{"id":22052,"date":"2024-01-10T15:51:42","date_gmt":"2024-01-10T10:21:42","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22052"},"modified":"2024-01-11T12:26:37","modified_gmt":"2024-01-11T06:56:37","slug":"inter-2nd-year-maths-2a-de-moivres-theorem-solutions-ex-2a","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-2nd-year-maths-2a-de-moivres-theorem-solutions-ex-2a\/","title":{"rendered":"Inter 2nd Year Maths 2A De Moivre\u2019s Theorem Solutions Ex 2(a)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions<\/a> Inter 2nd Year Maths 2A De Moivre\u2019s Theorem Solutions Exercise 2(a) will help students to clear their doubts quickly.<\/p>\n

Intermediate 2nd Year Maths 2A De Moivre\u2019s Theorem Solutions Exercise 2(a)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nIf n is an integer then show that (1 + i)2n<\/sup> + (1 – i)2n<\/sup> = 2n+1<\/sup> cos \\(\\frac{n \\pi}{2}\\)
\nSolution:
\n\"Inter
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the values of the following:
\n(i) (1 + i\u221a3)3<\/sup>
\nSolution:
\nLet 1 + i\u221a3 = r (cos \u03b8 + i sin \u03b8)
\nEquating real and imaginary parts
\nr cos \u03b8 = 1 and r sin \u03b8 = \u221a3
\n\u21d2 r2<\/sup> (1) = 1 + 3 = 4
\n\u21d2 r = 2
\ncos \u03b8 = \\(\\frac{1}{2}\\) and sin \u03b8 = \\(\\frac{\\sqrt{3}}{2}\\)
\n\u21d2 \u03b8 = \\(\\frac{\\pi}{3}\\)
\n\"Inter<\/p>\n

(ii) (1 – i)8<\/sup>
\nSolution:
\n\"Inter<\/p>\n

(iii) (1 + i)16<\/sup>
\nSolution:
\n\"Inter<\/p>\n

(iv) \\(\\left(\\frac{\\sqrt{3}}{2}+\\frac{i}{2}\\right)^{5}-\\left(\\frac{\\sqrt{3}}{2}-\\frac{i}{2}\\right)^{5}\\)
\nSolution:
\n\"Inter<\/p>\n

\"Inter<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nIf \u03b1, \u03b2 are the roots of the equation x2<\/sup> – 2x + 4 = 0 then for any n \u2208 N show that \u03b1n<\/sup> + \u03b2n<\/sup> = 2n+1<\/sup> cos(\\(\\frac{n \\pi}{3}\\))
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 2.
\nIf cos \u03b1 + cos \u03b2 + cos \u03b3 = 0 = sin \u03b1 + sin \u03b2 + sin \u03b3 then show that
\n(i) cos 3\u03b1 + cos 3\u03b2 + cos 3\u03b3 = 3 cos(\u03b1 + \u03b2 + \u03b3)
\n(ii) sin 3\u03b1 + sin 3\u03b2 + sin 3\u03b3 = 3 sin(\u03b1 + \u03b2 + \u03b3)
\n(iii) cos(\u03b1 + \u03b2) + cos(\u03b2 + \u03b3) + cos(\u03b3 + \u03b1) = 0
\nSolution:
\nGiven that cos \u03b1 + cos \u03b2 + cos \u03b3 = 0 and sin \u03b1 + sin \u03b2 + sin \u03b3 = 0
\nFrom the hypothesis, we have
\n(cos \u03b1 + cos \u03b2 + cos \u03b3) + i (sin \u03b1 + sin \u03b2 + sin \u03b3) = 0
\ni.e., (cos \u03b1 + i sin \u03b1) + (cos \u03b2 + i sin \u03b2) + (cos \u03b3 + i sin \u03b3) = 0
\nWe know that if a + b + c = 0, then a3<\/sup> + b3<\/sup> + c3<\/sup> = 3abc
\nHence (cos \u03b1 + i sin \u03b1)3<\/sup> + (cos \u03b2 + i sin \u03b2)3<\/sup> + (cos \u03b3 + i sin \u03b3)3<\/sup> = 3 (cos \u03b1 + i sin \u03b1) (cos \u03b2 + i sin \u03b2) + (cos \u03b3 + i sin \u03b3)
\ni.e., cos 3\u03b1 + i sin 3\u03b1 + cos 3\u03b2 + i sin 3\u03b2 + cos 3\u03b3 + i sin 3\u03b3 = 3[cos(\u03b1 + \u03b2 + \u03b3) + i sin(\u03b1 + \u03b2 + \u03b3)] …….(1)
\nOn equating the real parts on both sides of equation (1) we get
\n(i) cos 3\u03b1 + cos 3\u03b2 + cos 3\u03b3 = 3 cos(\u03b1 + \u03b2 + \u03b3)
\nOn equation thenmaginary parts on both sides of equation (1) we get,
\n(ii) sin 3\u03b1 + sin 3\u03b2 + sin 3\u03b3 = 3 sin(\u03b1 + \u03b2 + \u03b3)
\n(iii) Let a = cos \u03b1 + i sin \u03b1 then \\(\\frac{1}{a}\\) = cos \u03b1 – i sin \u03b1
\nb = cos \u03b2 + i sin \u03b2 and \\(\\frac{1}{b}\\) = cos \u03b2 – i sin \u03b2
\nc = cos \u03b3 + i sin \u03b3 and \\(\\frac{1}{c}\\) = cos \u03b3 – i sin \u03b3
\nNow \\(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\) = (cos \u03b1 + cos \u03b2 + cos \u03b3) – i (sin \u03b1 + sin \u03b2 + sin \u03b3)
\n\u21d2 \\(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\) = 0 – i(0)
\n\u21d2 \\(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\) = 0
\n\u21d2 bc + ca + ab = 0
\n\u21d2 [cos(\u03b2 + \u03b3) + i sin(\u03b2 + \u03b3)] + [cos(\u03b3 + \u03b1) + i sin(\u03b3 + \u03b1)] + [cos(\u03b1 + \u03b2) + i sin(\u03b1 + \u03b2)] = 0
\n\u21d2 [cos(\u03b1 + \u03b2) + cos(\u03b2 + \u03b3) + cos(\u03b3 + \u03b1)] + i [sin(\u03b1 + \u03b2) + sin(\u03b2 + \u03b3) + sin(\u03b3 + \u03b1)] = 0
\nBy equation real parts on both sides
\ncos (\u03b1 + \u03b2) + cos (\u03b2 + \u03b3) + cos (\u03b3 + \u03b1) = 0<\/p>\n

\"Inter<\/p>\n

Question 3.
\nIf n is an integer and z = cis \u03b8, (\u03b8 \u2260 (2n + 1) \u03c0\/2), then show that \\(\\frac{z^{2 n}-1}{z^{2 n}+1}\\) = i tan n\u03b8.
\nSolution:
\nz = cis \u03b8 = cos \u03b8 + i sin \u03b8, \u03b8 \u2260 (2n + 1) \u03c0\/2
\n\"Inter
\n(\u2235 i2<\/sup> = -1)
\n= i tan(n\u03b8)
\n= R.H.S<\/p>\n

Question 4.
\nIf (1 + x)n<\/sup> = a0<\/sub> + a1<\/sub>x + a2<\/sub>x2<\/sup> + ……… + anxn, then show that
\n(i) a0<\/sub> – a2<\/sub> + a4<\/sub> – a6<\/sub> + …….. = \\(2^{n \/ 2} \\cos \\frac{n \\pi}{4}\\)
\n(ii) a1<\/sub> – a3<\/sub> + a5<\/sub> – a7<\/sub> + ……… = \\(2^{n \/ 2} \\sin \\frac{n \\pi}{4}\\)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A De Moivre\u2019s Theorem Solutions Exercise 2(a) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2A De Moivre\u2019s Theorem Solutions Exercise 2(a) I. Question 1. If n is an integer then show that (1 + i)2n + (1 … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22052"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22052"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22052\/revisions"}],"predecessor-version":[{"id":22053,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22052\/revisions\/22053"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22052"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22052"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22052"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}