Dados [tex]\alpha,\,\beta\in\mathbb{R},[/tex] valem:
[tex]\cos^2\alpha=1-\mathrm{sen}^2\,\alpha [/tex]
[tex]\cos^2\alpha-\mathrm{sen}^2\alpha=\cos(2\alpha)[/tex]
[tex]2\,\mathrm{sen}\,\alpha\cos\alpha=\mathrm{sen}(2\alpha)[/tex]
[tex]\mathrm{sen}\,\alpha-\mathrm{sen}\,\beta=2\,\mathrm{sen}\!\left(\dfrac{\alpha-\beta}{2}\right)\!\cos\!\left(\dfrac{\alpha+\beta}{2}\right)[/tex]
[tex]\mathrm{sen}\,\alpha=\cos\!\left(\dfrac{\pi}{2}-\alpha\right)[/tex]
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[Proposição] Sejam [tex]x,\,y\in\mathbb{R}.[/tex] Se [tex]\mathrm{sen}^2\,x=\mathrm{sen}\,y\cos y,[/tex] então,
[tex]\cos(2x)=2\cos^2\!\left(\dfrac{\pi}{4}+y\right).[/tex]
[Demonstração]
Por hipótese, temos
[tex]\mathrm{sen}^2\,x=\mathrm{sen}\,y\cos y\qquad\mathrm{(i)}[/tex]
Pela Identidade Trigonométrica Fundamental, podemos escrever
[tex]\Longleftrightarrow\quad \cos^2 x=1-\mathrm{sen}\,y\cos y\qquad\mathrm{(ii)}[/tex]
Efetuando a subtração de (ii) e (i) membro a membro, segue que
[tex]\Longrightarrow\quad \cos^2 x-\mathrm{sen}^2\,x=(1-\mathrm{sen}\,y\cos y)-\mathrm{sen}\,y\cos y\\\\ \Longleftrightarrow\quad \cos(2x)=1-2\,\mathrm{sen}\,y\cos y\\\\ \Longleftrightarrow\quad \cos(2x)=1-\mathrm{sen}(2y)[/tex]
Escreva [tex]1=\mathrm{sen}\,\dfrac{\pi}{2}:[/tex]
[tex]\Longleftrightarrow\quad \cos(2x)=\mathrm{sen}\,\dfrac{\pi}{2}-\mathrm{sen}(2y)\\\\ \Longleftrightarrow\quad \cos(2x)=2\,\mathrm{sen}\!\bigg(\dfrac{\frac{\pi}{2}-2y}{2}\bigg)\!\cos\!\bigg(\dfrac{\frac{\pi}{2}+2y}{2}\bigg)\\\\ \Longleftrightarrow\quad \cos(2x)=2\,\mathrm{sen}\!\left(\dfrac{\pi}{4}-y\right)\!\cos\!\left(\dfrac{\pi}{4}+y\right)\\\\ \Longleftrightarrow\quad \cos(2x)=2\,\cos\!\bigg[\dfrac{\pi}{2}-\bigg(\dfrac{\pi}{4}-y\bigg)\bigg]\!\cos\!\left(\dfrac{\pi}{4}+y\right)\\\\ \Longleftrightarrow\quad \cos(2x)=2\,\cos\!\bigg[\dfrac{\pi}{2}-\dfrac{\pi}{4}+y\bigg]\!\cos\!\left(\dfrac{\pi}{4}+y\right)\\\\ \Longleftrightarrow\quad \cos(2x)=2\cos\!\bigg(\dfrac{\pi}{4}+y\bigg)\!\cos\!\left(\dfrac{\pi}{4}+y\right)\\\\ \Longleftrightarrow\quad \cos(2x)=2\cos^2\!\bigg(\dfrac{\pi}{4}+y\bigg)\qquad\qquad\blacksquare[/tex]
Dúvidas? Comente.
Bons estudos! :-)
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Identidades trigonométricas utilizadas
Dados [tex]\alpha,\,\beta\in\mathbb{R},[/tex] valem:
[tex]\cos^2\alpha=1-\mathrm{sen}^2\,\alpha [/tex]
[tex]\cos^2\alpha-\mathrm{sen}^2\alpha=\cos(2\alpha)[/tex]
[tex]2\,\mathrm{sen}\,\alpha\cos\alpha=\mathrm{sen}(2\alpha)[/tex]
[tex]\mathrm{sen}\,\alpha-\mathrm{sen}\,\beta=2\,\mathrm{sen}\!\left(\dfrac{\alpha-\beta}{2}\right)\!\cos\!\left(\dfrac{\alpha+\beta}{2}\right)[/tex]
[tex]\mathrm{sen}\,\alpha=\cos\!\left(\dfrac{\pi}{2}-\alpha\right)[/tex]
─────
[Proposição] Sejam [tex]x,\,y\in\mathbb{R}.[/tex] Se [tex]\mathrm{sen}^2\,x=\mathrm{sen}\,y\cos y,[/tex] então,
[tex]\cos(2x)=2\cos^2\!\left(\dfrac{\pi}{4}+y\right).[/tex]
[Demonstração]
Por hipótese, temos
[tex]\mathrm{sen}^2\,x=\mathrm{sen}\,y\cos y\qquad\mathrm{(i)}[/tex]
Pela Identidade Trigonométrica Fundamental, podemos escrever
[tex]\Longleftrightarrow\quad \cos^2 x=1-\mathrm{sen}\,y\cos y\qquad\mathrm{(ii)}[/tex]
Efetuando a subtração de (ii) e (i) membro a membro, segue que
[tex]\Longrightarrow\quad \cos^2 x-\mathrm{sen}^2\,x=(1-\mathrm{sen}\,y\cos y)-\mathrm{sen}\,y\cos y\\\\ \Longleftrightarrow\quad \cos(2x)=1-2\,\mathrm{sen}\,y\cos y\\\\ \Longleftrightarrow\quad \cos(2x)=1-\mathrm{sen}(2y)[/tex]
Escreva [tex]1=\mathrm{sen}\,\dfrac{\pi}{2}:[/tex]
[tex]\Longleftrightarrow\quad \cos(2x)=\mathrm{sen}\,\dfrac{\pi}{2}-\mathrm{sen}(2y)\\\\ \Longleftrightarrow\quad \cos(2x)=2\,\mathrm{sen}\!\bigg(\dfrac{\frac{\pi}{2}-2y}{2}\bigg)\!\cos\!\bigg(\dfrac{\frac{\pi}{2}+2y}{2}\bigg)\\\\ \Longleftrightarrow\quad \cos(2x)=2\,\mathrm{sen}\!\left(\dfrac{\pi}{4}-y\right)\!\cos\!\left(\dfrac{\pi}{4}+y\right)\\\\ \Longleftrightarrow\quad \cos(2x)=2\,\cos\!\bigg[\dfrac{\pi}{2}-\bigg(\dfrac{\pi}{4}-y\bigg)\bigg]\!\cos\!\left(\dfrac{\pi}{4}+y\right)\\\\ \Longleftrightarrow\quad \cos(2x)=2\,\cos\!\bigg[\dfrac{\pi}{2}-\dfrac{\pi}{4}+y\bigg]\!\cos\!\left(\dfrac{\pi}{4}+y\right)\\\\ \Longleftrightarrow\quad \cos(2x)=2\cos\!\bigg(\dfrac{\pi}{4}+y\bigg)\!\cos\!\left(\dfrac{\pi}{4}+y\right)\\\\ \Longleftrightarrow\quad \cos(2x)=2\cos^2\!\bigg(\dfrac{\pi}{4}+y\bigg)\qquad\qquad\blacksquare[/tex]
Dúvidas? Comente.
Bons estudos! :-)