Answer to Question #137944 in Linear Algebra for mr prince

Question

(a) Express cos7 θ in terms of multiples of angles.

(b) Express cos4 θ sin3 θ in terms of multiples of angles.

(c) Using complex numbers, prove that the, angles A, B and C of a planar triangle satisfy the relations (i) cos2 A + cos2 B + cos2 C = 1 − 2 cos A cos B cos C

(ii) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C.

Answer 1

$\displaystyle(a)\\\textsf{We know from our knowledge on Complex Numbers that,} \\ \begin{aligned} \cos(n\theta) + i\sin(n\theta) &= (\cos\theta + i\sin\theta)^n\\ \cos(7\theta) + i\sin(7\theta) &= (\cos\theta + i\sin\theta)^7\\ &= \cos^7\theta + 7\cos^6\theta(i\sin\theta) + 21\cos^5\theta(i\sin\theta)^2 + 35\cos^4\theta(i\sin\theta)^3 \\&+ 35\cos^3\theta(i\sin\theta)^4 + 21\cos^2\theta(i\sin\theta)^5 + 7\cos\theta(i\sin\theta)^6 + (i\sin\theta)^7\\ &= \cos^7\theta + 7i\cos^6\theta\sin\theta - 21\cos^5\theta\sin^2\theta - 35i\cos^4\theta\sin^3\theta \\&+ 35\cos^3\theta\sin^4\theta + 21i\cos^2\theta\sin^5\theta - 7\cos\theta\sin^6\theta - i\sin^7\theta \end{aligned}\\ \textsf{Comparing the real parts}\\ \begin{aligned} \cos(7\theta) = \cos^7\theta - 21\cos^5\theta\sin^2\theta \\&+ 35\cos^3\theta\sin^4\theta - 7\cos\theta\sin^6\theta \end{aligned}\\ (b)\\ \begin{aligned} \cos(n\theta) + i\sin(n\theta) &= (\cos\theta + i\sin\theta)^n\\ \cos(4\theta) + i\sin(4\theta) &= (\cos\theta + i\sin\theta)^4\\ &= \cos^4\theta + 4\cos^3\theta(i\sin\theta) + 6\cos^2\theta(-\sin^2\theta)\\ &\hspace{0.3cm}+ 4\cos^3\theta(-i\sin^3\theta) + \sin^4\theta \end{aligned}\\ \textsf{Comparing the real parts}\\ \cos(4\theta) = \cos^4\theta + \sin^4\theta - 6\cos^2\theta\sin^2\theta \hspace{1cm} (2)\\ \textsf{Also}\\ \begin{aligned} \cos(3\theta) + i\sin(3\theta) &= (\cos\theta + \sin\theta)^3 \\ &= \cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(-\sin^2\theta) - i\sin^3\theta \hspace{1cm} (3) \end{aligned}\\ \textsf{Comparing the imaginary parts}\\ \sin(3\theta) = 3\cos^2\theta\sin\theta - \sin^3\theta \\ \textsf{To derive the value of}\, \cos(4\theta)\sin(3\theta), \textsf{Multiply the result in}\\ \textsf{equation}\, (3) \,\textsf{by the result in equation}\, (2) \\ \begin{aligned} \therefore \cos(4\theta)\sin(3\theta) &= (\cos^4\theta + \sin^4\theta - 6\cos^2\theta\sin^2\theta)(3\cos^2\theta\sin\theta - \sin^3\theta)\\ &= 3\cos^6\theta\sin\theta + 3\cos^2\theta\sin^5\theta - 18\cos^4\theta\sin^3\theta - \\&\cos^4\theta\sin^3\theta - \sin^7\theta + 6\cos^2\theta\sin^5\theta \\ &= 3\cos^6\theta\sin\theta + 9\cos^2\theta\sin^5\theta - 19\cos^4\theta \sin^3\theta - \sin^7\theta \end{aligned} \\ \implies \cos(4\theta)\sin(3\theta) = 3\cos^6\theta\sin\theta + 9\cos^2\theta\sin^5\theta - 19\cos^4\theta \sin^3\theta - \sin^7\theta \\ (c)(i)\\\textsf{Prove that}\\ \sin(2A) + \sin(2B) + \sin(2C) = 4\sin A \sin B \sin C \\ \textsf{Since the figure is a planar triangle}\\ A + B + C = \pi\\ \textsf{It is known that,}\\ \sin P + \sin Q = 2\sin\left(\frac{P + Q}{2}\right)\cos\left(\frac{P - Q}{2}\right) \\ \begin{aligned} \sin(2A) + \sin(2B) + \sin(2C) &= 2\sin(A + B)\cos(A - B) + \sin(2C)\\ &= 2\sin(\pi - C)\cos(A - B) + 2\sin(C)\cos(C)\\ &= 2\sin(C)\cos(A - B) + 2\sin(C)\cos(C)\\ &= 2\sin(C)(\cos(A - B) + \cos(C)) \end{aligned} \\ \textsf{It is also known that,}\\ \cos P + \cos Q = 2\cos\left(\frac{P + Q}{2}\right)\cos\left(\frac{P - Q}{2}\right) \\ \begin{aligned} \implies \sin(2A) + \sin(2B) + \sin(2C) &= 2\sin(C)\left(2\cos\left(\frac{A + C - B}{2}\right)\cos\left(\frac{A -B - C}{2}\right) \right)\\ &= 4\sin C \cos\left(\frac{\pi}{2} - B\right) \cos\left(\frac{\pi}{2} - A\right) \\ &= 4\sin C \sin B \sin A\\ &= 4\sin A \sin B \sin C \end{aligned}\\ (ii)\\\begin{aligned} \cos(2A) + \cos(2B) + \cos(2C) \\&= 2\cos\left(\frac{2(A + B)}{2}\right)\cos\left(\frac{2(A - B)}{2}\right) + \cos(2C) \\&= 2\cos(\pi - C)\cos(A - B) + 2\cos^2 C - 1 \\&= 2\cos C (\cos C - \cos(A - B)) - 1 \end{aligned}\\ \textsf{It is also known that,}\\ \cos P - \cos Q = 2\sin\left(\frac{P + Q}{2}\right)\sin\left(\frac{P - Q}{2}\right) \\ \begin{aligned} \cos(2A) + \cos(2B) + \cos(2C) \\&= 2\cos C \left(2\sin\left(\frac{A + C - B}{2}\right)\sin\left(\frac{C - A+ B}{2}\right)\right) - 1 \\&= 2\cos C \left(2\sin\left(\frac{\pi}{2} - B\right)\sin\left(\frac{\pi}{2} - A\right)\right) - 1 \\&= 2\cos C(2\cos B \times 2\cos A) - 1 = 4\cos A \cos B \cos C - 1 \end{aligned}$