Prove that cos20-cos40-cos60-cos80-1-16

To Prove: cos20°·cos40°·cos60°.cos80°=1/16cos20°×cos40°×cos60°×cos80°=1/16.L.H.S=cos20°×cos40°×cos60°×cos80°.We know that, cos60°=1/2.Substituting the value cos ...

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Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

Prove that co...

Question

Prove that $\cos 20 ° \cdot \cos 40 ° \cdot \cos 60 ° . \cos 80 ° = \frac{1}{16}$

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Solution

To Prove: $\cos 20 ° \cdot \cos 40 ° \cdot \cos 60 ° . \cos 80 ° = \frac{1}{16}$
$\cos 20 ° \times \cos 40 ° \times \cos 60 ° \times \cos 80 ° = \frac{1}{16}$ .
$L . H . S = \cos 20 ° \times \cos 40 ° \times \cos 60 ° \times \cos 80 °$ .
We know that, $\cos 60 ° = \frac{1}{2}$ .
Substituting the value $\cos 60 ° = \frac{1}{2}$ in $L . H . S .$
$L . H . S = \cos 20 ° \times \cos 40 ° \times \frac{1}{2} \times \cos 80 °$
Multiplying and dividing the equation by $2$ .
$= \frac{1}{2} \times \frac{2}{2} (\cos 20 ° \times \cos 40 ° \times \cos 80 °)$
We know that, $2 \cos a \times \cos b = \cos (a + b) + \cos (a - b)$
$= \frac{1}{2 \times 2} (2 \times \cos 20 ° \times \cos 40 ° \times \cos 80 °)$
$= \frac{1}{4} [\cos (20 ° + 80 °) + \cos (20 ° - 80 °)] . \cos 40 °$
$= \frac{1}{4} [\cos 100 ° + \cos (- 60 °)] . \cos 40 °$
$= \frac{1}{4} [\cos 100 ° + \frac{1}{2}] . \cos 40 °$ $[\cos (- 60 °) = \frac{1}{2}]$
$= \frac{1}{4} \cos 100 ° \times \cos 40 ° + \frac{1}{4} \times \frac{1}{2} \times \cos 40 °$
$= \frac{1}{4} (\cos 40 ° \times \cos 100 °) + \frac{1}{8} \times \cos 40 °$
Multiplying and dividing the equation by $2$ .
$= \frac{2}{2} \times \frac{1}{4} (\cos 40 ° \times \cos 100 °) + \frac{2}{2} \times \frac{1}{8} \times \cos 40 °$
$= \frac{1}{8} (2 \cos 40 ° . \cos 100 °) + \frac{2}{2} \times \frac{1}{8} \times \cos 40 °$ .
We know that, $2 \cos a \times \cos b = \cos (a + b) + \cos (a - b)$
$= \frac{1}{8} [\cos (40 ° + 100 °) . \cos (40 ° - 100 °)] + \frac{1}{8} \times \cos 40 °$
$= \frac{1}{8} [\cos 140 ° + \cos (- 60 °)] + \frac{1}{8} \times \cos 40 °$
$= \frac{1}{8} [\cos 140 ° + \frac{1}{2}] + \frac{1}{8} \times \cos 40 °$
$= \frac{1}{8} \cos 140 ° + \frac{1}{8} \times \frac{1}{2} + \frac{1}{8} \times \cos 40 °$
$= \frac{1}{8} \cos 140 ° + \frac{1}{16} + \frac{1}{8} \times \cos 40 °$
$= \frac{1}{16} + \frac{1}{8} (\cos 40 ° + \cos 140 °)$
$= \frac{1}{16} + \frac{1}{8} (2 \cos 90 ° . \cos (- 50 °))$ ………….(As, $2 \cos a \times \cos b = \cos (a + b) + \cos (a - b)$ , As, $90 ° + 50 ° = 140 °, 40 ° - 90 ° = - 50 °$ )
we know that, $\cos 90 ° = 0$
$= \frac{1}{16} + \frac{1}{8} ((0) . \cos (- 50 °))$
$= \frac{1}{16} + 0$
$= \frac{1}{16}$
$= R . H . S .$
Therefore, it is proved that $\cos 20 ° \cdot \cos 40 ° \cdot \cos 60 ° . \cos 80 ° = \frac{1}{16}$ .

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Prove that cos20°·cos40°·cos60°.cos80°=116

Prove that $\cos 20 ° \cdot \cos 40 ° \cdot \cos 60 ° . \cos 80 ° = \frac{1}{16}$