€”2pq 2pq p 2 —q 2 (i), (ii) j: o p2=q2 '201 0 cf 1, 203 'co) '2 c cos o sin o...

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Page 1: €”2Pq 2Pq p 2 —q 2 (i), (ii) J: O p2=q2 '201 0 CF 1, 203 'co) '2 c COS O sin O — sin 0 cos cos30 sin 30 x3cos30 x 3sin 30 — sin 30 COS 30

Page 2: €”2Pq 2Pq p 2 —q 2 (i), (ii) J: O p2=q2 '201 0 CF 1, 203 'co) '2 c COS O sin O — sin 0 cos cos30 sin 30 x3cos30 x 3sin 30 — sin 30 COS 30

Page 3: €”2Pq 2Pq p 2 —q 2 (i), (ii) J: O p2=q2 '201 0 CF 1, 203 'co) '2 c COS O sin O — sin 0 cos cos30 sin 30 x3cos30 x 3sin 30 — sin 30 COS 30

Page 4: €”2Pq 2Pq p 2 —q 2 (i), (ii) J: O p2=q2 '201 0 CF 1, 203 'co) '2 c COS O sin O — sin 0 cos cos30 sin 30 x3cos30 x 3sin 30 — sin 30 COS 30

Page 5: €”2Pq 2Pq p 2 —q 2 (i), (ii) J: O p2=q2 '201 0 CF 1, 203 'co) '2 c COS O sin O — sin 0 cos cos30 sin 30 x3cos30 x 3sin 30 — sin 30 COS 30

madinpoly.commadinpoly.com/pdf/solved/3/phy_march_2013.pdfp2 +Q2 +2PQ = (P+Q)2 R = when e-1800 ; cos 180--1 ; Sin 180 = R +Q2 -2PQ = (P-Q)2 a = or 1800 o PART-C (Answer one full question

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mrgleason.weebly.commrgleason.weebly.com/uploads/1/3/9/7/13977727/day_15_-_degrees-all... · sin 300 sin 80 sin 500 cos 2130 2) 4) 6) 8) cos sin 3100 cos 500 sin 1690 sin 2300 cos

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Scanned with CamScanner · 2019. 12. 1. · cos4 x — sin4 x = cos 2x = I + sin2x (sin x + cos x)2 I + cos I()y = 2 cos2 5y 59. 6 sin — cos — 61. sin 59 cos 39 63. 5 cos(—5ß)

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> plot(cos(x) + sin(x), x=0..Pi); plot(tan(x), x=-Pi..Pi ... · > plot3d({sin(x*y), x + 2*y},x=-Pi..Pi,y=-Pi..Pi); ↵ c1:= [cos(x)-2*cos(0.4*y),sin(x)-2*sin(0.4*y),y]: ↵ c2:= [cos(x)+2*cos(0.4*y),sin(x)+2*sin(0.4*y),y]:

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