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Maple, en una ecuación, interpreta directamente los ángulos de las razones trigonométricas como las incógnitas a calcular. Por ejemplo,
> solve(sin(2*x)=tan(x));
π
4
,−
3π
4
,0,π
MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacqaHapaCaeaacaaI0aaaaiaacYcacqGHsisldaWcaaqaaiaaiodacqaHapaCaeaacaaI0aaaaiaacYcacaaIWaGaaiilaiabec8aWbaa@4134@
Obsérvese que la respuesta sólo da los argumentos que se encuentran en el intervalo
[
−π,π
]
MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaacqGHsislcqaHapaCcaGGSaGaeqiWdahacaGLBbGaayzxaaaaaa@3CF6@
Más aún,
> solve(cos(Pi/6+x)=sin(x));
π
6
MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacqaHapaCaeaacaaI2aaaaaaa@387A@
omitiendo la respuesta
−
5π
6
MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaSaaaeaacaaI1aGaeqiWdahabaGaaGOnaaaacqGH9aqpdaWcaaqaaiabec8aWbqaaiaaiAdaaaGaey4kaSIaeqiWdahaaa@4058@
, toda vez que
−
5π
6
=
π
6
+π
MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaSaaaeaacaaI1aGaeqiWdahabaGaaGOnaaaacqGH9aqpdaWcaaqaaiabec8aWbqaaiaaiAdaaaGaey4kaSIaeqiWdahaaa@4058@
está incluído en la solución general
−
5π
6
=
π
6
+kπ
MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaSaaaeaacaaI1aGaeqiWdahabaGaaGOnaaaacqGH9aqpdaWcaaqaaiabec8aWbqaaiaaiAdaaaGaey4kaSIaeqiWdahaaa@4058@
Maple nos proporciona tal solución general, previo aviso y a petición del usuario:
> _EnvAllSolutions := true:
Ahora > solve(cos(Pi/6+x)=sin(x));
1
6
π+π_Z1∼
MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaaabaGaaGOnaaaacqaHapaCcqGHRaWkcqaHapaCcaGGFbGaamOwaiaaigdacqWI8iIoaaa@3F7A@
donde la expresión
_Z1∼
MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaaabaGaaGOnaaaacqaHapaCcqGHRaWkcqaHapaCcaGGFbGaamOwaiaaigdacqWI8iIoaaa@3F7A@
es fácilmente interpretable por el navegante; para mayor
precisión . La ventana es el resultado del comando
> ?solve[details]
Para volver al uso normal de solve
, deshacemos el comando _EnvAllSolutions
mediante
> _EnvAllSolutions := false:
Así,
> solve( sin(x) = cos(x) - 1, x );
−
π
2
,0
MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0YaaSaaaeaacqaHapaCaeaacaaIYaaaaiaacYcacaaIWaaaaa@3ACD@
> solve(sec(x) = 2*(1 + cos (x)));
π−arccos(
1
2
+
3
2
),arccos(
3
2
−
1
2
)
MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqiWdaNaeyOeI0IaciyyaiaackhacaGGJbGaai4yaiaac+gacaGGZbWaaeWaaeaadaWcaaqaaiaaigdaaeaacaaIYaaaaiabgUcaRmaalaaabaWaaOaaaeaacaaIZaaaleqaaaGcbaGaaGOmaaaaaiaawIcacaGLPaaacaGGSaGaciyyaiaackhacaGGJbGaai4yaiaac+gacaGGZbWaaeWaaeaadaWcaaqaamaakaaabaGaaG4maaWcbeaaaOqaaiaaikdaaaGaeyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaaaaaiaawIcacaGLPaaaaaa@4FBC@
Maple resuelve sistemas de ecuaciones trigonométricas
> solve({cos(2*x+y)=1/2,sin(2*x)-sin(y)=sqrt(3)/2},{x,y});
{
x=0,y= −
π
3
},{
x=
π
2
,y= −
2π
3
},{
x=
π
3
,y= π
},{
x=
π
6
,y=0
}
MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6801@
> solve({sin(x)*sin(y)=-1/2,cos(x)*cos(y)=-1/2},{x,y});
{
y= arctan(
2_Z
2
−1
,
2_Z
2
−1
),x=arctan(
−
2_Z
2
−1
,−
2_Z
2
−1
)
}
MathType@MTEF@5@5@+=feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiWaaeaacaWG5bGaeyypa0JaaeiiaiaabggacaqGYbGaae4yaiaabshacaqGHbGaaeOBamaabmaabaWaaOaaaeaacaqGYaGaae4xaiaabQfadaahaaWcbeqaaiaabkdaaaGccqGHsislcaqGXaaaleqaaOGaaiilamaakaaabaGaaeOmaiaab+facaqGAbWaaWbaaSqabeaacaqGYaaaaOGaeyOeI0IaaeymaaWcbeaaaOGaayjkaiaawMcaaiaabYcacaWG4bGaeyypa0JaaeyyaiaabkhacaqGJbGaaeiDaiaabggacaqGUbWaaeWaaeaacqGHsisldaGcaaqaaiaabkdacaqGFbGaaeOwamaaCaaaleqabaGaaeOmaaaakiabgkHiTiaabgdaaSqabaGccaGGSaGaeyOeI0YaaOaaaeaacaqGYaGaae4xaiaabQfadaahaaWcbeqaaiaabkdaaaGccqGHsislcaqGXaaaleqaaaGccaGLOaGaayzkaaaacaGL7bGaayzFaaaaaa@6371@
La resolución de la última ecuación y el último sistema ha involucrado las funciones circulares inversas, de las que hablaremos en próximas ventanas.
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