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Grade: 12

                        

Find the general solution of the equation Sin 5 x – cos 5 x = 1/cosx – 1/sinx (sinx ≠ cosx )

9 years ago

Answers : (1)

Jitender Singh
IIT Delhi
askIITians Faculty
158 Points
							Ans:Hello student, please find answer to your question
sin^{5}x - cos^{5}x = \frac{1}{cosx}-\frac{1}{sinx}
(sinx - cosx)(sin^{4}x+sin^{3}x.cosx+cos^{3}x.sinx+sin^{2}x.cos^{2}x+cos^{4}x) = \frac{sinx-cosx}{sinx.cosx}sinx \neq cosx
sinx.cosx(sin^{4}x+sinx.cosx(sin^{2}x+cos^{2}x)+sin^{2}x.cos^{2}x+cos^{4}x) = 1sinx.cosx(sin^{4}x+sinx.cosx+sin^{2}x.cos^{2}x+cos^{4}x) = 1
sinx.cosx(sin^{4}x+cos^{4}x+sinx.cosx+sin^{2}x.cos^{2}x) = 1
sin^{4}x+cos^{4}x = (sin^{2}x-\sqrt{2}sinx.cosx+cos^{2}x)(sin^{2}x+\sqrt{2}sinx.cosx+cos^{2}x)sin^{4}x+cos^{4}x = (1-\sqrt{2}sinx.cosx)(1+\sqrt{2}sinx.cosx)
sin^{4}x+cos^{4}x = 1-2sin^{2}x.cos^{2}x
sinx.cosx(sin^{4}x+cos^{4}x+sinx.cosx+sin^{2}x.cos^{2}x) = 1
sinx.cosx(1-2sin^{2}x.cos^{2}x+sinx.cosx+sin^{2}x.cos^{2}x) = 1
sinx.cosx(1-sin^{2}x.cos^{2}x+sinx.cosx) = 1
\Rightarrow {2\pi n-2tan^{-1}(1-\sqrt{2}), 2\pi n-2tan^{-1}(1+\sqrt{2})}
6 years ago
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