The period of the function fx-sin x-sin 2x-sin 4x-sin 5xcos x-cos 2x-cos 4x-cos 5x is

The correct option is D π3 We have f( x ) =sin #160; x+sin #160; 2x+sin #160; 4x+sin #160; 5x cos #160; x+cos #160; 2x+cos #160; ...

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The period of...

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The period of the function
$f (x) = \frac{sin x + sin 2 x + sin 4 x + sin 5 x}{cos x + cos 2 x + cos 4 x + cos 5 x}$ is

\frac{π}{3}

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\frac{π}{4}

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π

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N o n e o f t h e s e

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y = \frac{cos x + cos 2 x + cos 3 x + cos 4 x + cos 5 x + cos 6 x + cos 7 x}{sin x + sin 2 x + sin 3 x + sin 4 x + sin 5 x + sin 6 x + sin 7 x},

\cos x + \cos 2 x + \cos 3 x = 0

\cos x + \cos 3 x - \cos 2 x = 0

\sin x + \sin 5 x = \sin 3 x

\cos x \cos 2 x \cos 3 x = \frac{1}{4}

\cos x + \sin x = \cos 2 x + \sin 2 x

\sin x + \sin 2 x + \sin 3 = 0

\sin x + \sin 2 x + \sin 3 x + \sin 4 x = 0

\sin 3 x - \sin x = 4 \cos^{2} x - 2

\sin 2 x - \sin 4 x + \sin 6 x = 0

{sin}^{5} x - {cos}^{5} x = \frac{1}{cos x} - \frac{1}{sin x}

sin x \neq cos x)

\frac{\sin x}{e^{x}}

{e^{1 - x}}^{2}

f (x) = \frac{x}{2} - \sin \frac{π x}{6} on [- 1, 0]

f (x) = \frac{6 x}{π} - 4 \sin^{2} x on [0, π / 6]

[0, \frac{π}{2}]

f (x) = s i n^{2} x + s i n^{2} (x + \frac{π}{3}) + c o s x c o s (x + \frac{π}{3}) a n d g (\frac{5}{4}) = 1

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