Find the value of $sin (270^{o})$ .

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Solution

Now,
$sin (270^{o})$ $= sin (3 \times 90^{o} + 0^{o})$

$sin (270^{o})$ $= - cos 0^{o}$

$sin (270^{o})$ $= - 1$ .

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Similar questions

cos (270^{o} + θ) cos (90^{o} - θ) - sin (270^{o} - θ) cos θ

cos θ + s i n (270^{o} + θ) - sin (270^{o} - θ) + cos (180^{o} + θ) = 0

sin (270^{o} - x) = cos 292^{o}

x

({0, 360}^{o}) .

\frac{sin (270^{o} + x) {cos}^{3} (720^{o} - x) - sin (270^{o} - x) {sin}^{3} (540^{o} + x)}{sin (90^{o} + x) sin (- x) - {cos}^{2} (180^{o} - x)} + \frac{c o t (270^{o} - x)}{c o s e c^{2} (450^{o} + x)}

(666)^{2}

(66666)^{2}

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