Sin230⁰ Cos245⁰ - 3 sin2 30⁰ + 2 cos2 90⁰

1.	Sin230⁰ Cos245⁰ - 3 sin2 30⁰ + 2 cos2 90⁰
Answer» sin²30° cos²45° - 3 sin² 30° + 2 cos² 90° = \((\frac{1}{2})^2 (\frac{1}{\sqrt2})^2 \) - 3 \((\frac{1}{2})^2\) + 2 (0)² = \(\frac{1}{4} \times \frac{1}{2}\) - 3 x \(\frac{1}{4}\) + 0 = \(\frac{1}{8}\) - \(\frac{3}{4}\) = \(\frac{1-3}{8}\) = \(\frac{-2}{8}\) = \(\frac{-1}{4}\)

Answer»

sin²30° cos²45° - 3 sin² 30° + 2 cos² 90°

= \((\frac{1}{2})^2 (\frac{1}{\sqrt2})^2 \) - 3 \((\frac{1}{2})^2\) + 2 (0)²

= \(\frac{1}{4} \times \frac{1}{2}\) - 3 x \(\frac{1}{4}\) + 0

= \(\frac{1}{8}\) - \(\frac{3}{4}\)

= \(\frac{1-3}{8}\)

= \(\frac{-2}{8}\)

= \(\frac{-1}{4}\)

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