sin 45300 - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

The given value is $\sin 4530^{0}$

We know that $4530 = 12 \cdot (360) + 210$

Use the formula $s i n (n \cdot 360 ° + θ) = s i n (θ)$

Hence,

$\begin{array}{rcl} \sin 4530 ° & = & \sin (12 (360 °) + 210 °) \\ = & \sin (210 °) \\ = & \sin (180 ° + 30 °) \end{array}$

Use the formula $s i n (180 ° + θ) = - s i n (θ)$

$= - \sin 30 ° = - \frac{1}{2}$

Therrefore, $\sin 4530^{0} = - \frac{1}{2}$

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