How do you find the value of ${sin}^{2} (225^{\circ})$ $sin^2(225^circ)$ ?

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Answer 1 · 2016-11-21T08:35:24

${sin}^{2} 225^{o} = \frac{1}{2}$ $sin^2 225^o=1/2$

As $sin (180 + A) = - sin A$ $sin(180+A)=-sinA$

${sin 225}^{o} = sin (180^{\oplus} 45^{o}) = - {sin 45}^{o}$ $sin225^o=sin(180^o+45^o)=-sin45^o$

but ${sin 45}^{o} = \frac{1}{\sqrt{2}}$ $sin45^o=1/sqrt2$ and hence ${sin 225}^{o} = - \frac{1}{\sqrt{2}}$ $sin225^o=-1/sqrt2$

Hence, ${sin}^{2} 225^{o} = {(- \frac{1}{\sqrt{2}})}^{2} = \frac{1}{2}$ $sin^2 225^o=(-1/sqrt2)^2=1/2$

How do you find the value of sin2(225∘)sin^2(225^circ)?