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Precalculus
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Simplify using the double-angle formula: x3 - 2x3 sin2 (x + 1). A. x3 cos (x + 1) B. x3 cos (2x + 1) C. x3 cos 2(x + 1) D. x3 cos2 (x + 1)
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I can't help you with this one, but I'll try to find someone else who can :) alright?
\[x^3 - 2x^3 \sin^2 (x + 1)\] First factor out \(x^3\) \[x^3(1-2\sin^2(x+1))\] let u=x+1 \[x^3(1-2\sin^2(u))\] Using identity \[\cos(2u)=1-2\sin^2(u),\]we have \[x^3\cos(2u)\] Then \[x^3\cos(2(x+1))\]
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