June 2009 Markscheme

EDEXCEL CORE MATHEMATICS C3 (6665) – JUNE 2009FINAL MARK SCHEMEQuestionNumber6. (a) ( )SchemeMarksA = B ⇒ cos A + A = cos2A = cos Acos A − sin AsinAM12 2cos2A = cos A − sin A and2 2cos A sin A 1+ = gives2 2 2cos2A = 1− sin A − sin A = 1 − 2sin A (as required) A1 (2)(b) C1 = C2⇒23sin 2 4sin 2cos2x = x − xM1⎛1−cos2x⎞3sin 2x= 4⎜⎟ − 2cos 2x⎝ 2 ⎠( )3sin 2x = 2 1− cos2x − 2cos2x3sin 2x = 2 − 2cos2x − 2cos2xM13sin 2x+ 4cos2x= 2A1 (3)(c) 3sin 2x + 4cos2x = Rcos( 2x − α )3sin 2x + 4cos2x = Rcos 2xcosα+ Rsin 2xsinαEquate sin 2 x : 3 = RsinαEquate cos 2 x : 4 = R cosα2 2R = 3 + 4 ; = 25 = 5B1tanα= ⇒ α = 36.86989765...34Hence, 3sin 2x 4cos2x 5cos( 2x36.87)(d) 3sin 2x+ 4cos2x= 25cos( 2x − 36.87)= 2oM1 A1+ = − A1 (3)2cos( 2x − 36.87)= M15( )2x − 36.87 = 66.42182...( )2x − 36.87 = 360 − 66.42182...ooHence, x = 51.64591... o, 165.22409... o A1 A1 (4)A1(12 marks)5