Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1

Marks : 6 M

Year : Dec 2013

Let, $ x\;=\; cos\theta+isin\theta \; \; \; \; \dfrac{1}{x} \;=\; cos\theta-isin\theta \\ \; \\ \; \\ \therefore x^n \;=\; (cos\theta+isin\theta)^n \;=\; cosn\theta+isinn\theta \\ \; \\ \; \\ Also \; \dfrac{1}{x^n} \;=\; (cos\theta-isin\theta)^n \;=\; cosn\theta-isinn\theta \\ \{ \because DeMoivre's \; Theorem \} \\ \; \\ \; \\ x^n + \dfrac{1}{x^n} \;=\; 2cosn\theta \; \; \; \; \ldots (i) \\ \; \\ \; \\ x^n - \dfrac{1}{x^n} \;=\; 2isin \;n\theta \; \; \; \; \ldots (ii) \\ \; \\ \; \\ \therefore sin^4\theta cos^3 \theta \;=\; (sin\theta)^4 (cos\theta)^3 \\ \; \\ \; \\ = \Bigg[ \dfrac{1}{2i} \bigg(x-\dfrac{1}{x} \bigg) \Bigg]^4 \Bigg[ \dfrac{1}{2} \bigg(x+\dfrac{1}{x}) \Bigg]^3 \; \; from \; (i) \; and \; (ii) \\ \; \\ \; \\ = \dfrac{1}{16i^6} \bigg(x-\dfrac{1}{x} \bigg)^4 \cdot \dfrac{1}{8} \bigg(x+\dfrac{1}{x} \bigg)^3 \; \; \; Expanding \; Binomally \\ \; \\ \; \\ = \dfrac{1}{128} \bigg[ x^4 - 4x^3 . \dfrac{1}{x} + 6 x^2 . \dfrac{1}{x^2} - 4 x . \dfrac{1}{x^3} + \dfrac{1}{x^4} \bigg] \bigg[ x^3 + 3x^2 . \dfrac{1}{x} + 3 x . \dfrac{1}{x^2} + \dfrac{1}{x^3} \bigg] \\ \; \\ \; \\ = \dfrac{1}{128} \Bigg[ \bigg(x^4+\dfrac{1}{x^4} \bigg) -4 \bigg(x^2+\dfrac{1}{x^2} \bigg) + 6 \Bigg] + 6\Bigg[ \bigg(x^3+\dfrac{1}{x^3} \bigg) + 3\bigg(x+\dfrac{1}{x} \bigg) \Bigg] \\ \; \\ \; \\ = \dfrac{1}{128} [ 2cos4\theta - 8 cos2\theta + 6 ] [2cos3\theta + 6 cos\theta] \; \; \; from \; (i) \\ \; \\ \; \\ = \dfrac{1}{32} (cos4\theta - 4cos2\theta +3) (cos3\theta+3cos\theta) \\ \; \\ \; \\ \therefore sin^4\theta cos^3 \theta \;=\; \dfrac{1}{32} [ cos4\theta cos3\theta + 3cos4\theta cos\theta -4cos4\theta cos3\theta - 12cos2\theta cos\theta + 3cos3\theta + 9 cos \theta ] \\ \; \\ We \; know \; that \; cos(A+B) + cos(A-B) \;=\; 2cosAcosB \\ \; \\ \; \\ \therefore sin^4\theta cos^3 \theta \;=\; \dfrac{1}{32} \\ \bigg[ \dfrac{1}{2} cos7\theta + \dfrac{1}{2} cos\theta + \dfrac{3}{2} cos5\theta + \dfrac{3}{2} cos3\theta - \dfrac{4}{2} cos5\theta - \dfrac{4}{2} cos\theta - \dfrac{12}{2} cos3\theta - \dfrac{12}{2} cos\theta + 3cos3\theta + 9cos\theta \bigg] \\ \; \\ \; \\ = \dfrac{1}{32} \bigg[ \dfrac{1}{2} cos7\theta + \dfrac{3}{2} cos\theta - \dfrac{1}{2} cos5\theta - \dfrac{3}{2} cos3\theta \bigg] \\ \; \\ \; \\ = \dfrac{1}{74} cos7\theta + \dfrac{3}{64} cos\theta - \dfrac{1}{64} cos5\theta - \dfrac{3}{64} cos3\theta \\ \; \\ \; \\ $

On comparing it with , we get $ acos\theta + b cos3\theta + c cos 5 \theta + d cos7 \theta $ , we get

$ \\ \; \\ a \;=\; \dfrac{3}{64} \; , \; b \;=\; \dfrac{-3}{64} \; , \; c \;=\; \dfrac{-1}{64} \; , \; d \;=\; \dfrac{1}{64} $