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Prove that:i ...

Question

Prove that:
(i) cos 10° cos 30° cos 50° cos 70° = $\frac{3}{16}$

(ii) cos 40° cos 80° cos 160° = $- \frac{1}{8}$

(iii) sin 20° sin 40° sin 80° = $\frac{\sqrt{3}}{8}$

(iv) cos 20° cos 40° cos 80° = $\frac{1}{8}$

(v) tan 20° tan 40° tan 60° tan 80° = 3

(vi) tan 20° tan 30° tan 40° tan 80° = 1

(vii) sin 10° sin 50° sin 60° sin 70° = $\frac{\sqrt{3}}{16}$

(viii) sin 20° sin 40° sin 60° sin 80° = $\frac{3}{16}$

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Solution

(i)

$LHS = \cos 10 ° \cos 30 ° \cos 50 ° \cos 70 ° = \frac{1}{2} [2 \cos 10 ° \cos 50 °] \cos 30 ° \cos 70 ° = \frac{1}{2} [\cos (10 ° + 50 °) + \cos (10 ° - 50 °)] \cos 30 ° \cos 70 ° \{∵ 2 \cos A \cos B = \cos (A + B) - \cos (A - B)\} = \frac{1}{2} [\cos 60 ° + \cos (- 40 °)] \cos 30 ° \cos 70 ° = \frac{1}{2} [\frac{1}{2} + \cos 40 °] (\frac{\sqrt{3}}{2}) \times \cos 70 °$

$= \frac{\sqrt{3}}{4} \cos 70 ° [\frac{1}{2} + \cos 40 °] = \frac{\sqrt{3}}{8} \cos 70 ° + \frac{\sqrt{3}}{4} [\cos 70 ° \cos 40 °] = \frac{\sqrt{3}}{8} \cos 70 ° + \frac{\sqrt{3}}{8} [2 \cos 70 ° \cos 40 °] = \frac{\sqrt{3}}{8} \cos 70 ° + \frac{\sqrt{3}}{8} [\cos (70 ° + 40 °) + \cos (70 ° - 40 °)] = \frac{\sqrt{3}}{8} \cos 70 ° + \frac{\sqrt{3}}{8} [\cos 110 ° + \cos 30 °] = \frac{\sqrt{3}}{8} \cos 70 ° + \frac{\sqrt{3}}{8} [\cos (180 ° - 70 °) + \frac{\sqrt{3}}{2}] = \frac{\sqrt{3}}{2} \cos 70 ° - \frac{\sqrt{3}}{8} \cos 70 ° + \frac{3}{16} [∵ \cos (180 ° - 70 °) = - \cos 70 °] = \frac{3}{16} = RHS$

(ii)
$LHS = \cos 40 ° \cos 80 ° \cos 160 ° = \frac{1}{2} [2 \cos 40 ° \cos 80 °] \cos 160 ° = \frac{1}{2} [\cos (40 ° + 80 °) + \cos (40 ° - 80 °)] \cos 160 ° = \frac{1}{2} [\cos 120 ° + \cos (- 40 °)] \cos 160 ° = \frac{1}{2} \cos (160 °) [- \frac{1}{2} + \cos 40 °] = - \frac{1}{4} \cos 160 ° + \frac{1}{2} \cos 160 ° \cos 40 °$
$= - \frac{1}{4} \cos 160 ° + \frac{1}{4} [2 \cos 160 ° \cos 40 °] = - \frac{1}{4} \cos 160 ° + \frac{1}{4} [\cos (160 ° + 40 °) + \cos (160 ° - 40 °)] = - \frac{1}{4} \cos 160 ° + \frac{1}{4} [\cos 200 ° + \cos 120 °] = - \frac{1}{4} \cos 160 ° + \frac{1}{4} [\cos (360 ° - 160 °) - \frac{1}{2}] = - \frac{1}{4} \cos 160 ° + \frac{1}{4} \cos 160 ° - \frac{1}{8} [∵ \cos (360 ° - 160 °) = \cos 160 °] = - \frac{1}{8} = RHS$

(iii)
$LHS = \sin 20 ° \sin 40 ° \sin 80 ° = \frac{1}{2} [2 \sin 20 ° \sin 40 °] \sin 80 ° = \frac{1}{2} [\cos (20 ° - 40 °) - \cos (20 ° + 40 °)] \sin 80 ° = \frac{1}{2} [\cos 20 ° - \frac{1}{2}] \sin 80 ° = \frac{1}{2} \sin 80 ° [\cos 20 ° - \frac{1}{2}] = \frac{1}{2} \sin 80 ° \cos 20 ° - \frac{1}{4} \sin 80 ° = \frac{1}{2} \sin (90 ° - 10 °) \cos 20 ° - \frac{1}{4} \sin 80 ° = \frac{1}{2} \cos 10 ° \cos 20 ° - \frac{1}{4} \sin 80 °$

$= \frac{1}{4} [2 \cos 10 ° c os 20 °] - \frac{1}{4} \sin 80 ° = \frac{1}{4} [\cos (10 ° + 20 °) + \cos (10 ° - 20 °)] - \frac{1}{4} \sin 80 ° = \frac{1}{4} [\cos 30 ° + \cos (- 10 °)] - \frac{1}{4} \sin 80 ° = \frac{1}{4} [\cos 30 ° + \cos (90 ° - 80 °)] - \frac{1}{4} \sin 80 ° = \frac{\sqrt{3}}{8} + \frac{1}{4} \sin 80 ° - \frac{1}{4} \sin 80 ° [∵ \cos (90 ° - 80 °) = \sin 80 °] = \frac{\sqrt{3}}{8} = RHS$

(iv)

$LHS = \cos 20 ° \cos 40 ° \cos 80 ° = \frac{1}{2} [2 \cos 20 ° \cos 40 °] \cos 80 ° = \frac{1}{2} [\cos (20 ° + 40 °) + \cos (20 ° - 40 °)] \cos 80 ° = \frac{1}{2} [\cos 60 ° + \cos (- 20 °)] \cos 80 ° = \frac{1}{2} \cos 80 ° [\frac{1}{2} + \cos 20 °] = \frac{1}{4} c o s 80 ° + \frac{1}{2} \cos 80 ° \cos 20 °$

$= \frac{1}{4} \cos 80 ° + \frac{1}{4} [2 \cos 80 ° \cos 20 °] = \frac{1}{4} \cos 80 ° + \frac{1}{4} [\cos (80 ° + 20 °) + \cos (80 ° - 20 °)] = \frac{1}{4} \cos 80 ° + \frac{1}{4} [\cos 100 ° + \cos 60 °] = \frac{1}{4} \cos 80 ° + \frac{1}{4} [\cos (180 ° - 80 °) + \frac{1}{2}] = \frac{1}{4} \cos 80 ° - \frac{1}{4} \cos 80 ° + \frac{1}{8} \{∵ \cos (180 ° - 80 °) = - \cos 80 °\} = \frac{1}{8} = RHS$

(v)
LHS = tan 20° tan 40° tan 60° tan 80°
$= \tan 60 ° \frac{\sin 20 ° \sin 40 ° \sin 80 °}{\cos 20 ° \cos 40 ° \cos 80 °} = \sqrt{3} \times \frac{\frac{1}{2} [2 \sin 20 ° \sin 40 °] \sin 80 °}{\frac{1}{2} [2 \cos 20 ° \cos 40 °] \cos 80 °} = \sqrt{3} \times \frac{\frac{1}{2} [\cos (20 ° - 40 °) - \cos (20 ° + 40 °)] \sin 80 °}{\frac{1}{2} [\cos (20 ° + 40 °) + \cos (20 ° - 40 °)] \cos 80 °} = \sqrt{3} \times \frac{\frac{1}{2} [\cos (- 20 °) - \cos 60 °] \sin 80 °}{\frac{1}{2} [\cos 60 ° + \cos (- 20 °)] \cos 80 °} = \sqrt{3} \times \frac{\frac{1}{2} \sin 80 ° [\cos 20 ° - \frac{1}{2}]}{\frac{1}{2} \cos 80 ° [\frac{1}{2} + \cos 20 °]} = \sqrt{3} \times \frac{\frac{1}{2} \sin 80 ° \cos 20 ° - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{2} \cos 80 ° \cos 20 °} = \sqrt{3} \times \frac{\frac{1}{2} \sin (90 ° - 10 °) \cos 20 ° - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{2} \cos 80 ° \cos 20 °} = \sqrt{3} \times \frac{\frac{1}{2} \cos 10 ° \cos 20 ° - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{2} \cos 80 ° \cos 20 °}$

$= \sqrt{3} \times \frac{\frac{1}{4} [2 \cos 10 ° \cos 20 °] - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{4} [2 \cos 80 ° \cos 20 °]} = \sqrt{3} \times \frac{\frac{1}{4} [\cos (10 ° + 20 °) + \cos (10 ° - 20 °)] - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{4} [\cos (80 ° + 20 °) + \cos (80 ° - 20 °)]} = \sqrt{3} \times \frac{\frac{1}{4} [\cos 30 ° + \cos (- 10 °)] - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{4} [\cos 100 ° + \cos 60 °]} = \sqrt{3} \times \frac{\frac{1}{4} [\cos 30 ° + \cos (90 ° - 80 °)] - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{4} [\cos (180 ° - 80 °) + \frac{1}{2}]} = \sqrt{3} \times \frac{\frac{\sqrt{3}}{8} + \frac{1}{4} \sin 80 ° - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° - \frac{1}{4} \cos 80 ° + \frac{1}{8}} [\cos (90 ° - 80 °) = \sin 80 °, and \cos (180 ° - 80 °) = - \cos (80 °)] = \sqrt{3} \times \frac{\frac{\sqrt{3}}{8}}{\frac{1}{8}} = 3 = RHS$

(vi)
LHS = tan 20° tan 30° tan 60° tan 80°
$= \tan 30 ° \frac{\sin 20 ° \sin 40 ° \sin 80 °}{\cos 20 ° \cos 40 ° \cos 80 °} = \frac{1}{\sqrt{3}} \times \frac{\frac{1}{2} [2 \sin 20 ° \sin 40 °] \sin 80 °}{\frac{1}{2} [2 \cos 20 ° \cos 40 °] \cos 80 °} = \frac{1}{\sqrt{3}} \times \frac{\frac{1}{2} [\cos (20 ° - 40 °) - \cos (20 ° + 40 °)] \sin 80 °}{\frac{1}{2} [\cos (20 ° + 40 °) + \cos (20 ° - 40 °)] \cos 80 °} = \frac{1}{\sqrt{3}} \times \frac{\frac{1}{2} [\cos 20 ° - \frac{1}{2}] \sin 80 °}{\frac{1}{2} [\cos 60 ° + \cos (- 20 °)] \cos 80 °} = \frac{1}{\sqrt{3}} \times \frac{\frac{1}{2} \sin 80 ° [\cos 20 ° - \frac{1}{2}]}{\frac{1}{2} \cos 80 ° [\frac{1}{2} + \cos 20 °]} = \frac{1}{\sqrt{3}} \times \frac{\frac{1}{2} \sin 80 ° \cos 20 ° - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{2} \cos 80 ° \cos 20 °} = \frac{1}{\sqrt{3}} \times \frac{\frac{1}{2} \sin (90 ° - 10 °) \cos 20 ° - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{2} \cos 80 ° \cos 20 °} = \frac{1}{\sqrt{3}} \times \frac{\frac{1}{2} \cos 10 ° \cos 20 ° - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{2} \cos 80 ° \cos 20 °}$

$= \sqrt{3} \times \frac{\frac{1}{4} [2 \cos 10 ° \cos 20 °] - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{4} [2 \cos 80 ° \cos 20 °]} = \sqrt{3} \times \frac{\frac{1}{4} [\cos (10 ° + 20 °) + \cos (10 ° - 20 °)] - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{4} [\cos (80 ° + 20 °) + \cos (80 ° - 20 °)]} = \sqrt{3} \times \frac{\frac{1}{4} [\cos 30 ° + \cos (- 10 °)] - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{4} [\cos 100 ° + \cos 60 °]} = \sqrt{3} \times \frac{\frac{1}{4} [\cos 30 ° + \cos (90 ° - 80 °)] - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° + \frac{1}{4} [\cos (180 ° - 80 °) + \frac{1}{2}]} = \sqrt{3} \times \frac{\frac{\sqrt{3}}{8} + \frac{1}{4} \sin 80 ° - \frac{1}{4} \sin 80 °}{\frac{1}{4} \cos 80 ° - \frac{1}{4} \cos (80 °) + \frac{1}{8}} \{∵ \cos (90 ° - 80 °) = \sin 80 °, \cos (180 ° - 80 °) = - \cos 80 °\} = \sqrt{3} \times \frac{\frac{\sqrt{3}}{8}}{\frac{1}{8}} = 3 = RHS$

(vii)

$LHS = \sin 10 ° \sin 50 ° \sin 60 ° \sin 70 ° = \frac{1}{2} \sin 60 ° [2 \sin 10 ° \sin 50 °] \sin 70 ° = \frac{1}{2} \times \frac{\sqrt{3}}{2} [\cos (10 ° - 50 °) - \cos (10 ° + 50 °)] \sin 70 ° = \frac{\sqrt{3}}{4} [\cos (- 40 °) - \frac{1}{2}] \sin 70 ° = \frac{\sqrt{3}}{4} \sin 70 ° [\cos 40 ° - \frac{1}{2}] = \frac{\sqrt{3}}{4} \sin 70 ° \cos 40 ° - \frac{\sqrt{3}}{8} \sin 70 ° = \frac{\sqrt{3}}{4} \sin (90 ° - 20 °) \cos 40 ° - \frac{\sqrt{3}}{8} \sin 70 ° = \frac{\sqrt{3}}{4} \cos 20 ° \cos 40 ° - \frac{\sqrt{3}}{8} \sin 70 °$

$= \frac{\sqrt{3}}{8} [2 \cos 20 ° \cos 40 °] - \frac{\sqrt{3}}{8} \sin 70 ° = \frac{\sqrt{3}}{8} [\cos (20 ° + 40 °) + \cos (20 ° - 40 °)] - \frac{\sqrt{3}}{8} \sin 70 ° = \frac{\sqrt{3}}{8} [\cos 60 ° + \cos (- 20 °)] - \frac{\sqrt{3}}{8} \sin 70 ° = \frac{\sqrt{3}}{8} [\cos 60 ° + \cos (90 ° - 70 °)] - \frac{\sqrt{3}}{8} \sin 70 ° = \frac{\sqrt{3}}{16} + \frac{\sqrt{3}}{8} \sin 70 ° - \frac{\sqrt{3}}{8} \sin 70 ° [∵ \cos (90 ° - 70 °) = \sin 70 °] = \frac{\sqrt{3}}{16} = RHS$

(viii)

$LHS = \sin 20 ° \sin 40 ° \sin 60 ° \sin 80 ° = \frac{1}{2} \sin 60 ° [2 \sin 20 ° \sin 40 °] \sin 80 ° = \frac{1}{2} \times \frac{\sqrt{3}}{2} [\cos (20 ° - 40 °) - \cos (20 ° + 40 °)] \sin 80 ° = \frac{\sqrt{3}}{4} [\cos 20 ° - \frac{1}{2}] \sin 80 ° = \frac{\sqrt{3}}{4} \sin 80 ° [\cos 20 ° - \frac{1}{2}] = \frac{\sqrt{3}}{4} \sin 80 ° \cos 20 ° - \frac{\sqrt{3}}{8} \sin 80 ° = \frac{\sqrt{3}}{4} \sin (90 ° - 10 °) \cos 20 ° - \frac{\sqrt{3}}{8} \sin 80 ° = \frac{\sqrt{3}}{4} \cos 10 ° \cos 20 ° - \frac{\sqrt{3}}{8} \sin (80 °)$
$= \frac{\sqrt{3}}{8} [2 \cos 10 ° \cos 20 °] - \frac{\sqrt{3}}{8} \sin 80 ° = \frac{\sqrt{3}}{8} [\cos (10 ° + 20 °) + \cos (10 ° - 20 °)] - \frac{\sqrt{3}}{8} \sin 80 ° = \frac{\sqrt{3}}{8} [\cos 30 ° + \cos (- 10 °)] - \frac{\sqrt{3}}{8} \sin 80 ° = \frac{\sqrt{3}}{8} [\cos 30 ° + \cos (90 ° - 80 °)] - \frac{\sqrt{3}}{8} \sin 80 ° = \frac{3}{16} + \frac{\sqrt{3}}{8} \sin 80 ° - \frac{\sqrt{3}}{8} \sin 80 ° [∵ \cos (90 ° - 80 °) = \sin 80 °] = \frac{3}{16} = RHS$

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