Can anyone see a way to simplify one of these expressions? cos2θsin⁡ϕ+sin2θcos⁡ϕ or sin2θsin⁡ϕ−cos2θcos⁡ϕ Ive

Question

Can anyone see a way to simplify one of these expressions?  cos2θsin⁡ϕ+sin2θcos⁡ϕ  or  sin2θsin⁡ϕ−cos2θcos⁡ϕ  Ive

encolatgehu · Accepted Answer

cos2sin⁡ϕ+sin2θcos⁡ϕ=  =(1−sin2(θ))sin⁡ϕ+sin2θcos⁡ϕ=   =sin⁡ϕ+sin2(θ)(cos⁡(ϕ)−sin⁡ϕ)=   =sin⁡ϕ+sin2(θ)(cos⁡(ϕ)+cos⁡(ϕ+π4))=   =sin⁡ϕ+2sin2(θ)(cos⁡(ϕ+π4)cos⁡(π4))   =sin⁡ϕ+2sin2(θ)cos⁡(ϕ+π4)

redhotdevil13l3 · Accepted Answer

The expressions are pretty simple. You could writef=cos2θsin⁡ϕ+sin2θcos⁡ϕ=12(sin⁡(π4+ϕ)−cos⁡2θsin⁡(π4−ϕ))g=sin2θsin⁡ϕ−cos2cos⁡ϕ=−12(sin⁡(π4−ϕ)+cos⁡2θsin⁡(π4+ϕ))but those are not what I&#039;d call &quot;simpler&quot;. (That said, if your context lends some significance to the quantity ϕ±π4, then there could be some benefit here.)Depending upon your needs, it could be useful to write them asf=cos2θcos⁡ϕ(tan2θ+tan⁡ϕ)g=−cos2θcos⁡ϕ(1−tan2θtan⁡ϕ)Those may-or-may-not seem better, but note that the ratio f/g looks an awful lot like (the negative of) the angle-addition formula for tangent ... if only we had tan2θ=tan⁡ψ for some ψ. But in that case, we&#039;d havecos2θ=11+tan2θ=11+tan⁡ψ=cos⁡ψsin⁡ψ+cos⁡ψ=cos⁡ψ2sin⁡(ψ+π4)sin2θ=sin⁡ψ2sin⁡(ψ+π4)whereupon, we&#039;d obtainf=sin⁡(ψ+ϕ)2sin⁡(ψ+π4)     g=−cos⁡(ψ+ϕ)2sin⁡(ψ+π4)Those are also not necessarily &quot;simpler&quot; than the original forms, and the tan⁡2θ→tan⁡ψ re-parameterization may not be appropriate for your particular needsWe couuld even take it further, defining ψ′=ψ+ππ4 and ϕ′=ϕ−π4:f=sin⁡(ψ′+ϕ′)2sin⁡ψ′     g=−cos⁡()ψ′+ϕ′2sin⁡ψ′but there&#039;s a danger of veering too far outside the context of your investigation.

Vasquez · Accepted Answer

Hint: cos⁡t=eit+e−it2,sin⁡t=eit−e−it2i,t=θ, ϕ

Can anyone see a way to simplify one of these

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