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Question Number 97142 by mhmd last updated on 06/Jun/20

prove that cos(mx)cos(ny)=((cos(mx+ny)+cos(mx−ny))/2) ? help me sir ?

$${prove}\:{that}\:{cos}\left({mx}\right){cos}\left({ny}\right)=\frac{{cos}\left({mx}+{ny}\right)+{cos}\left({mx}−{ny}\right)}{\mathrm{2}}\:\:? \\ $$$${help}\:{me}\:{sir}\:? \\ $$

Commented by mr W last updated on 06/Jun/20

just apply the formulae for cos (a+b) and cos (a−b).

$${just}\:{apply}\:{the}\:{formulae}\:{for} \\ $$$$\mathrm{cos}\:\left({a}+{b}\right)\:{and}\:\mathrm{cos}\:\left({a}−{b}\right). \\ $$

Commented by mhmd last updated on 06/Jun/20

are you can send steb sir ?

$${are}\:{you}\:{can}\:{send}\:{steb}\:{sir}\:? \\ $$

Commented by mr W last updated on 06/Jun/20

i can, but i don′t want to do! if you don′t know the formulae for cos (a+b) and cos (a−b) or how to apply them, it means you have not learnt trigonometry yet, then you won′t be able to understand the answer even if it is step by step. in this case it has totally just one step.

$${i}\:{can},\:{but}\:{i}\:{don}'{t}\:{want}\:{to}\:{do}! \\ $$$${if}\:{you}\:{don}'{t}\:{know}\:{the}\:{formulae}\:{for}\: \\ $$$$\mathrm{cos}\:\left({a}+{b}\right)\:{and}\:\mathrm{cos}\:\left({a}−{b}\right)\:{or}\:{how}\:{to} \\ $$$${apply}\:{them},\:{it}\:{means}\:{you}\:{have}\:{not} \\ $$$${learnt}\:{trigonometry}\:{yet},\:{then} \\ $$$${you}\:{won}'{t}\:{be}\:{able}\:{to}\:{understand} \\ $$$${the}\:{answer}\:{even}\:{if}\:{it}\:{is} \\ $$$${step}\:{by}\:{step}.\:{in}\:{this}\:{case}\:{it}\:{has} \\ $$$${totally}\:{just}\:{one}\:{step}. \\ $$

Answered by mathmax by abdo last updated on 06/Jun/20

we have cos(mx+ny) =cos(mx)cos(ny)−sin(nx)sin(ny) (1) cos(mx−ny) =cos(mx)cos(ny)+sin(nx)sin(ny) (2) (1)+(2) ⇒cos(mx+ny)+cos(mx−ny) =2cos(mx)cos(ny) ⇒ cos(mx)cos(ny) =(1/2)(cos(mx+ny)+cos(mx−ny)). another way by use cos(mx) =((e^(imx) +e^(−imx) )/2) and cos(ny) =((e^(iny) +e^(−iny) )/2)

$$\mathrm{we}\:\mathrm{have}\:\mathrm{cos}\left(\mathrm{mx}+\mathrm{ny}\right)\:=\mathrm{cos}\left(\mathrm{mx}\right)\mathrm{cos}\left(\mathrm{ny}\right)−\mathrm{sin}\left(\mathrm{nx}\right)\mathrm{sin}\left(\mathrm{ny}\right)\:\left(\mathrm{1}\right) \\ $$$$\mathrm{cos}\left(\mathrm{mx}−\mathrm{ny}\right)\:=\mathrm{cos}\left(\mathrm{mx}\right)\mathrm{cos}\left(\mathrm{ny}\right)+\mathrm{sin}\left(\mathrm{nx}\right)\mathrm{sin}\left(\mathrm{ny}\right)\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:\Rightarrow\mathrm{cos}\left(\mathrm{mx}+\mathrm{ny}\right)+\mathrm{cos}\left(\mathrm{mx}−\mathrm{ny}\right)\:=\mathrm{2cos}\left(\mathrm{mx}\right)\mathrm{cos}\left(\mathrm{ny}\right)\:\Rightarrow \\ $$$$\mathrm{cos}\left(\mathrm{mx}\right)\mathrm{cos}\left(\mathrm{ny}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left(\mathrm{mx}+\mathrm{ny}\right)+\mathrm{cos}\left(\mathrm{mx}−\mathrm{ny}\right)\right). \\ $$$$\mathrm{another}\:\mathrm{way}\:\mathrm{by}\:\mathrm{use}\:\:\mathrm{cos}\left(\mathrm{mx}\right)\:=\frac{\mathrm{e}^{\mathrm{imx}} \:+\mathrm{e}^{−\mathrm{imx}} }{\mathrm{2}}\:\mathrm{and}\:\mathrm{cos}\left(\mathrm{ny}\right)\:=\frac{\mathrm{e}^{\mathrm{iny}} \:+\mathrm{e}^{−\mathrm{iny}} }{\mathrm{2}} \\ $$

Answered by 1549442205 last updated on 07/Jun/20

The formulas to transform in trigonometry as: i) cos𝛟+cos𝛙=2cos((𝛟+𝛙)/2)cos((𝛟−𝛙)/2) ii) cos𝛟−cos𝛙=−2sin((𝛟+𝛙)/2)sin((𝛟−𝛙)/2) iii)sin𝛟+sin𝛙=2sin((𝛟+𝛙)/2)cos((𝛟−𝛙)/2) iv)sin𝛟−sin𝛙=2cos((𝛟+𝛙)/2)sin((𝛟−𝛙)/2)

$$\mathrm{The}\:\mathrm{formulas}\:\mathrm{to}\:\mathrm{transform}\:\mathrm{in}\:\mathrm{trigonometry}\:\mathrm{as}: \\ $$$$\left.\boldsymbol{\mathrm{i}}\right)\:\boldsymbol{\mathrm{cos}\varphi}+\boldsymbol{\mathrm{cos}\psi}=\mathrm{2}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\varphi}+\boldsymbol{\psi}}{\mathrm{2}}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\varphi}−\boldsymbol{\psi}}{\mathrm{2}} \\ $$$$\left.\boldsymbol{\mathrm{ii}}\right)\:\boldsymbol{\mathrm{cos}\varphi}−\boldsymbol{\mathrm{cos}\psi}=−\mathrm{2}\boldsymbol{\mathrm{sin}}\frac{\boldsymbol{\varphi}+\boldsymbol{\psi}}{\mathrm{2}}\boldsymbol{\mathrm{sin}}\frac{\boldsymbol{\varphi}−\boldsymbol{\psi}}{\mathrm{2}} \\ $$$$\left.\boldsymbol{\mathrm{iii}}\right)\boldsymbol{\mathrm{sin}\varphi}+\boldsymbol{\mathrm{sin}\psi}=\mathrm{2}\boldsymbol{\mathrm{sin}}\frac{\boldsymbol{\varphi}+\boldsymbol{\psi}}{\mathrm{2}}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\varphi}−\boldsymbol{\psi}}{\mathrm{2}} \\ $$$$\left.\boldsymbol{\mathrm{iv}}\right)\boldsymbol{\mathrm{sin}\varphi}−\boldsymbol{\mathrm{sin}\psi}=\mathrm{2}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\varphi}+\boldsymbol{\psi}}{\mathrm{2}}\boldsymbol{\mathrm{sin}}\frac{\boldsymbol{\varphi}−\boldsymbol{\psi}}{\mathrm{2}} \\ $$

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