The correct way of simplifying 4cos22α−4cos2α+3sin2α4cos2(5π2−α)−sin22(α−π)Using half angle formula, I wrote it as follows4cos22α−4cos2α+3sin2α4cos2(5π2−α)−sin22(α−π)=4cos⁡4α−7cos⁡2α+32−2cos⁡2α−1−cos⁡4α2But I can't figure out what to do next. In fact, I tried other ways of expressing it, but again had numbers that couldn't be simplified.

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The correct way of simplifying 4cos22α−4cos2α+3sin2α4cos2(5π2−α)−sin22(α−π)Using half angle formula, I wrote it as follows4cos22α−4cos2α+3sin2α4cos2(5π2−α)−sin22(α−π)=4cos⁡4α−7cos⁡2α+32−2cos⁡2α−1−cos⁡4α2But I can&#039;t figure out what to do next. In fact, I tried other ways of expressing it, but again had numbers that couldn&#039;t be simplified.

tabido8uvt · Accepted Answer

It&#039;s4(2cos2α−1)2−4cos2α+3−3cos2α4(1−cos2α)−4(1−cos2α)cos2αor(1−cos2α)(7−16cos2α)4(1−cos2α)−4(1−cos2α)cos2αor7−16cos2α4sin2αor7−8(1+cos⁡2α}{2(1−cos⁡2α)}or1+8cos⁡2α2cos⁡2α−2

The correct way of simplifying \(\displaystyle{\frac{{{4}{{\cos}^{{2}}{2}}\alpha-{4}{{\cos}^{{2}}\alpha}+{3}{{\sin}^{{2}}\alpha}}}{{{4}{{\cos}^{{2}}{\left({\frac{{{5}\pi}}{{{2}}}}-\alpha\right)}}-{{\sin}^{{2}}{2}}{\left(\alpha-\pi\right)}}}}\) Using half

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