Prove that-i sin60- cos30-cos60- sin30-1 -ii sin 4 - 9-7 cos- 9-7-cos 4 - 9-7 sin- 9-7-32iii sin 3 - 8-5 cos- 8-5-cos 3 - 8-5 sin- 8-5-1

(i) LHS #160;= #160;sin60 #176; #952; #160;cos30 #176;+ #952; #160;+cos60 #176; #952; #160;sin30 #176;+ #952; #160; #160; #160 ...

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Standard XII

Mathematics

Composition of Trigonometric Functions and Inverse Trigonometric Functions

Prove that:i ...

Question

Prove that:
(i) sin (60° − θ) cos (30° + θ) + cos (60° − θ) sin (30° + θ) = 1.
(ii) $\sin (\frac{4 π}{9} + 7) \cos (\frac{π}{9} + 7) - \cos (\frac{4 π}{9} + 7) \sin (\frac{π}{9} + 7) = \frac{\sqrt{3}}{2}$
(iii) $\sin (\frac{3 π}{8} - 5) \cos (\frac{π}{8} + 5) + \cos (\frac{3 π}{8} - 5) \sin (\frac{π}{8} + 5) = 1$

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Solution

(i)
$LHS = \sin (60 ° - θ) \cos (30 ° + θ) + \cos (60 ° - θ) \sin (30 ° + θ) = \sin [(60 ° - θ) + (30 ° + θ)] (U sing the formula \sin A \cos B + \cos A \sin B = \sin (A + B) and taking A = 60 ° - θ a n d B = 30 ° + θ) = \sin 90 ° = 1 = RHS Hence proved .$

(ii)
$\sin (\frac{4 π}{9} + 7) \cos (\frac{π}{9} + 7) - \cos (\frac{4 π}{9} + 7) \sin (\frac{π}{9} + 7) = \sin [(\frac{4 π}{9} + 7) - (\frac{π}{9} + 7)] [\sin A \cos B - \cos A \sin B = \sin (A - B)] = \sin \frac{3 π}{9} = \sin \frac{π}{3} = \frac{\sqrt{3}}{2}$

(iii)
$\sin (\frac{3 π}{8} - 5) \cos (\frac{π}{8} + 5) + \cos (\frac{3 π}{8} - 5) \sin (\frac{π}{8} + 5) = \sin [(\frac{3 π}{8} - 5) + (\frac{π}{8} + 5)] [\sin A \cos B + \cos A \sin B = \sin (A + B)] = \sin \frac{4 π}{8} = \sin \frac{π}{2} = 1$

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Similar questions

Q. Prove that:
(i)

\sin (\frac{π}{3} - x) \cos (\frac{π}{6} + x) + \cos (\frac{π}{3} - x) \sin (\frac{π}{6} + x) = 1

(ii)

\sin (\frac{4 π}{9} + 7) \cos (\frac{π}{9} + 7) - \cos (\frac{4 π}{9} + 7) \sin (\frac{π}{9} + 7) = \frac{\sqrt{3}}{2}

(iii)

\sin (\frac{3 π}{8} - 5) \cos (\frac{π}{8} + 5) + \cos (\frac{3 π}{8} - 5) \sin (\frac{π}{8} + 5) = 1

Prove that:
(i) $s i n (60^{\circ} - θ) c o s (30^{\circ} + θ) + c o s (60^{\circ} - θ) s i n (30^{\circ} + θ) = 1$
(ii) $s i n (\frac{4 π}{7} + 7) c o s (\frac{π}{9} + 7) - c o s (\frac{4 π}{9} + 7) s i n (\frac{π}{9} + 7) = \frac{\sqrt{3}}{2}$
(iii) $s i n (\frac{3 π}{8} - 5) c o s (\frac{π}{8} + 5) + c o s (\frac{3 π}{8} - 5) s i n (\frac{π}{8} + 5) = 1$

Q. Show that:
(i)

\frac{1 - \sin 60 °}{\cos 60 °} = \frac{\tan 60 ° - 1}{\tan 60 ° + 1}

(ii)

\frac{\cos 30 ° + \sin 60 °}{1 + \sin 30 ° + \cos 60 °} = \cos 30 °

Show that:

(i) $\frac{1 - sin 60^{\circ}}{cos 60^{\circ}} = \frac{tan 60^{\circ} - 1}{tan 60^{\circ} + 1}$

(ii) $\frac{cos 30^{\circ} + sin 60^{\circ}}{1 + sin 30^{\circ} + cos 60^{\circ}} = cos 30^{\circ}$

Q. If

cot θ = \frac{3}{4}

, prove that

\sqrt{\frac{sec θ - cosec θ}{sec θ + cosec θ}} = \frac{1}{\sqrt{7}}

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Prove that: (i) sin (60° − θ) cos (30° + θ) + cos (60° − θ) sin (30° + θ) = 1. (ii) sin4π9+7cosπ9+7-cos4π9+7sinπ9+7=32 (iii) sin3π8-5cosπ8+5+cos3π8-5sinπ8+5=1