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Question Number 152035 by puissant last updated on 25/Aug/21

Show that 2sin7θcos3θ=sin10θ+sin4θ.

$${Show}\:{that}\:\mathrm{2}{sin}\mathrm{7}\theta{cos}\mathrm{3}\theta={sin}\mathrm{10}\theta+{sin}\mathrm{4}\theta. \\ $$

Answered by som(math1967) last updated on 25/Aug/21

2sinAcosB=sin(A+B)+sin(A−B)  ∴2sin7𝛉cos3𝛉=sin(7𝛉+3𝛉)+sin(7𝛉−3𝛉)                =sin10𝛉+sin4𝛉

$$\mathrm{2}{sinA}\boldsymbol{{cosB}}=\boldsymbol{{sin}}\left(\boldsymbol{{A}}+\boldsymbol{{B}}\right)+\boldsymbol{{sin}}\left(\boldsymbol{{A}}−\boldsymbol{{B}}\right) \\ $$$$\therefore\mathrm{2}\boldsymbol{{sin}}\mathrm{7}\boldsymbol{\theta{cos}}\mathrm{3}\boldsymbol{\theta}=\boldsymbol{{sin}}\left(\mathrm{7}\boldsymbol{\theta}+\mathrm{3}\boldsymbol{\theta}\right)+\boldsymbol{{sin}}\left(\mathrm{7}\boldsymbol{\theta}−\mathrm{3}\boldsymbol{\theta}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\boldsymbol{{sin}}\mathrm{10}\boldsymbol{\theta}+{s}\boldsymbol{{in}}\mathrm{4}\boldsymbol{\theta} \\ $$

Commented by puissant last updated on 25/Aug/21

thank you sir..

$${thank}\:{you}\:{sir}.. \\ $$

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