Solve cos2 48-sin2 12-

The correct option is B √(5)+18 cos ^248^∘ sin^212 = cos^2A sin^2B #160;=cos(A+B) cos(A B)= cos^2A sin^2B #160;= cos(48+12)co ...

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Range of Trigonometric Ratios from 0 to 90 Degrees

Solve cos2 ...

Question

Solve $c o s^{2} 48^{\circ} - s i n^{2} 12^{\circ}$

\frac{\sqrt{5} - 1}{4}

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\frac{\sqrt{5} + 1}{8}

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\frac{\sqrt{3} - 1}{4}

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\frac{\sqrt{3} + 1}{2 \sqrt{2}}

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Solution

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{cos}^{2} 48^{\circ} - {sin}^{2} 12^{\circ} = \frac{\sqrt{5} + 1}{8}

Prove that $({cos}^{2} 48^{\circ} - {sin}^{2} 12^{\circ}) = \frac{(\sqrt{5} + 1)}{8}$

4 \frac{1}{2} \times 4 \frac{1}{3} - 8 \frac{1}{3} \div 5 \frac{2}{3} = ? .

[\frac{4}{5} + {\frac{3}{5} + (\frac{2}{5} - \frac{1}{6})} + \frac{1}{2}] - \frac{2}{3}

\frac{8^{- 1} \times 5^{3}}{2^{- 4}}

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