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Sign of Trigonometric Ratios in Different Quadrants

Cos 2 π / 15 ...

Question

Cos2π/15cos4π/15cos8π/15cos16π/15=1/16

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Solution

cos(2π/15) cos(4π/15) cos(8π/15) cos(14π/15)
= cos(2π/15) cos(4π/15) cos(8π/15) cos(π - π/15)
= cos(2π/15) cos(4π/15) cos(8π/15) * -cos(π/15)
= -cos(π/15) cos(2π/15) cos(4π/15) cos(8π/15)
= -16sin(π/15) cos(π/15) cos(2π/15) cos(4π/15) cos(8π/15) / [ 16 sin(π/15) ]
= -8 * [ 2sin(π/15) cos(π/15) ] cos(2π/15) cos(4π/15) cos(8π/15) / [ 16 sin(π/15) ]
= -8 sin(2π/15) cos(2π/15) cos(4π/15) cos(8π/15) / [ 16 sin(π/15) ]
= -4 * [ 2 sin(2π/15) cos(2π/15) ] cos(4π/15) cos(8π/15) / [ 16 sin(π/15) ]
= -4 sin(4π/15) cos(4π/15) cos(8π/15) / [ 16 sin(π/15) ]
= -2 * [2 sin(4π/15) cos(4π/15) ] cos(8π/15) / [ 16 sin(π/15) ]
= -2 sin(8π/15) cos(8π/15) / [ 16 sin(π/15) ]
= -sin(16π/15) / [ 16 sin(π/15) ]
= -sin(π + π/15) / [ 16 sin(π/15) ]
= - * -sin(π/15) / [ 16 sin(π/15) ]
= 1/16

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Similar questions

Prove: $cos \frac{2 π}{15} cos \frac{4 π}{15} cos \frac{8 π}{15} cos \frac{16 π}{15} = \frac{1}{16}$

Prove that :
$cos \frac{2 π}{15} cos \frac{4 π}{15} cos \frac{8 π}{15} cos \frac{16 π}{15} = \frac{1}{16}$

sin x \neq 0

cos x cos 2 x cos 4 x cos 8 x = \frac{sin (2^{4} x)}{2^{4} sin x}

and hence prove that

cos \frac{2 π}{15} cos \frac{4 π}{15} cos \frac{8 π}{15} cos \frac{16 π}{15} = \frac{1}{16}

c o s \frac{2 π}{15} c o s \frac{4 π}{15} c o s \frac{8 π}{15} c o s \frac{14 π}{15}

cos \frac{2 π}{15} cos \frac{4 π}{15} cos \frac{8 π}{15} cos \frac{16 π}{15} = \frac{1}{16}

.
2. One value of

A

which satisfies the equation

{sin}^{4} A - 2 {sin}^{2} A - 1

lies between 0 and

2 π

If 1,2 are true then enter 1 else 0.

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