB
cos(60-60) = cos 60 . cos 60 + sin60 . sin 60
cos 0 = ( 1/2 × 1/2 ) + (✓3/2 × ✓3/2)
1 = 1/4 + 3/4
1 = 1
therefore, cos( A - B) = cos A . cosB + sinA . sinB
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I think it is right answer answer
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