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Question Number 30856 by Joel578 last updated on 27/Feb/18

Commented by mrW2 last updated on 28/Feb/18

Question is wrong. It should be:  sin^4  A=...

$${Question}\:{is}\:{wrong}.\:{It}\:{should}\:{be}: \\ $$$$\mathrm{sin}^{\mathrm{4}} \:{A}=... \\ $$

Answered by MJS last updated on 27/Feb/18

none of these  (1/2)cos2A=cos^2 A−(1/2)=U  (1/8)cos4A=(1/8)−sin^2 Acos^2 A=V  U+V=cos^4 A−(3/8)  U−V=(3/8)−sin^4 A

$$\mathrm{none}\:\mathrm{of}\:\mathrm{these} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos2}{A}=\mathrm{cos}^{\mathrm{2}} {A}−\frac{\mathrm{1}}{\mathrm{2}}={U} \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos4}{A}=\frac{\mathrm{1}}{\mathrm{8}}−\mathrm{sin}^{\mathrm{2}} {A}\mathrm{cos}^{\mathrm{2}} {A}={V} \\ $$$${U}+{V}=\mathrm{cos}^{\mathrm{4}} {A}−\frac{\mathrm{3}}{\mathrm{8}} \\ $$$${U}−{V}=\frac{\mathrm{3}}{\mathrm{8}}−\mathrm{sin}^{\mathrm{4}} {A} \\ $$

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