Respuestas
Respuesta dada por:
1
Trataremos de abreviar procedimiento
1) Sea
![a=\sin f\cos x\\
b=\cos f\cos x\\
c=\sin x a=\sin f\cos x\\
b=\cos f\cos x\\
c=\sin x](https://tex.z-dn.net/?f=a%3D%5Csin+f%5Ccos+x%5C%5C%0Ab%3D%5Ccos+f%5Ccos+x%5C%5C%0Ac%3D%5Csin+x)
entonces en el numerador tenemos algo como
![N=(a + b + c)^5 + (a - b - c)^5 - (a + b - c)^5 - (a - b + c)^5 N=(a + b + c)^5 + (a - b - c)^5 - (a + b - c)^5 - (a - b + c)^5](https://tex.z-dn.net/?f=N%3D%28a+%2B+b+%2B+c%29%5E5+%2B+%28a+-+b+-+c%29%5E5+-+%28a+%2B+b+-+c%29%5E5+-+%28a+-+b+%2B+c%29%5E5)
2) Pongamos
y = b + c
z = b - c
entonces
![N=(a + y)^5 + (a - y)^5 - (a + z)^5 - (a - z)^5\\ \\
N=2(a^5+10a^3y^2+5ay^4)-2(a^5+10a^3z^2+5az^4)\\ \\
N=2a(a^4+10a^2y^2+5y^4)-2a(a^4+10a^2z^2+5z^4)\\ \\
N=2a(10a^2y^2+5y^4-10a^2z^2-5z^4)\\ \\
N=10a[(y^4-z^4)+2a^2(y^2-z^2)]\\ \\
N=10a(y^2-z^2)(y^2+z^2+2a^2)\\ \\
N=10a(y-z)(y+z)(2b^2+2c^2+2a^2)\\ \\
N=10a(2c)(2b)(2b^2+2c^2+2a^2)\\ \\
\boxed{N=80abc(a^2+b^2+c^2)} N=(a + y)^5 + (a - y)^5 - (a + z)^5 - (a - z)^5\\ \\
N=2(a^5+10a^3y^2+5ay^4)-2(a^5+10a^3z^2+5az^4)\\ \\
N=2a(a^4+10a^2y^2+5y^4)-2a(a^4+10a^2z^2+5z^4)\\ \\
N=2a(10a^2y^2+5y^4-10a^2z^2-5z^4)\\ \\
N=10a[(y^4-z^4)+2a^2(y^2-z^2)]\\ \\
N=10a(y^2-z^2)(y^2+z^2+2a^2)\\ \\
N=10a(y-z)(y+z)(2b^2+2c^2+2a^2)\\ \\
N=10a(2c)(2b)(2b^2+2c^2+2a^2)\\ \\
\boxed{N=80abc(a^2+b^2+c^2)}](https://tex.z-dn.net/?f=N%3D%28a+%2B+y%29%5E5+%2B+%28a+-+y%29%5E5+-+%28a+%2B+z%29%5E5+-+%28a+-+z%29%5E5%5C%5C+%5C%5C%0AN%3D2%28a%5E5%2B10a%5E3y%5E2%2B5ay%5E4%29-2%28a%5E5%2B10a%5E3z%5E2%2B5az%5E4%29%5C%5C+%5C%5C%0AN%3D2a%28a%5E4%2B10a%5E2y%5E2%2B5y%5E4%29-2a%28a%5E4%2B10a%5E2z%5E2%2B5z%5E4%29%5C%5C+%5C%5C%0AN%3D2a%2810a%5E2y%5E2%2B5y%5E4-10a%5E2z%5E2-5z%5E4%29%5C%5C+%5C%5C%0AN%3D10a%5B%28y%5E4-z%5E4%29%2B2a%5E2%28y%5E2-z%5E2%29%5D%5C%5C+%5C%5C%0AN%3D10a%28y%5E2-z%5E2%29%28y%5E2%2Bz%5E2%2B2a%5E2%29%5C%5C+%5C%5C%0AN%3D10a%28y-z%29%28y%2Bz%29%282b%5E2%2B2c%5E2%2B2a%5E2%29%5C%5C+%5C%5C%0AN%3D10a%282c%29%282b%29%282b%5E2%2B2c%5E2%2B2a%5E2%29%5C%5C+%5C%5C%0A%5Cboxed%7BN%3D80abc%28a%5E2%2Bb%5E2%2Bc%5E2%29%7D)
3) En el denominador tenemos
por ende
![R=\dfrac{80abc(a^2+b^2+c^2)}{16abc}=5(a^2+b^2+c^2)\\ \\ \\
R=5[(\sin f\cos x)^2+(\cos f\cos x)^2+(\sin x)^2]\\ \\
R=5[\cos^2x+\sin^2x]\\ \\
\boxed{\boxed{R=5}} R=\dfrac{80abc(a^2+b^2+c^2)}{16abc}=5(a^2+b^2+c^2)\\ \\ \\
R=5[(\sin f\cos x)^2+(\cos f\cos x)^2+(\sin x)^2]\\ \\
R=5[\cos^2x+\sin^2x]\\ \\
\boxed{\boxed{R=5}}](https://tex.z-dn.net/?f=R%3D%5Cdfrac%7B80abc%28a%5E2%2Bb%5E2%2Bc%5E2%29%7D%7B16abc%7D%3D5%28a%5E2%2Bb%5E2%2Bc%5E2%29%5C%5C+%5C%5C+%5C%5C%0AR%3D5%5B%28%5Csin+f%5Ccos+x%29%5E2%2B%28%5Ccos+f%5Ccos+x%29%5E2%2B%28%5Csin+x%29%5E2%5D%5C%5C+%5C%5C+%0AR%3D5%5B%5Ccos%5E2x%2B%5Csin%5E2x%5D%5C%5C+%5C%5C%0A%5Cboxed%7B%5Cboxed%7BR%3D5%7D%7D)
1) Sea
entonces en el numerador tenemos algo como
2) Pongamos
y = b + c
z = b - c
entonces
3) En el denominador tenemos
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