Prove that-i 2 sin 5 - 12 sin- 12-12ii 2 cos 5 - 12 cos- 12-12iii 2 sin 5 - 12 cos- 12-3-22

(i) LHS #160;= #160; #160;2sin #160;5 #960;12 #160;sin #160; #960;12 #160;= #160;cos #160;5 #960;12 #160; #160; #960;12 #160; ...

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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

Prove that:i ...

Question

Prove that:
(i) $2 \sin \frac{5 π}{12} \sin \frac{π}{12} = \frac{1}{2}$

(ii) $2 \cos \frac{5 π}{12} \cos \frac{π}{12} = \frac{1}{2}$

(iii) $2 \sin \frac{5 π}{12} \cos \frac{π}{12} = \frac{\sqrt{3} + 2}{2}$

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Solution

(i)
$LHS = 2 (\sin \frac{5 π}{12}) (\sin \frac{π}{12}) = \cos (\frac{5 π}{12} - \frac{π}{12}) - \cos (\frac{5 π}{12} + \frac{π}{12}) [∵ 2 \sin A \sin B = \cos (A - B) - \cos (A + B)] = \cos \frac{π}{3} - \cos \frac{π}{2} = \frac{1}{2} - 0 = \frac{1}{2} RHS = \frac{1}{2} Hence, LHS = RHS$

(ii)
$LHS = 2 (\cos \frac{5 π}{12}) (\cos \frac{π}{12}) = \cos (\frac{5 π}{12} + \frac{π}{12}) + \cos (\frac{5 π}{12} - \frac{π}{12}) [∵ 2 \cos A \cos B = \cos (A + B) + \cos (A - B)] = \cos \frac{π}{2} + \cos \frac{π}{3} = 0 + \frac{1}{2} = \frac{1}{2} RHS = \frac{1}{2} Hence, LHS = RHS$

(iii)
$LHS = 2 (\sin \frac{5 π}{12}) (\cos \frac{π}{12}) = \sin (\frac{5 π}{12} + \frac{π}{12}) + \sin (\frac{5 π}{12} - \frac{π}{12}) [∵ 2 \sin A \cos B = \sin (A + B) + \sin (A - B)] = \sin \frac{π}{2} + \sin \frac{π}{3} = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2} RHS = \frac{2 + \sqrt{3}}{2} Hence, LHS = RHS$

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Similar questions

Prove that:
(i) $2 s i n \frac{5 π}{12} s i n \frac{π}{12} = \frac{1}{2}$
(ii) $2 c o s \frac{5 π}{12} c o s \frac{π}{12}$
(iii) $2 s i n \frac{5 π}{12} c o s \frac{5 π}{12} = \frac{\sqrt{3} + 2}{2}$

Q. Prove that:
(i)

\tan 4 π - \cos \frac{3 π}{2} - \sin \frac{5 π}{6} \cos \frac{2 π}{3} = \frac{1}{4}

(ii)

\sin \frac{13 π}{3} \sin \frac{8 π}{3} + \cos \frac{2 π}{3} \sin \frac{5 π}{6} = \frac{1}{2}

(iii)

\sin \frac{13 π}{3} \sin \frac{2 π}{3} + \cos \frac{4 π}{3} \sin \frac{13 π}{6} = \frac{1}{2}

(iv)

\sin \frac{10 π}{3} \cos \frac{13 π}{6} + \cos \frac{8 π}{3} \sin \frac{5 π}{6} = - 1

(v)

\tan \frac{5 π}{4} \cot \frac{9 π}{4} + \tan \frac{17 π}{4} \cot \frac{15 π}{4} = 0

Q. Prove that:
(i) tan 720° − cos 270° − sin 150° cos 120° =

\frac{1}{4}

(ii) sin 780° sin 480° + cos 120° sin 150° =

\frac{1}{2}

(iii) sin 780° sin 120° + cos 240° sin 390 =

\frac{1}{2}

(iv) sin 600° cos 390° + cos 480° sin 150° = −1
(v) tan 225° cot 405° + tan 765° cot 675° = 0

Q. Prove that:
(i) tan 225° cot 405° + tan 765° cot 675° = 0
(ii)

\sin \frac{8 π}{3} \cos \frac{23 π}{6} + \cos \frac{13 π}{3} \sin \frac{35 π}{6} = \frac{1}{2}

(iii) cos 24° + cos 55° + cos 125° + cos 204° + cos 300° =

\frac{1}{2}

(iv) tan (−225°) cot (−405°) −tan (−765°) cot (675°) = 0
(v) cos 570° sin 510° + sin (−330°) cos (−390°) = 0
(vi)

\tan \frac{11 π}{3} - 2 \sin \frac{4 π}{6} - \frac{3}{4} {cosec}^{2} \frac{π}{4} + 4 \cos^{2} \frac{17 π}{6} = \frac{3 - 4 \sqrt{3}}{2}

(vii)

3 \sin \frac{π}{6} \sec \frac{π}{3} - 4 \sin \frac{5 π}{6} \cot \frac{π}{4} = 1

Q. Prove that:
(i)

c o s^{2} 45^{\circ} - s i n^{2} 15^{\circ} = \frac{\sqrt{3}}{4}

(ii)

S i n^{2} (n + 1) A - s i n^{2} n A = s i n (2 n + 1) A s i n A

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Prove that: (i) 2sin5π12sinπ12=12 (ii) 2cos5π12cosπ12=12 (iii) 2sin5π12cosπ12=3+22

Prove that:
(i) $2 \sin \frac{5 π}{12} \sin \frac{π}{12} = \frac{1}{2}$

(ii) $2 \cos \frac{5 π}{12} \cos \frac{π}{12} = \frac{1}{2}$

(iii) $2 \sin \frac{5 π}{12} \cos \frac{π}{12} = \frac{\sqrt{3} + 2}{2}$