Aspire Faculty ID #15188 · Topic: JAMIA MCA 2017 · Just now
JAMIA MCA 2017

$\sin6^\circ\,\sin42^\circ\,\sin66^\circ\,\sin78^\circ$ is equal to …

Solution

Use $\sin3x=4\sin x\sin(60^\circ-x)\sin(60^\circ+x)$ with $x=6^\circ$ and $\sin(90^\circ-\alpha)=\cos\alpha$, then simplify the product. Value $= \tfrac{1}{16}$.

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