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Question Number 126720 by mathocean1 last updated on 23/Dec/20
show that cos(a)+cos(b)=2cos(((a+b)/2))cos(((a−b)/2))
$$ \\ $$$${show}\:{that} \\ $$$${cos}\left({a}\right)+{cos}\left({b}\right)=\mathrm{2}{cos}\left(\frac{{a}+{b}}{\mathrm{2}}\right){cos}\left(\frac{{a}−{b}}{\mathrm{2}}\right) \\ $$
Answered by mathmax by abdo last updated on 23/Dec/20
let ((a+b)/2)=u and ((a−b)/2)=v ⇒ { ((a+b=2u)),((a−b=2v ⇒)) :} { ((a=u+v)),((b=u−v ⇒coaa +cosb=cos(u+v)+cos(u−v))) :} =cosu cosv−sinu sinv+cosu cosv +sinu sinv =2cosu cosv =2cos(((a+b)/2))cos(((a−b)/2)) the equality is proved
$$\mathrm{let}\:\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}=\mathrm{u}\:\mathrm{and}\:\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2}}=\mathrm{v}\:\Rightarrow\begin{cases}{\mathrm{a}+\mathrm{b}=\mathrm{2u}}\\{\mathrm{a}−\mathrm{b}=\mathrm{2v}\:\Rightarrow}\end{cases} \\ $$$$\begin{cases}{\mathrm{a}=\mathrm{u}+\mathrm{v}}\\{\mathrm{b}=\mathrm{u}−\mathrm{v}\:\:\Rightarrow\mathrm{coaa}\:+\mathrm{cosb}=\mathrm{cos}\left(\mathrm{u}+\mathrm{v}\right)+\mathrm{cos}\left(\mathrm{u}−\mathrm{v}\right)}\end{cases} \\ $$$$=\mathrm{cosu}\:\mathrm{cosv}−\mathrm{sinu}\:\mathrm{sinv}+\mathrm{cosu}\:\mathrm{cosv}\:+\mathrm{sinu}\:\mathrm{sinv} \\ $$$$=\mathrm{2cosu}\:\mathrm{cosv}\:=\mathrm{2cos}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\mathrm{a}−\mathrm{b}}{\mathrm{2}}\right)\:\mathrm{the}\:\mathrm{equality}\:\mathrm{is}\:\mathrm{proved} \\ $$
Answered by mr W last updated on 24/Dec/20
θ=((θ+ϕ)/2)+((θ−ϕ)/2) ϕ=((θ+ϕ)/2)−((θ−ϕ)/2) cos θ=cos (((θ+ϕ)/2)+((θ−ϕ)/2))=cos ((θ+ϕ)/2) cos ((θ−ϕ)/2)−sin ((θ+ϕ)/2) sin ((θ−ϕ)/2) cos ϕ=cos (((θ+ϕ)/2)−((θ−ϕ)/2))=cos ((θ+ϕ)/2) cos ((θ−ϕ)/2)+sin ((θ+ϕ)/2) sin ((θ−ϕ)/2) ⇒cos θ+cos ϕ=2 cos ((θ+ϕ)/2) cos ((θ−ϕ)/2)
$$\theta=\frac{\theta+\varphi}{\mathrm{2}}+\frac{\theta−\varphi}{\mathrm{2}} \\ $$$$\varphi=\frac{\theta+\varphi}{\mathrm{2}}−\frac{\theta−\varphi}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\theta=\mathrm{cos}\:\left(\frac{\theta+\varphi}{\mathrm{2}}+\frac{\theta−\varphi}{\mathrm{2}}\right)=\mathrm{cos}\:\frac{\theta+\varphi}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta−\varphi}{\mathrm{2}}−\mathrm{sin}\:\frac{\theta+\varphi}{\mathrm{2}}\:\mathrm{sin}\:\frac{\theta−\varphi}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\varphi=\mathrm{cos}\:\left(\frac{\theta+\varphi}{\mathrm{2}}−\frac{\theta−\varphi}{\mathrm{2}}\right)=\mathrm{cos}\:\frac{\theta+\varphi}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta−\varphi}{\mathrm{2}}+\mathrm{sin}\:\frac{\theta+\varphi}{\mathrm{2}}\:\mathrm{sin}\:\frac{\theta−\varphi}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\:\theta+\mathrm{cos}\:\varphi=\mathrm{2}\:\mathrm{cos}\:\frac{\theta+\varphi}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta−\varphi}{\mathrm{2}} \\ $$

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