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How do I find the range of $y = a sin x + b cos x$
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How do I find the range of $$ f(x) = a sin x + b cos x $$
My textbook says, $$R_f equiv left[ - sqrta^2+ b^2 , sqrta^2 + b^2 right] $$
But How? Why? How do I prove this?
functions trigonometry
asked Jul 18 at 10:10
William
801214
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How do I find the range of $$ f(x) = a sin x + b cos x $$
My textbook says, $$R_f equiv left[ - sqrta^2+ b^2 , sqrta^2 + b^2 right] $$
But How? Why? How do I prove this?
functions trigonometry
asked Jul 18 at 10:10
William
801214
Have you heard of R-formula?
– Karn Watcharasupat
Jul 18 at 10:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How do I find the range of $$ f(x) = a sin x + b cos x $$
My textbook says, $$R_f equiv left[ - sqrta^2+ b^2 , sqrta^2 + b^2 right] $$
But How? Why? How do I prove this?
functions trigonometry
asked Jul 18 at 10:10
William
801214
How do I find the range of $$ f(x) = a sin x + b cos x $$
My textbook says, $$R_f equiv left[ - sqrta^2+ b^2 , sqrta^2 + b^2 right] $$
But How? Why? How do I prove this?
functions trigonometry
asked Jul 18 at 10:10
William
801214
asked Jul 18 at 10:10
William
801214
asked Jul 18 at 10:10
William
801214
asked Jul 18 at 10:10
William
801214
801214
Have you heard of R-formula?
– Karn Watcharasupat
Jul 18 at 10:12
add a comment |Â
Have you heard of R-formula?
– Karn Watcharasupat
Jul 18 at 10:12
Have you heard of R-formula?
– Karn Watcharasupat
Jul 18 at 10:12
Have you heard of R-formula?
– Karn Watcharasupat
Jul 18 at 10:12
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
Suppose
beginalign
asin x + bcos x
&= Rcosphisin x+Rsinphicos x\
&= Rsin(x+phi)
endalign
Then
beginalign
Rcosphi&=atag1\Rsinphi&=btag2
endalign
$(1)^2+(2)^2$,
$$R^2cos^2phi+R^2sin^2phi=R^2(cos^2phi+sin^2)=R^2=a^2+b^2$$
By convention,
$$R=sqrta^2+b^2$$ but actually it can be negative, it doesn't matter.
So
beginalign
asin x + bcos x
&= sqrta^2+b^2sin(x+phi)
endalign
Since $-1lesin x le 1$,
the range is $$left[-sqrta^2+b^2,sqrta^2+b^2right]$$
To find $phi$,
$(2)/(1)$,
$$fracRsinphiRcosphi=tanphi=frac ba$$
So
$$phi=arctan frac ba$$
answered Jul 18 at 10:18
Karn Watcharasupat
3,8192426
add a comment |Â
up vote
0
down vote
HINT:
Notice that $sin^2x+cos^2x=1implies sin x=sqrt1-cos^2x$ so find the stationary points of the function $$f(x)=asqrt1-cos^2x+bcos x$$
answered Jul 18 at 10:13
TheSimpliFire
9,67361951
add a comment |Â
up vote
0
down vote
Let $y=asin x+bcos x$
Using Weierstrass substitution,
we have $$ t^2(b+y)-2at+y-b=0$$ where $t=tandfrac x2$
As $t$ is real, the discriminant must be $ge0$
i.e.,
$$(2a)^2ge4(b+y)(b-y)iff y^2le a^2+b^2$$
answered Jul 18 at 12:12
lab bhattacharjee
215k14152264
See also: mathworld.wolfram.com/SecondDerivativeTest.html
– lab bhattacharjee
Jul 18 at 12:14
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Suppose
beginalign
asin x + bcos x
&= Rcosphisin x+Rsinphicos x\
&= Rsin(x+phi)
endalign
Then
beginalign
Rcosphi&=atag1\Rsinphi&=btag2
endalign
$(1)^2+(2)^2$,
$$R^2cos^2phi+R^2sin^2phi=R^2(cos^2phi+sin^2)=R^2=a^2+b^2$$
By convention,
$$R=sqrta^2+b^2$$ but actually it can be negative, it doesn't matter.
So
beginalign
asin x + bcos x
&= sqrta^2+b^2sin(x+phi)
endalign
Since $-1lesin x le 1$,
the range is $$left[-sqrta^2+b^2,sqrta^2+b^2right]$$
To find $phi$,
$(2)/(1)$,
$$fracRsinphiRcosphi=tanphi=frac ba$$
So
$$phi=arctan frac ba$$
answered Jul 18 at 10:18
Karn Watcharasupat
3,8192426
add a comment |Â
up vote
3
down vote
Suppose
beginalign
asin x + bcos x
&= Rcosphisin x+Rsinphicos x\
&= Rsin(x+phi)
endalign
Then
beginalign
Rcosphi&=atag1\Rsinphi&=btag2
endalign
$(1)^2+(2)^2$,
$$R^2cos^2phi+R^2sin^2phi=R^2(cos^2phi+sin^2)=R^2=a^2+b^2$$
By convention,
$$R=sqrta^2+b^2$$ but actually it can be negative, it doesn't matter.
So
beginalign
asin x + bcos x
&= sqrta^2+b^2sin(x+phi)
endalign
Since $-1lesin x le 1$,
the range is $$left[-sqrta^2+b^2,sqrta^2+b^2right]$$
To find $phi$,
$(2)/(1)$,
$$fracRsinphiRcosphi=tanphi=frac ba$$
So
$$phi=arctan frac ba$$
answered Jul 18 at 10:18
Karn Watcharasupat
3,8192426
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Suppose
beginalign
asin x + bcos x
&= Rcosphisin x+Rsinphicos x\
&= Rsin(x+phi)
endalign
Then
beginalign
Rcosphi&=atag1\Rsinphi&=btag2
endalign
$(1)^2+(2)^2$,
$$R^2cos^2phi+R^2sin^2phi=R^2(cos^2phi+sin^2)=R^2=a^2+b^2$$
By convention,
$$R=sqrta^2+b^2$$ but actually it can be negative, it doesn't matter.
So
beginalign
asin x + bcos x
&= sqrta^2+b^2sin(x+phi)
endalign
Since $-1lesin x le 1$,
the range is $$left[-sqrta^2+b^2,sqrta^2+b^2right]$$
To find $phi$,
$(2)/(1)$,
$$fracRsinphiRcosphi=tanphi=frac ba$$
So
$$phi=arctan frac ba$$
answered Jul 18 at 10:18
Karn Watcharasupat
3,8192426
Suppose
beginalign
asin x + bcos x
&= Rcosphisin x+Rsinphicos x\
&= Rsin(x+phi)
endalign
Then
beginalign
Rcosphi&=atag1\Rsinphi&=btag2
endalign
$(1)^2+(2)^2$,
$$R^2cos^2phi+R^2sin^2phi=R^2(cos^2phi+sin^2)=R^2=a^2+b^2$$
By convention,
$$R=sqrta^2+b^2$$ but actually it can be negative, it doesn't matter.
So
beginalign
asin x + bcos x
&= sqrta^2+b^2sin(x+phi)
endalign
Since $-1lesin x le 1$,
the range is $$left[-sqrta^2+b^2,sqrta^2+b^2right]$$
To find $phi$,
$(2)/(1)$,
$$fracRsinphiRcosphi=tanphi=frac ba$$
So
$$phi=arctan frac ba$$
answered Jul 18 at 10:18
Karn Watcharasupat
3,8192426
answered Jul 18 at 10:18
Karn Watcharasupat
3,8192426
answered Jul 18 at 10:18
Karn Watcharasupat
3,8192426
answered Jul 18 at 10:18
Karn Watcharasupat
3,8192426
3,8192426
add a comment |Â
add a comment |Â
up vote
0
down vote
HINT:
Notice that $sin^2x+cos^2x=1implies sin x=sqrt1-cos^2x$ so find the stationary points of the function $$f(x)=asqrt1-cos^2x+bcos x$$
answered Jul 18 at 10:13
TheSimpliFire
9,67361951
add a comment |Â
up vote
0
down vote
HINT:
Notice that $sin^2x+cos^2x=1implies sin x=sqrt1-cos^2x$ so find the stationary points of the function $$f(x)=asqrt1-cos^2x+bcos x$$
answered Jul 18 at 10:13
TheSimpliFire
9,67361951
add a comment |Â
up vote
0
down vote
up vote
0
down vote
HINT:
Notice that $sin^2x+cos^2x=1implies sin x=sqrt1-cos^2x$ so find the stationary points of the function $$f(x)=asqrt1-cos^2x+bcos x$$
answered Jul 18 at 10:13
TheSimpliFire
9,67361951
HINT:
Notice that $sin^2x+cos^2x=1implies sin x=sqrt1-cos^2x$ so find the stationary points of the function $$f(x)=asqrt1-cos^2x+bcos x$$
answered Jul 18 at 10:13
TheSimpliFire
9,67361951
answered Jul 18 at 10:13
TheSimpliFire
9,67361951
answered Jul 18 at 10:13
TheSimpliFire
9,67361951
answered Jul 18 at 10:13
TheSimpliFire
9,67361951
9,67361951
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $y=asin x+bcos x$
Using Weierstrass substitution,
we have $$ t^2(b+y)-2at+y-b=0$$ where $t=tandfrac x2$
As $t$ is real, the discriminant must be $ge0$
i.e.,
$$(2a)^2ge4(b+y)(b-y)iff y^2le a^2+b^2$$
answered Jul 18 at 12:12
lab bhattacharjee
215k14152264
See also: mathworld.wolfram.com/SecondDerivativeTest.html
– lab bhattacharjee
Jul 18 at 12:14
add a comment |Â
up vote
0
down vote
Let $y=asin x+bcos x$
Using Weierstrass substitution,
we have $$ t^2(b+y)-2at+y-b=0$$ where $t=tandfrac x2$
As $t$ is real, the discriminant must be $ge0$
i.e.,
$$(2a)^2ge4(b+y)(b-y)iff y^2le a^2+b^2$$
answered Jul 18 at 12:12
lab bhattacharjee
215k14152264
See also: mathworld.wolfram.com/SecondDerivativeTest.html
– lab bhattacharjee
Jul 18 at 12:14
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $y=asin x+bcos x$
Using Weierstrass substitution,
we have $$ t^2(b+y)-2at+y-b=0$$ where $t=tandfrac x2$
As $t$ is real, the discriminant must be $ge0$
i.e.,
$$(2a)^2ge4(b+y)(b-y)iff y^2le a^2+b^2$$
answered Jul 18 at 12:12
lab bhattacharjee
215k14152264
Let $y=asin x+bcos x$
Using Weierstrass substitution,
we have $$ t^2(b+y)-2at+y-b=0$$ where $t=tandfrac x2$
As $t$ is real, the discriminant must be $ge0$
i.e.,
$$(2a)^2ge4(b+y)(b-y)iff y^2le a^2+b^2$$
answered Jul 18 at 12:12
lab bhattacharjee
215k14152264
answered Jul 18 at 12:12
lab bhattacharjee
215k14152264
answered Jul 18 at 12:12
lab bhattacharjee
215k14152264
answered Jul 18 at 12:12
lab bhattacharjee
215k14152264
215k14152264
See also: mathworld.wolfram.com/SecondDerivativeTest.html
– lab bhattacharjee
Jul 18 at 12:14
add a comment |Â
See also: mathworld.wolfram.com/SecondDerivativeTest.html
– lab bhattacharjee
Jul 18 at 12:14
See also: mathworld.wolfram.com/SecondDerivativeTest.html
– lab bhattacharjee
Jul 18 at 12:14
See also: mathworld.wolfram.com/SecondDerivativeTest.html
– lab bhattacharjee
Jul 18 at 12:14
add a comment |Â
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Have you heard of R-formula?
– Karn Watcharasupat
Jul 18 at 10:12