Mixing
Simplifying $(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x$
$begingroup$
$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x =$
Right Answer: $0$, but I could not solve this question. Help me please.
trigonometry weierstrass-factorization
edited Feb 3 at 12:52
N. F. Taussig
45.4k103358
asked Feb 3 at 12:47
GeoerkeamGeoerkeam
33
$endgroup$
add a comment |
$begingroup$
$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x =$
Right Answer: $0$, but I could not solve this question. Help me please.
trigonometry weierstrass-factorization
edited Feb 3 at 12:52
N. F. Taussig
45.4k103358
asked Feb 3 at 12:47
GeoerkeamGeoerkeam
33
$endgroup$
1
$begingroup$
Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 3 at 12:55
$begingroup$
$cos^2x=1-sin^2x$
$endgroup$
– Lord Shark the Unknown
Feb 3 at 12:56
$begingroup$
$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x = sin^4x (sin^2x - 1) + cos^4x (cos^2x - 1) + sin^2x cos^2x. $$ Can you take it from here?
$endgroup$
– HerrWarum
Feb 3 at 12:57
add a comment |
$begingroup$
$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x =$
Right Answer: $0$, but I could not solve this question. Help me please.
trigonometry weierstrass-factorization
edited Feb 3 at 12:52
N. F. Taussig
45.4k103358
asked Feb 3 at 12:47
GeoerkeamGeoerkeam
33
$endgroup$
$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x =$
Right Answer: $0$, but I could not solve this question. Help me please.
trigonometry weierstrass-factorization
trigonometry weierstrass-factorization
edited Feb 3 at 12:52
N. F. Taussig
45.4k103358
asked Feb 3 at 12:47
GeoerkeamGeoerkeam
33
edited Feb 3 at 12:52
N. F. Taussig
45.4k103358
asked Feb 3 at 12:47
GeoerkeamGeoerkeam
33
edited Feb 3 at 12:52
N. F. Taussig
45.4k103358
edited Feb 3 at 12:52
N. F. Taussig
45.4k103358
edited Feb 3 at 12:52
N. F. Taussig
45.4k103358
45.4k103358
asked Feb 3 at 12:47
GeoerkeamGeoerkeam
33
asked Feb 3 at 12:47
GeoerkeamGeoerkeam
33
asked Feb 3 at 12:47
GeoerkeamGeoerkeam
33
33
1
$begingroup$
Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 3 at 12:55
$begingroup$
$cos^2x=1-sin^2x$
$endgroup$
– Lord Shark the Unknown
Feb 3 at 12:56
$begingroup$
$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x = sin^4x (sin^2x - 1) + cos^4x (cos^2x - 1) + sin^2x cos^2x. $$ Can you take it from here?
$endgroup$
– HerrWarum
Feb 3 at 12:57
add a comment |
1
$begingroup$
Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 3 at 12:55
$begingroup$
$cos^2x=1-sin^2x$
$endgroup$
– Lord Shark the Unknown
Feb 3 at 12:56
$begingroup$
$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x = sin^4x (sin^2x - 1) + cos^4x (cos^2x - 1) + sin^2x cos^2x. $$ Can you take it from here?
$endgroup$
– HerrWarum
Feb 3 at 12:57
1
1
$begingroup$
Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 3 at 12:55
$begingroup$
Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 3 at 12:55
$begingroup$
$cos^2x=1-sin^2x$
$endgroup$
– Lord Shark the Unknown
Feb 3 at 12:56
$begingroup$
$cos^2x=1-sin^2x$
$endgroup$
– Lord Shark the Unknown
Feb 3 at 12:56
$begingroup$
$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x = sin^4x (sin^2x - 1) + cos^4x (cos^2x - 1) + sin^2x cos^2x. $$ Can you take it from here?
$endgroup$
– HerrWarum
Feb 3 at 12:57
$begingroup$
$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x = sin^4x (sin^2x - 1) + cos^4x (cos^2x - 1) + sin^2x cos^2x. $$ Can you take it from here?
$endgroup$
– HerrWarum
Feb 3 at 12:57
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x $$
$$= sin^6x + cos^6x - sin^4x -cos^4x+sin^2xcos^2x$$
$$= cos^6x(tan^6x + 1 - tan^4xsec^2{x} -sec^2x+tan^2xsec^2x)$$
$$= cos^6x(tan^6x + 1 - tan^4x(1+tan^2x) -(1+tan^2x)+tan^2x(1+tan^2x))$$
$$= cos^6x(tan^6x + 1 - tan^4x-tan^6x -1-tan^2x+tan^2x+tan^4x)$$
$$= cos^6x(0)$$
$$= 0$$
answered Feb 3 at 12:58
Peter ForemanPeter Foreman
7,5341320
$endgroup$
add a comment |
$begingroup$
With $c:=cos^2 x,,s:=sin^2 x$, it suffices to note $$c^3+s^3-c^2-s^2+cs=(c+s-1)(c^2-cs+s^2).$$
answered Feb 3 at 13:11
J.G.J.G.
33.5k23252
$endgroup$
$begingroup$
Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
$endgroup$
– Elie Louis
Feb 4 at 12:51
$begingroup$
@ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
$endgroup$
– J.G.
Feb 4 at 13:00
add a comment |
$begingroup$
hint: $sin^2 x + cos^2 x = 1$; if you square that, you get
$$
sin^4 x + 2sin^2 x cos^2 x + cos^4 x = 1,
$$
which you can turn into
$$
sin^4 x + cos^4 x = 1 - 2sin^2 x cos^2 x
$$
and that lets you get rid of the 4th powers in the left hand side.
Can you run with that idea?
answered Feb 3 at 12:57
John HughesJohn Hughes
65.5k24293
$endgroup$
add a comment |
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3 Answers
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active
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3 Answers
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$begingroup$
$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x $$
$$= sin^6x + cos^6x - sin^4x -cos^4x+sin^2xcos^2x$$
$$= cos^6x(tan^6x + 1 - tan^4xsec^2{x} -sec^2x+tan^2xsec^2x)$$
$$= cos^6x(tan^6x + 1 - tan^4x(1+tan^2x) -(1+tan^2x)+tan^2x(1+tan^2x))$$
$$= cos^6x(tan^6x + 1 - tan^4x-tan^6x -1-tan^2x+tan^2x+tan^4x)$$
$$= cos^6x(0)$$
$$= 0$$
answered Feb 3 at 12:58
Peter ForemanPeter Foreman
7,5341320
$endgroup$
add a comment |
$begingroup$
$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x $$
$$= sin^6x + cos^6x - sin^4x -cos^4x+sin^2xcos^2x$$
$$= cos^6x(tan^6x + 1 - tan^4xsec^2{x} -sec^2x+tan^2xsec^2x)$$
$$= cos^6x(tan^6x + 1 - tan^4x(1+tan^2x) -(1+tan^2x)+tan^2x(1+tan^2x))$$
$$= cos^6x(tan^6x + 1 - tan^4x-tan^6x -1-tan^2x+tan^2x+tan^4x)$$
$$= cos^6x(0)$$
$$= 0$$
answered Feb 3 at 12:58
Peter ForemanPeter Foreman
7,5341320
$endgroup$
add a comment |
$begingroup$
$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x $$
$$= sin^6x + cos^6x - sin^4x -cos^4x+sin^2xcos^2x$$
$$= cos^6x(tan^6x + 1 - tan^4xsec^2{x} -sec^2x+tan^2xsec^2x)$$
$$= cos^6x(tan^6x + 1 - tan^4x(1+tan^2x) -(1+tan^2x)+tan^2x(1+tan^2x))$$
$$= cos^6x(tan^6x + 1 - tan^4x-tan^6x -1-tan^2x+tan^2x+tan^4x)$$
$$= cos^6x(0)$$
$$= 0$$
answered Feb 3 at 12:58
Peter ForemanPeter Foreman
7,5341320
$endgroup$
$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x $$
$$= sin^6x + cos^6x - sin^4x -cos^4x+sin^2xcos^2x$$
$$= cos^6x(tan^6x + 1 - tan^4xsec^2{x} -sec^2x+tan^2xsec^2x)$$
$$= cos^6x(tan^6x + 1 - tan^4x(1+tan^2x) -(1+tan^2x)+tan^2x(1+tan^2x))$$
$$= cos^6x(tan^6x + 1 - tan^4x-tan^6x -1-tan^2x+tan^2x+tan^4x)$$
$$= cos^6x(0)$$
$$= 0$$
answered Feb 3 at 12:58
Peter ForemanPeter Foreman
7,5341320
answered Feb 3 at 12:58
Peter ForemanPeter Foreman
7,5341320
answered Feb 3 at 12:58
Peter ForemanPeter Foreman
7,5341320
answered Feb 3 at 12:58
Peter ForemanPeter Foreman
7,5341320
7,5341320
add a comment |
add a comment |
$begingroup$
With $c:=cos^2 x,,s:=sin^2 x$, it suffices to note $$c^3+s^3-c^2-s^2+cs=(c+s-1)(c^2-cs+s^2).$$
answered Feb 3 at 13:11
J.G.J.G.
33.5k23252
$endgroup$
$begingroup$
Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
$endgroup$
– Elie Louis
Feb 4 at 12:51
$begingroup$
@ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
$endgroup$
– J.G.
Feb 4 at 13:00
add a comment |
$begingroup$
With $c:=cos^2 x,,s:=sin^2 x$, it suffices to note $$c^3+s^3-c^2-s^2+cs=(c+s-1)(c^2-cs+s^2).$$
answered Feb 3 at 13:11
J.G.J.G.
33.5k23252
$endgroup$
$begingroup$
Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
$endgroup$
– Elie Louis
Feb 4 at 12:51
$begingroup$
@ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
$endgroup$
– J.G.
Feb 4 at 13:00
add a comment |
$begingroup$
With $c:=cos^2 x,,s:=sin^2 x$, it suffices to note $$c^3+s^3-c^2-s^2+cs=(c+s-1)(c^2-cs+s^2).$$
answered Feb 3 at 13:11
J.G.J.G.
33.5k23252
$endgroup$
With $c:=cos^2 x,,s:=sin^2 x$, it suffices to note $$c^3+s^3-c^2-s^2+cs=(c+s-1)(c^2-cs+s^2).$$
answered Feb 3 at 13:11
J.G.J.G.
33.5k23252
answered Feb 3 at 13:11
J.G.J.G.
33.5k23252
answered Feb 3 at 13:11
J.G.J.G.
33.5k23252
answered Feb 3 at 13:11
J.G.J.G.
33.5k23252
33.5k23252
$begingroup$
Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
$endgroup$
– Elie Louis
Feb 4 at 12:51
$begingroup$
@ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
$endgroup$
– J.G.
Feb 4 at 13:00
add a comment |
$begingroup$
Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
$endgroup$
– Elie Louis
Feb 4 at 12:51
$begingroup$
@ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
$endgroup$
– J.G.
Feb 4 at 13:00
$begingroup$
Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
$endgroup$
– Elie Louis
Feb 4 at 12:51
$begingroup$
Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
$endgroup$
– Elie Louis
Feb 4 at 12:51
$begingroup$
@ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
$endgroup$
– J.G.
Feb 4 at 13:00
$begingroup$
@ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
$endgroup$
– J.G.
Feb 4 at 13:00
add a comment |
$begingroup$
hint: $sin^2 x + cos^2 x = 1$; if you square that, you get
$$
sin^4 x + 2sin^2 x cos^2 x + cos^4 x = 1,
$$
which you can turn into
$$
sin^4 x + cos^4 x = 1 - 2sin^2 x cos^2 x
$$
and that lets you get rid of the 4th powers in the left hand side.
Can you run with that idea?
answered Feb 3 at 12:57
John HughesJohn Hughes
65.5k24293
$endgroup$
add a comment |
$begingroup$
hint: $sin^2 x + cos^2 x = 1$; if you square that, you get
$$
sin^4 x + 2sin^2 x cos^2 x + cos^4 x = 1,
$$
which you can turn into
$$
sin^4 x + cos^4 x = 1 - 2sin^2 x cos^2 x
$$
and that lets you get rid of the 4th powers in the left hand side.
Can you run with that idea?
answered Feb 3 at 12:57
John HughesJohn Hughes
65.5k24293
$endgroup$
add a comment |
$begingroup$
hint: $sin^2 x + cos^2 x = 1$; if you square that, you get
$$
sin^4 x + 2sin^2 x cos^2 x + cos^4 x = 1,
$$
which you can turn into
$$
sin^4 x + cos^4 x = 1 - 2sin^2 x cos^2 x
$$
and that lets you get rid of the 4th powers in the left hand side.
Can you run with that idea?
answered Feb 3 at 12:57
John HughesJohn Hughes
65.5k24293
$endgroup$
hint: $sin^2 x + cos^2 x = 1$; if you square that, you get
$$
sin^4 x + 2sin^2 x cos^2 x + cos^4 x = 1,
$$
which you can turn into
$$
sin^4 x + cos^4 x = 1 - 2sin^2 x cos^2 x
$$
and that lets you get rid of the 4th powers in the left hand side.
Can you run with that idea?
answered Feb 3 at 12:57
John HughesJohn Hughes
65.5k24293
answered Feb 3 at 12:57
John HughesJohn Hughes
65.5k24293
answered Feb 3 at 12:57
John HughesJohn Hughes
65.5k24293
answered Feb 3 at 12:57
John HughesJohn Hughes
65.5k24293
65.5k24293
add a comment |
add a comment |
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$begingroup$
Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 3 at 12:55
$begingroup$
$cos^2x=1-sin^2x$
$endgroup$
– Lord Shark the Unknown
Feb 3 at 12:56
$begingroup$
$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x = sin^4x (sin^2x - 1) + cos^4x (cos^2x - 1) + sin^2x cos^2x. $$ Can you take it from here?
$endgroup$
– HerrWarum
Feb 3 at 12:57