3cos260º + 2cot230º – 5sin245º = ?

Foundation

Mathematics Foundation

Introduction to Trigonometry

Question

3cos²60º + 2cot²30º – 5sin²45º = ?

$\frac{13}{6}$

$\frac{17}{4}$

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3cos²60º + 2cot²30º – 5sin²45º

$= 3 \times {(\frac{1}{2})}^{2} + 2 \times {(\sqrt{3})}^{2} - 5 \times {(\frac{1}{\sqrt{2}})}^{2} = \frac{3}{4} + 6 - \frac{5}{2} = \frac{17}{4}$