Sin 2 42-cos 2 78-5-1-8

L.H.S. = sin^2 42^∘ cos^2 78^∘ = sin^2 (90 48) cos^2 (90 12) =cos^2 48^∘ sin^2 12^∘ =cos (48 + 12). cos(48 12) [ ∵ cos (A+B). cos (A B) = cos^2 A sin^2 ...

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Byju's Answer

Standard IX

Mathematics

Values of Trigonometric Ratios

sin 2 42∘-cos...

Question

$s i n^{2} 42^{\circ} - c o s^{2} 78^{\circ} = \frac{\sqrt{5} + 1}{8}$

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Solution

$L . H . S . = s i n^{2} 42^{\circ} - c o s^{2} 78^{\circ} = s i n^{2} (90 - 48) - c o s^{2} (90 - 12) = c o s^{2} 48^{\circ} - s i n^{2} 12^{\circ} = c o s (48 + 12) . c o s (48 - 12) [∵ c o s (A + B) . c o s (A - B) = c o s^{2} A - s i n^{2} B] = c o s 60^{\circ} . c o s 36^{\circ} = \frac{1}{2} . \frac{\sqrt{5} + 1}{4} [∵ c o s 36^{\circ} = \frac{\sqrt{5} + 1}{4}] = \frac{\sqrt{5} + 1}{8}$
= RHS

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Similar questions

Q. Prove that:

\sin^{2} 42 ° - \cos^{2} 78 = \frac{\sqrt{5} + 1}{8}

Q. Find the value of

{cos}^{2} 12^{\circ} + {cos}^{2} 78^{\circ}

Q. Without using trigonometric tables, prove that:
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(v) sin²48° + sin²42° = 1
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Find :
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ii) $\frac{1}{5} \times \frac{3}{4}$

iii) $\frac{5}{8} \times \frac{8}{15}$

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sin2 42∘−cos2 78∘=√5+18

L.H.S.=sin2 42∘−cos2 78∘=sin2 (90−48)−cos2 (90−12)=cos2 48∘−sin2 12∘=cos (48+12). cos(48−12)[∵ cos (A+B). cos (A−B)=cos2 A−sin2 B]=cos 60∘. cos 36∘=12.√5+14 [∵ cos 36∘=√5+14]=√5+18 = RHS

$s i n^{2} 42^{\circ} - c o s^{2} 78^{\circ} = \frac{\sqrt{5} + 1}{8}$