Q. Solve : $\sin^{8} x - \cos^{8} x$ .

Dear student

consider, \sin^{8} x - \cos^{8} x = (\sin^{4} x - \cos^{4} x) (\sin^{4} x + \cos^{4} x) [∵ a^{2} - b^{2} = (a - b) (a + b)] = (\sin^{2} x - \cos^{2} x) (\sin^{2} x + \cos^{2} x) ({(\sin^{2} x)}^{2} + {(\cos^{2} x)}^{2}) = (\sin^{2} x - \cos^{2} x) (1) [{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \cos^{2} x \sin^{2} x] [∵ a^{2} + b^{2} = (a + b)^{2} - 2 ab and \sin^{2} A + \cos^{2} A = 1] = (- \cos 2 x) (1 - 2 \cos^{2} x \sin^{2} x) [∵ \cos 2 A = \cos^{2} A - \sin^{2} A] = \cos 2 x (2 \cos^{2} x \sin^{2} x - 1) = \cos 2 x (\frac{(2 sinx cosx)^{2}}{2} - 1) = \cos 2 x (\frac{\sin^{2} 2 x}{2} - 1) [∵ \sin 2 A = 2 sinA cosA] = \cos 2 x (\frac{1 - \cos 4 x}{4} - 1) [∵ \cos 2 A = 1 - 2 \sin^{2} A] = \cos 2 x (\frac{1 - \cos 4 x - 4}{4}) = - \frac{1}{4} \cos 2 x (3 + \cos 4 x) = - \frac{1}{4} (3 \cos 2 x + \cos 2 x \cos 4 x) = - \frac{1}{4} (3 \cos 2 x + \frac{\cos (\frac{2 x + 4 x}{2}) + \cos (\frac{2 x - 4 x}{2})}{2}) [∵ 2 cosAcosB = \cos (A + B) + \cos (A - B) and \cos (- A) = cosA] = - \frac{1}{4} (3 \cos 2 x + \frac{\cos 6 x + \cos 2 x}{2}) = - \frac{1}{8} (7 \cos 2 x + \cos 6 x) So, 8 \sin^{8} x - \cos^{8} x = - (7 \cos 2 x + \cos 6 x)

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Q. Solve : sin 8 x - cos 8 x .