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Question Number 148454 by mathdanisur last updated on 28/Jul/21

$${sin}^6\mathit{\alpha }+{co}^6\mathit{\alpha }=\frac{3}{4}\Rightarrow 6cos4\mathit{\alpha }=?$$$$$$

Answered by Ar Brandon last updated on 28/Jul/21

$${\sin }^6\alpha +{\cos }^6\alpha =\frac{3}{4}$$$$({\sin }^2\alpha +{\cos }^2\alpha )[{({\sin }^2\alpha +{\cos }^2\alpha )}^2-{3\sin }^2\alpha {\cos }^2\alpha ]=\frac{3}{4}$$$$1-{3\sin }^2\alpha {\cos }^2\alpha =\frac{3}{4}\Rightarrow {\sin }^2\alpha {\cos }^2\alpha =\frac{1}{12}$$$$6\cos 4\alpha =6({\cos }^{2}2\alpha -{\sin }^{2}2\alpha )$$$$6\cos 4\alpha =6[({\cos }^4\alpha -{2\sin }^2\alpha {\cos }^2\alpha +{\sin }^4\alpha )-{4\sin }^2\alpha {\cos }^2\alpha ]$$$$=6[{({\cos }^2\alpha +{\sin }^2\alpha )}^2-{8\sin }^2\alpha {\cos }^2\alpha ]$$$$=6(1-{8\sin }^2\alpha {\cos }^2\alpha )=6(1-\frac{8}{12})=2$$

Commented by mathdanisur last updated on 28/Jul/21

$$ThankyouSercool$$

Answered by Olaf_Thorendsen last updated on 28/Jul/21

$${\sin }^6\alpha ={\left(\frac{e^{i\alpha }-e^{-i\alpha }}{2i}\right)}^6$$$${\sin }^6\alpha =-\frac{1}{64}(e^{6i\alpha }-6e^{i4\alpha }+15e^{i2\alpha }-20$$$$+15e^{-i2\alpha }-6e^{-i4\alpha }+e^{-i6\alpha })$$$${\sin }^6\alpha =-\frac{1}{32}(\cos 6\alpha -6\cos 4\alpha +15\cos 2\alpha -10)$$$$$$$${\cos }^6\alpha ={\left(\frac{e^{i\alpha }+e^{-i\alpha }}{2}\right)}^6$$$${\cos }^6\alpha =\frac{1}{64}(e^{6i\alpha }+6e^{i4\alpha }+15e^{i2\alpha }+20$$$$+15e^{-i2\alpha }+6e^{-i4\alpha }+e^{-i6\alpha })$$$${\cos }^6\alpha =\frac{1}{32}(\cos 6\alpha +6\cos 4\alpha +15\cos 2\alpha +10)$$$$$$$${\sin }^6\alpha +{\cos }^6\alpha =\frac{3}{4}$$$$\frac{1}{32}(12\cos 4\alpha +20)=\frac{3}{4}$$$$12\cos 4\alpha =\frac{3}{4}\times 32-20=4$$$$6\cos 4\alpha =2$$

Commented by mathdanisur last updated on 28/Jul/21

$$ThankyouSercool$$

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