sin 47°cos 43° + cos 47°sin 43° = ?
(a) sin 4°
(b) cos 4°
(c) 1
(d) 0

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Solution

$(c) 1 We have: (\sin 47^{0} \cos 43^{0} + \cos 47^{0} \sin 43^{0}) = \sin 47^{0} \cos (90^{0} - 47^{0}) + \cos 47^{0} \sin (90^{0} - 47^{0}) = \sin 47^{0} \sin 47^{0} + \cos 47^{0} \cos 47^{0} [∵ \cos (90^{0} - θ) = \sin θ and \sin (90^{0} - θ) = \cos θ] = \sin^{2} 47^{0} + \cos^{2} 47^{0} = 1$

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Similar questions

sin 43^{\circ} cos 47^{\circ} + cos 43^{\circ} sin 47^{\circ} =

sin $47^{o}$ cos $43^{o}$ + cos $47^{o}$ sin $43^{o}$ = ?

(a) sin $4^{o}$ (b) cos $4^{o}$ (c) 1 (d) 0

{(\frac{\sin 47 °}{\cos 43 °})}^{2} + {(\frac{\cos 43 °}{\sin 47 °})}^{2} - 4 \cos^{2} 45 °

Simplify: ${(\frac{sin 47^{\circ}}{cos 43^{\circ}})}^{2} + {(\frac{cos 43^{\circ}}{sin 47^{\circ}})}^{2} - 4 {cos}^{2} 45^{\circ}$

s i n

43^{\circ}

\times

\frac{1}{c o s 47^{\circ}}

\times

s e c

63^{\circ}

\times

\frac{1}{c o s e c 27^{\circ}}

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