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(Índia) Trigonometria
Enviado: 28 Mai 2021, 11:47
por Deleted User 23699
Se [tex3]\frac{sen^4\theta }{a}+\frac{cos^4\theta }{b}=\frac{1}{a+b}[/tex3]
, então prove que
[tex3]\frac{sen^8\theta }{a^3}+\frac{cos^8\theta }{b^3}=\frac{1}{(a+b)^3}[/tex3]
Re: (Índia) Trigonometria
Enviado: 28 Mai 2021, 13:32
por Ittalo25
[tex3]bsen^4\theta +acos^4\theta =\frac{ab}{a+b}[/tex3]
[tex3]b^2sen^4\theta +a^2cos^4\theta + ab \cdot (sen^4 \theta+cos^4 \theta)=ab[/tex3]
[tex3]b^2sen^4\theta +a^2cos^4\theta + ab \cdot (1-2sen^2 \theta\cdot cos^2 \theta)=ab[/tex3]
[tex3]b^2sen^4\theta +a^2cos^4\theta - 2ab \cdot sen^2 \theta\cdot cos^2 \theta=0[/tex3]
[tex3](bsen^2\theta-a cos^2\theta)^2=0[/tex3]
[tex3]bsen^2 \theta = a cos^2\theta[/tex3]
[tex3]bsen^2 \theta = a (1-sen^2\theta)[/tex3]
[tex3]sen^2 \theta = \frac{a}{a+b} \rightarrow cos^2\theta = \frac{b}{a+b}[/tex3]
[tex3]\frac{sen^8 \theta}{a^3} = \frac{a}{(a+b)^4} \rightarrow \frac{cos^8\theta}{b^3} = \frac{b}{(a+b)^4}[/tex3]
[tex3]\frac{sen^8 \theta}{a^3}+\frac{cos^8\theta}{b^3} = \frac{a+b}{(a+b)^4} = \boxed{\frac{1}{(a+b)^3}}[/tex3]