Question

SE (sen^2)x = (cos^2)x +P, entao (sen^4)x + (cos^4)x sera igual a::::

Answer 1

$sen^{2}(x)=cos^{2}(x)+P$

Pela relação fundamental da trigonometria, temos que sen²x + cos²x = 1, então cos²x = 1 - sen²x

$sen^{2}(x)=cos^{2}(x)+P\\\\sen^{2}(x)=1-sen^{2}(x)+P\\\\sen^{2}x+sen^{2}x=1+P\\\\2\cdot sen^{2}x=1+P\\\\\\\boxed{\boxed{sen^{2}(x)=\dfrac{1}{2}(1+P)}}$

Daí, podemos achar quanto vale cos²x:

$sen^{2}(x)=\dfrac{1}{2}(1+P)\\\\\\1-cos^{2}(x)=\dfrac{1}{2}(1+P)~~~~~~~*(-1)\\\\\\cos^{2}(x)-1=-\dfrac{1}{2}(1+P)\\\\\\cos^{2}(x)=-\dfrac{1}{2}(1+P)+1\\\\\\cos^{2}(x)=\dfrac{-1-P+2}{2}\\\\\boxed{\boxed{cos^{2}(x)=\dfrac{1}{2}(1-P)}}$
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$sen^{4}(x)+cos^{4}(x)=[sen^{2}(x)]^{2}+[cos^{2}(x)]^{2}\\\\\\sen^{4}(x)+cos^{4}(x)=\left[\dfrac{1}{2}(1+P)\right]^{2}+\left[\dfrac{1}{2}(1-P)\right]^{2}\\\\\\sen^{4}(x)+cos^{4}(x)=\dfrac{1}{4}(1+P)^{2}+\dfrac{1}{4}(1-P)^{2}\\\\\\sen^{4}(x)+cos^{4}(x)=\dfrac{(1+P)^{2}+(1-P)^{2}}{4}\\\\\\sen^{4}(x)+cos^{4}(x)=\dfrac{1^{2}+2\cdot1\cdot P+P^{2}+1^{2}-2\cdot1\cdot P+P^{2}}{4}\\\\\\sen^{4}(x)+cos^{4}(x)=\dfrac{1+2P+P^{2}+1-2P+P^{2}}{4}\\\\\\sen^{4}(x)+cos^{4}(x)=\dfrac{2+2P^{2}}{4}$

$sen^{4}(x)+cos^{4}(x)=\dfrac{2\cdot(1+P^{2})}{4}\\\\\\\boxed{\boxed{sen^{4}(x)+cos^{4}(x)=\dfrac{1}{2}(1+P^{2})}}$