Given $sin(t) + cos(t) = a$, derive an expression in '$a$' for $(cos(t))^4 + (sin(t))^4$
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I received this question from a student's trigonometry review assignment. After spending an embarrassing amount of time on it, I consulted others and learned that nobody has been able to solve this problem for multiple semesters. I wonder if there is a typo in the statement or if I just haven't been rigorous enough? Here is the question:
$$textGiven sint + cost = a, text find an equivalent expression for sin^4t + cos^4t text in terms of a.$$
Has anybody seen this one before? I tried (among other things) this (which is an approximation of an earlier attempt):
$(sint + cost)^4$ and using whatever identities I could remember;
$(sint + cost)^4 = sin^4t + cos^4(t) + 4sintcos^3t + 6sin^2tcos^2t + 4sin^3tcost$
so then
$beginalign
sin^4t + cos^4t &= (sint + cost)^4 - 4sintcos^3t - 6sin^2tcos^2t - 4sin^3tcost\
&= (sint + cost)^4 - 2sintcost(2cos^2t + 3sintcost + 2sin^2t)\
&= (sint + cost)^4 - 2sintcost(3sintcost + 4)\
&= a^4 - 2sintcost(3sintcost + 4)
endalign$
Couldn't get further than this, felt like I was overthinking it.
calculus algebra-precalculus trigonometry
edited Sep 5 at 4:57
Nosrati
22.4k61747
asked Sep 5 at 4:18
hatinacat2000
294
add a comment |Â
up vote
3
down vote
favorite
I received this question from a student's trigonometry review assignment. After spending an embarrassing amount of time on it, I consulted others and learned that nobody has been able to solve this problem for multiple semesters. I wonder if there is a typo in the statement or if I just haven't been rigorous enough? Here is the question:
$$textGiven sint + cost = a, text find an equivalent expression for sin^4t + cos^4t text in terms of a.$$
Has anybody seen this one before? I tried (among other things) this (which is an approximation of an earlier attempt):
$(sint + cost)^4$ and using whatever identities I could remember;
$(sint + cost)^4 = sin^4t + cos^4(t) + 4sintcos^3t + 6sin^2tcos^2t + 4sin^3tcost$
so then
$beginalign
sin^4t + cos^4t &= (sint + cost)^4 - 4sintcos^3t - 6sin^2tcos^2t - 4sin^3tcost\
&= (sint + cost)^4 - 2sintcost(2cos^2t + 3sintcost + 2sin^2t)\
&= (sint + cost)^4 - 2sintcost(3sintcost + 4)\
&= a^4 - 2sintcost(3sintcost + 4)
endalign$
Couldn't get further than this, felt like I was overthinking it.
calculus algebra-precalculus trigonometry
edited Sep 5 at 4:57
Nosrati
22.4k61747
asked Sep 5 at 4:18
hatinacat2000
294
first that $2cos^2+2sin^2=2$
– Nosrati
Sep 5 at 4:46
By the fundamental theorem of symmetric polynomials, once you've written $sin t cos t$ in terms of $a$ you can write any polynomial expression which is symmetric in $sin t$ and $cos t$ in terms of $a$.
– Micah
Sep 5 at 4:55
1
$$a^4 - 2sintcost(3sintcost + 2)=a^4 - 2dfraca^2-12(3dfraca^2-12 + 2)=colorbluedfrac-a^4+2a^2+12$$
– Nosrati
Sep 5 at 4:57
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I received this question from a student's trigonometry review assignment. After spending an embarrassing amount of time on it, I consulted others and learned that nobody has been able to solve this problem for multiple semesters. I wonder if there is a typo in the statement or if I just haven't been rigorous enough? Here is the question:
$$textGiven sint + cost = a, text find an equivalent expression for sin^4t + cos^4t text in terms of a.$$
Has anybody seen this one before? I tried (among other things) this (which is an approximation of an earlier attempt):
$(sint + cost)^4$ and using whatever identities I could remember;
$(sint + cost)^4 = sin^4t + cos^4(t) + 4sintcos^3t + 6sin^2tcos^2t + 4sin^3tcost$
so then
$beginalign
sin^4t + cos^4t &= (sint + cost)^4 - 4sintcos^3t - 6sin^2tcos^2t - 4sin^3tcost\
&= (sint + cost)^4 - 2sintcost(2cos^2t + 3sintcost + 2sin^2t)\
&= (sint + cost)^4 - 2sintcost(3sintcost + 4)\
&= a^4 - 2sintcost(3sintcost + 4)
endalign$
Couldn't get further than this, felt like I was overthinking it.
calculus algebra-precalculus trigonometry
edited Sep 5 at 4:57
Nosrati
22.4k61747
asked Sep 5 at 4:18
hatinacat2000
294
I received this question from a student's trigonometry review assignment. After spending an embarrassing amount of time on it, I consulted others and learned that nobody has been able to solve this problem for multiple semesters. I wonder if there is a typo in the statement or if I just haven't been rigorous enough? Here is the question:
$$textGiven sint + cost = a, text find an equivalent expression for sin^4t + cos^4t text in terms of a.$$
Has anybody seen this one before? I tried (among other things) this (which is an approximation of an earlier attempt):
$(sint + cost)^4$ and using whatever identities I could remember;
$(sint + cost)^4 = sin^4t + cos^4(t) + 4sintcos^3t + 6sin^2tcos^2t + 4sin^3tcost$
so then
$beginalign
sin^4t + cos^4t &= (sint + cost)^4 - 4sintcos^3t - 6sin^2tcos^2t - 4sin^3tcost\
&= (sint + cost)^4 - 2sintcost(2cos^2t + 3sintcost + 2sin^2t)\
&= (sint + cost)^4 - 2sintcost(3sintcost + 4)\
&= a^4 - 2sintcost(3sintcost + 4)
endalign$
Couldn't get further than this, felt like I was overthinking it.
calculus algebra-precalculus trigonometry
calculus algebra-precalculus trigonometry
edited Sep 5 at 4:57
Nosrati
22.4k61747
asked Sep 5 at 4:18
hatinacat2000
294
edited Sep 5 at 4:57
Nosrati
22.4k61747
asked Sep 5 at 4:18
hatinacat2000
294
edited Sep 5 at 4:57
Nosrati
22.4k61747
edited Sep 5 at 4:57
Nosrati
22.4k61747
edited Sep 5 at 4:57
Nosrati
22.4k61747
22.4k61747
asked Sep 5 at 4:18
hatinacat2000
294
asked Sep 5 at 4:18
hatinacat2000
294
asked Sep 5 at 4:18
hatinacat2000
294
294
first that $2cos^2+2sin^2=2$
– Nosrati
Sep 5 at 4:46
By the fundamental theorem of symmetric polynomials, once you've written $sin t cos t$ in terms of $a$ you can write any polynomial expression which is symmetric in $sin t$ and $cos t$ in terms of $a$.
– Micah
Sep 5 at 4:55
1
$$a^4 - 2sintcost(3sintcost + 2)=a^4 - 2dfraca^2-12(3dfraca^2-12 + 2)=colorbluedfrac-a^4+2a^2+12$$
– Nosrati
Sep 5 at 4:57
add a comment |Â
first that $2cos^2+2sin^2=2$
– Nosrati
Sep 5 at 4:46
By the fundamental theorem of symmetric polynomials, once you've written $sin t cos t$ in terms of $a$ you can write any polynomial expression which is symmetric in $sin t$ and $cos t$ in terms of $a$.
– Micah
Sep 5 at 4:55
1
$$a^4 - 2sintcost(3sintcost + 2)=a^4 - 2dfraca^2-12(3dfraca^2-12 + 2)=colorbluedfrac-a^4+2a^2+12$$
– Nosrati
Sep 5 at 4:57
first that $2cos^2+2sin^2=2$
– Nosrati
Sep 5 at 4:46
first that $2cos^2+2sin^2=2$
– Nosrati
Sep 5 at 4:46
By the fundamental theorem of symmetric polynomials, once you've written $sin t cos t$ in terms of $a$ you can write any polynomial expression which is symmetric in $sin t$ and $cos t$ in terms of $a$.
– Micah
Sep 5 at 4:55
By the fundamental theorem of symmetric polynomials, once you've written $sin t cos t$ in terms of $a$ you can write any polynomial expression which is symmetric in $sin t$ and $cos t$ in terms of $a$.
– Micah
Sep 5 at 4:55
1
1
$$a^4 - 2sintcost(3sintcost + 2)=a^4 - 2dfraca^2-12(3dfraca^2-12 + 2)=colorbluedfrac-a^4+2a^2+12$$
– Nosrati
Sep 5 at 4:57
$$a^4 - 2sintcost(3sintcost + 2)=a^4 - 2dfraca^2-12(3dfraca^2-12 + 2)=colorbluedfrac-a^4+2a^2+12$$
– Nosrati
Sep 5 at 4:57
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
11
down vote
accepted
$$(sin t + cos t )^2= a^2rightarrow sin tcos t=dfraca^2-12$$
then
$$cos^4t + sin^4t=(cos^2t + sin^2t)^2-2cos^2t sin^2t=1-2left(dfraca^2-12right)^2$$
answered Sep 5 at 4:25
Nosrati
22.4k61747
Very nice, don't know how I missed that first line, but that's always math in hindsight. Very smart limiting next expansion to squares.
– hatinacat2000
Sep 5 at 4:42
add a comment |Â
up vote
1
down vote
$$sin t+cos t=a\(sin t + cos t)^2=a^2\2sin t cos t=a^2-1
\(sin t + cos t)^4=a^4\ sin^4 t + cos^4 t+4sin t cos t (sin^2 t+cos^2 t)+6sin^2 t cos^2 t=a^4\ sin^4 t + cos^4 t +2(a^2-1)+frac 32(a^2-1)^2=a^4\
sin^4 t+cos^4 t=a^4-2(a^2-1)-frac 32(a^2-1)^2$$
answered Sep 5 at 4:27
Ross Millikan
281k23191358
add a comment |Â
up vote
0
down vote
Alternatively, note that:
$$beginalignsin t+cos t=a Rightarrow sinleft(t+fracpi4right)=frac asqrt2 Rightarrow t&=arcsin frac asqrt2-fracpi4\
sin t=sin left(arcsin frac asqrt2-fracpi4right)&=frac asqrt2cdot frac 1sqrt2-sqrt1-fraca^22cdot frac1sqrt2;\
&=frac12left(a-sqrt2-a^2right);\
cos t=cos left(arcsin frac asqrt2-fracpi4right)&=sqrt1-fraca^22cdot frac1sqrt2+frac asqrt2cdot frac 1sqrt2=\
&=frac12left(sqrt2-a^2+aright).endalign$$
Hence:
$$beginalignsin^4t+cos^4t&=frac216cdot left(a^4+6a^2left(2-a^2right)+(2-a^2)^2right)=\
&=-fraca^42+a^2+frac12. endalign$$
answered Sep 5 at 5:51
farruhota
15.4k2734
add a comment |Â
up vote
0
down vote
If $sin x+cos x=a,a^2=1+2sin xcos xiffsin xcos x=dfraca^2-12 $
$$sin^n+2x+cos^n+2x=sin^nx(1-cos^2x)+cos^nx(1-sin^2x)$$
If $I_m=sin^mx+cos^mx,I_2=1,I_0=2$
$$I_n+2=I_n-(sin xcos x)^2I_n-2$$
Here $n=2$
answered Sep 5 at 6:02
lab bhattacharjee
216k14153265
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
$$(sin t + cos t )^2= a^2rightarrow sin tcos t=dfraca^2-12$$
then
$$cos^4t + sin^4t=(cos^2t + sin^2t)^2-2cos^2t sin^2t=1-2left(dfraca^2-12right)^2$$
answered Sep 5 at 4:25
Nosrati
22.4k61747
Very nice, don't know how I missed that first line, but that's always math in hindsight. Very smart limiting next expansion to squares.
– hatinacat2000
Sep 5 at 4:42
add a comment |Â
up vote
11
down vote
accepted
$$(sin t + cos t )^2= a^2rightarrow sin tcos t=dfraca^2-12$$
then
$$cos^4t + sin^4t=(cos^2t + sin^2t)^2-2cos^2t sin^2t=1-2left(dfraca^2-12right)^2$$
answered Sep 5 at 4:25
Nosrati
22.4k61747
Very nice, don't know how I missed that first line, but that's always math in hindsight. Very smart limiting next expansion to squares.
– hatinacat2000
Sep 5 at 4:42
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
$$(sin t + cos t )^2= a^2rightarrow sin tcos t=dfraca^2-12$$
then
$$cos^4t + sin^4t=(cos^2t + sin^2t)^2-2cos^2t sin^2t=1-2left(dfraca^2-12right)^2$$
answered Sep 5 at 4:25
Nosrati
22.4k61747
$$(sin t + cos t )^2= a^2rightarrow sin tcos t=dfraca^2-12$$
then
$$cos^4t + sin^4t=(cos^2t + sin^2t)^2-2cos^2t sin^2t=1-2left(dfraca^2-12right)^2$$
answered Sep 5 at 4:25
Nosrati
22.4k61747
answered Sep 5 at 4:25
Nosrati
22.4k61747
answered Sep 5 at 4:25
Nosrati
22.4k61747
answered Sep 5 at 4:25
Nosrati
22.4k61747
22.4k61747
Very nice, don't know how I missed that first line, but that's always math in hindsight. Very smart limiting next expansion to squares.
– hatinacat2000
Sep 5 at 4:42
add a comment |Â
Very nice, don't know how I missed that first line, but that's always math in hindsight. Very smart limiting next expansion to squares.
– hatinacat2000
Sep 5 at 4:42
Very nice, don't know how I missed that first line, but that's always math in hindsight. Very smart limiting next expansion to squares.
– hatinacat2000
Sep 5 at 4:42
Very nice, don't know how I missed that first line, but that's always math in hindsight. Very smart limiting next expansion to squares.
– hatinacat2000
Sep 5 at 4:42
add a comment |Â
up vote
1
down vote
$$sin t+cos t=a\(sin t + cos t)^2=a^2\2sin t cos t=a^2-1
\(sin t + cos t)^4=a^4\ sin^4 t + cos^4 t+4sin t cos t (sin^2 t+cos^2 t)+6sin^2 t cos^2 t=a^4\ sin^4 t + cos^4 t +2(a^2-1)+frac 32(a^2-1)^2=a^4\
sin^4 t+cos^4 t=a^4-2(a^2-1)-frac 32(a^2-1)^2$$
answered Sep 5 at 4:27
Ross Millikan
281k23191358
add a comment |Â
up vote
1
down vote
$$sin t+cos t=a\(sin t + cos t)^2=a^2\2sin t cos t=a^2-1
\(sin t + cos t)^4=a^4\ sin^4 t + cos^4 t+4sin t cos t (sin^2 t+cos^2 t)+6sin^2 t cos^2 t=a^4\ sin^4 t + cos^4 t +2(a^2-1)+frac 32(a^2-1)^2=a^4\
sin^4 t+cos^4 t=a^4-2(a^2-1)-frac 32(a^2-1)^2$$
answered Sep 5 at 4:27
Ross Millikan
281k23191358
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$sin t+cos t=a\(sin t + cos t)^2=a^2\2sin t cos t=a^2-1
\(sin t + cos t)^4=a^4\ sin^4 t + cos^4 t+4sin t cos t (sin^2 t+cos^2 t)+6sin^2 t cos^2 t=a^4\ sin^4 t + cos^4 t +2(a^2-1)+frac 32(a^2-1)^2=a^4\
sin^4 t+cos^4 t=a^4-2(a^2-1)-frac 32(a^2-1)^2$$
answered Sep 5 at 4:27
Ross Millikan
281k23191358
$$sin t+cos t=a\(sin t + cos t)^2=a^2\2sin t cos t=a^2-1
\(sin t + cos t)^4=a^4\ sin^4 t + cos^4 t+4sin t cos t (sin^2 t+cos^2 t)+6sin^2 t cos^2 t=a^4\ sin^4 t + cos^4 t +2(a^2-1)+frac 32(a^2-1)^2=a^4\
sin^4 t+cos^4 t=a^4-2(a^2-1)-frac 32(a^2-1)^2$$
answered Sep 5 at 4:27
Ross Millikan
281k23191358
answered Sep 5 at 4:27
Ross Millikan
281k23191358
answered Sep 5 at 4:27
Ross Millikan
281k23191358
answered Sep 5 at 4:27
Ross Millikan
281k23191358
281k23191358
add a comment |Â
add a comment |Â
up vote
0
down vote
Alternatively, note that:
$$beginalignsin t+cos t=a Rightarrow sinleft(t+fracpi4right)=frac asqrt2 Rightarrow t&=arcsin frac asqrt2-fracpi4\
sin t=sin left(arcsin frac asqrt2-fracpi4right)&=frac asqrt2cdot frac 1sqrt2-sqrt1-fraca^22cdot frac1sqrt2;\
&=frac12left(a-sqrt2-a^2right);\
cos t=cos left(arcsin frac asqrt2-fracpi4right)&=sqrt1-fraca^22cdot frac1sqrt2+frac asqrt2cdot frac 1sqrt2=\
&=frac12left(sqrt2-a^2+aright).endalign$$
Hence:
$$beginalignsin^4t+cos^4t&=frac216cdot left(a^4+6a^2left(2-a^2right)+(2-a^2)^2right)=\
&=-fraca^42+a^2+frac12. endalign$$
answered Sep 5 at 5:51
farruhota
15.4k2734
add a comment |Â
up vote
0
down vote
Alternatively, note that:
$$beginalignsin t+cos t=a Rightarrow sinleft(t+fracpi4right)=frac asqrt2 Rightarrow t&=arcsin frac asqrt2-fracpi4\
sin t=sin left(arcsin frac asqrt2-fracpi4right)&=frac asqrt2cdot frac 1sqrt2-sqrt1-fraca^22cdot frac1sqrt2;\
&=frac12left(a-sqrt2-a^2right);\
cos t=cos left(arcsin frac asqrt2-fracpi4right)&=sqrt1-fraca^22cdot frac1sqrt2+frac asqrt2cdot frac 1sqrt2=\
&=frac12left(sqrt2-a^2+aright).endalign$$
Hence:
$$beginalignsin^4t+cos^4t&=frac216cdot left(a^4+6a^2left(2-a^2right)+(2-a^2)^2right)=\
&=-fraca^42+a^2+frac12. endalign$$
answered Sep 5 at 5:51
farruhota
15.4k2734
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Alternatively, note that:
$$beginalignsin t+cos t=a Rightarrow sinleft(t+fracpi4right)=frac asqrt2 Rightarrow t&=arcsin frac asqrt2-fracpi4\
sin t=sin left(arcsin frac asqrt2-fracpi4right)&=frac asqrt2cdot frac 1sqrt2-sqrt1-fraca^22cdot frac1sqrt2;\
&=frac12left(a-sqrt2-a^2right);\
cos t=cos left(arcsin frac asqrt2-fracpi4right)&=sqrt1-fraca^22cdot frac1sqrt2+frac asqrt2cdot frac 1sqrt2=\
&=frac12left(sqrt2-a^2+aright).endalign$$
Hence:
$$beginalignsin^4t+cos^4t&=frac216cdot left(a^4+6a^2left(2-a^2right)+(2-a^2)^2right)=\
&=-fraca^42+a^2+frac12. endalign$$
answered Sep 5 at 5:51
farruhota
15.4k2734
Alternatively, note that:
$$beginalignsin t+cos t=a Rightarrow sinleft(t+fracpi4right)=frac asqrt2 Rightarrow t&=arcsin frac asqrt2-fracpi4\
sin t=sin left(arcsin frac asqrt2-fracpi4right)&=frac asqrt2cdot frac 1sqrt2-sqrt1-fraca^22cdot frac1sqrt2;\
&=frac12left(a-sqrt2-a^2right);\
cos t=cos left(arcsin frac asqrt2-fracpi4right)&=sqrt1-fraca^22cdot frac1sqrt2+frac asqrt2cdot frac 1sqrt2=\
&=frac12left(sqrt2-a^2+aright).endalign$$
Hence:
$$beginalignsin^4t+cos^4t&=frac216cdot left(a^4+6a^2left(2-a^2right)+(2-a^2)^2right)=\
&=-fraca^42+a^2+frac12. endalign$$
answered Sep 5 at 5:51
farruhota
15.4k2734
answered Sep 5 at 5:51
farruhota
15.4k2734
answered Sep 5 at 5:51
farruhota
15.4k2734
answered Sep 5 at 5:51
farruhota
15.4k2734
15.4k2734
add a comment |Â
add a comment |Â
up vote
0
down vote
If $sin x+cos x=a,a^2=1+2sin xcos xiffsin xcos x=dfraca^2-12 $
$$sin^n+2x+cos^n+2x=sin^nx(1-cos^2x)+cos^nx(1-sin^2x)$$
If $I_m=sin^mx+cos^mx,I_2=1,I_0=2$
$$I_n+2=I_n-(sin xcos x)^2I_n-2$$
Here $n=2$
answered Sep 5 at 6:02
lab bhattacharjee
216k14153265
add a comment |Â
up vote
0
down vote
If $sin x+cos x=a,a^2=1+2sin xcos xiffsin xcos x=dfraca^2-12 $
$$sin^n+2x+cos^n+2x=sin^nx(1-cos^2x)+cos^nx(1-sin^2x)$$
If $I_m=sin^mx+cos^mx,I_2=1,I_0=2$
$$I_n+2=I_n-(sin xcos x)^2I_n-2$$
Here $n=2$
answered Sep 5 at 6:02
lab bhattacharjee
216k14153265
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $sin x+cos x=a,a^2=1+2sin xcos xiffsin xcos x=dfraca^2-12 $
$$sin^n+2x+cos^n+2x=sin^nx(1-cos^2x)+cos^nx(1-sin^2x)$$
If $I_m=sin^mx+cos^mx,I_2=1,I_0=2$
$$I_n+2=I_n-(sin xcos x)^2I_n-2$$
Here $n=2$
answered Sep 5 at 6:02
lab bhattacharjee
216k14153265
If $sin x+cos x=a,a^2=1+2sin xcos xiffsin xcos x=dfraca^2-12 $
$$sin^n+2x+cos^n+2x=sin^nx(1-cos^2x)+cos^nx(1-sin^2x)$$
If $I_m=sin^mx+cos^mx,I_2=1,I_0=2$
$$I_n+2=I_n-(sin xcos x)^2I_n-2$$
Here $n=2$
answered Sep 5 at 6:02
lab bhattacharjee
216k14153265
answered Sep 5 at 6:02
lab bhattacharjee
216k14153265
answered Sep 5 at 6:02
lab bhattacharjee
216k14153265
answered Sep 5 at 6:02
lab bhattacharjee
216k14153265
216k14153265
add a comment |Â
add a comment |Â
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first that $2cos^2+2sin^2=2$
– Nosrati
Sep 5 at 4:46
By the fundamental theorem of symmetric polynomials, once you've written $sin t cos t$ in terms of $a$ you can write any polynomial expression which is symmetric in $sin t$ and $cos t$ in terms of $a$.
– Micah
Sep 5 at 4:55
1
$$a^4 - 2sintcost(3sintcost + 2)=a^4 - 2dfraca^2-12(3dfraca^2-12 + 2)=colorbluedfrac-a^4+2a^2+12$$
– Nosrati
Sep 5 at 4:57