Trigonometry's power (3)

Calculus Level 3

d d x ( 1 4 cos 2 ( x ) + 6 cos 4 ( x ) 4 cos 6 ( x ) + cos 8 ( x ) sin 8 ( x ) 4 sin 6 ( x ) + 6 sin 4 ( x ) 4 sin 2 ( x ) + 1 ) = ? \frac{d}{dx}\left(\frac{1-4\cos^2(x)+6\cos^4(x)-4\cos^6(x)+\cos^8(x)}{\sin^8(x)-4\sin^6(x)+6\sin^4(x)-4\sin^2(x)+1}\right) =\, ?

7 tan 8 ( x ) sec 2 ( x ) 7\tan^8(x)\sec^2(x) 7 tan 8 ( x ) csc 2 ( x ) -7\tan^8(x)\csc^2(x) 8 tan 7 ( x ) sec 2 ( x ) 8\tan^7(x)\sec^2(x) 8 tan 7 ( x ) csc 2 ( x ) 8\tan^7(x)\csc^2(x) 7 cot 8 ( x ) sec 2 ( x ) -7\cot^8(x)\sec^2(x) 7 cot 8 ( x ) csc 2 ( x ) -7\cot^8(x)\csc^2(x)

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2 solutions

Chew-Seong Cheong
Jul 16, 2016

d d x ( 1 4 cos 2 x + 6 cos 4 x 4 cos 6 x + cos 8 x sin 8 x 4 sin 6 x + 6 sin 4 x 4 sin 2 x + 1 ) = d d x ( 1 cos 2 x ) 4 ( 1 sin 2 x ) 4 = d d x s i n 8 x cos 8 x = d d x tan 8 x = 8 tan 7 x sec 2 x \dfrac d{dx} \left( \dfrac {1-4\cos^2 x + 6 \cos^4 x -4 \cos^6 x + \cos^8 x}{\sin^8 x -4\sin^6 x + 6 \sin^4 x -4 \sin^2 x + 1} \right) \\ = \dfrac d{dx} \dfrac {(1-\cos^2 x)^4}{(1-\sin^2 x)^4} \\ = \dfrac d{dx} \dfrac {sin^8 x}{\cos^8 x} \\ = \dfrac d{dx} \tan^8 x \\ = \boxed{8 \tan^7 x \sec^2 x}

Ravneet Singh
Jul 16, 2016

using binomial theorem we can see that

1 4 cos 2 ( x ) + 6 cos 4 ( x ) 4 cos 6 ( x ) + cos 8 ( x ) = ( ( 1 c o s 2 ( x ) ) 4 = s i n 8 ( x ) \large 1-4\cos^2(x)+6\cos^4(x)-4\cos^6(x)+\cos^8(x) = ((1-cos^2(x))^4 = sin^8(x)

and

s i n 8 ( x ) 4 sin 6 ( x ) + 6 sin 4 ( x ) 4 sin 2 ( x ) + 1 = ( ( 1 s i n 2 ( x ) ) 4 = c o s 8 ( x ) \large sin^8(x)-4\sin^6(x)+6\sin^4(x)-4\sin^2(x)+1 = ((1-sin^2(x))^4 = cos^8(x)

therefore

d d x ( s i n 8 ( x ) cos 8 ( x ) ) = d d x t a n 8 ( x ) = 8 tan 7 ( x ) sec 2 ( x ) \large\frac{d}{dx}\left(\frac{sin^8(x)}{\cos^8(x)}\right) = \frac{d}{dx}{tan^8(x)} = 8\tan^7(x)\sec^2(x)

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