2 срочно мне нужны ответы помогите!!!...

July 2022 1 12 Report

вычислить sin 2a если sin a = 3/5 и п/2 <a<пcos a = -4/5 и п<a<3п/2 срочно мне нужны ответы помогите!!!

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$a)\sin\alpha=\frac35;\ \ \frac\pi2\ \textless \ \alpha\ \textless \ \pi;\\
b)\cos\alpha=-\frac45;\ \ \pi\ \textless \ \alpha\ \textless \ \frac{3\pi}2;\\
\sin2\alpha-?;\\
a)\sin2\alpha=2\cdot\sin\alpha\cdot\cos\alpha;\\
 \frac\pi2\ \textless \ \alpha\ \textless \ \pi==\ \textgreater \ \sin\alpha\ \textgreater \ 0\cup\cos\alpha\ \textless \ 0==\ \textgreater \ \\
\sin\alpha=\frac35,\cos\alpha=-\sqrt{1-\sin^2\alpha}=-\sqrt{1-\left(\frac{3}{5}\right)^2}=-\sqrt{1-\frac{9}{25}}=\\
=-\sqrt{\frac{25-9}{25}}=-\sqrt{\frac{16}{25}}=-\frac45;\\
\sin2\alpha=2\cdot\frac35\cdot\left(-\frac45\right)=-\frac{24}{25};\\

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