Proving $tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)$ (and trig identities in general)
The problem is $$tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)$$
I asked a question like this previously, and understood that one once someone helped me, but now I am back to not understanding with this problem. Could someone show me the steps and help me understand?
Also, I know it's a part of memorizing the trig identities and becoming familiar with them, but if anyone has any tips or something that can help me that would be great too.
algebra-precalculus trigonometry
edited yesterday
Blue
47.7k870151
asked May 8 '14 at 22:27
Ila Isabelle
50310
add a comment |
The problem is $$tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)$$
I asked a question like this previously, and understood that one once someone helped me, but now I am back to not understanding with this problem. Could someone show me the steps and help me understand?
Also, I know it's a part of memorizing the trig identities and becoming familiar with them, but if anyone has any tips or something that can help me that would be great too.
algebra-precalculus trigonometry
edited yesterday
Blue
47.7k870151
asked May 8 '14 at 22:27
Ila Isabelle
50310
Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
– David H
May 8 '14 at 22:33
tan^2(x)*sin^2(x)
– Ila Isabelle
May 8 '14 at 22:35
add a comment |
The problem is $$tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)$$
I asked a question like this previously, and understood that one once someone helped me, but now I am back to not understanding with this problem. Could someone show me the steps and help me understand?
Also, I know it's a part of memorizing the trig identities and becoming familiar with them, but if anyone has any tips or something that can help me that would be great too.
algebra-precalculus trigonometry
edited yesterday
Blue
47.7k870151
asked May 8 '14 at 22:27
Ila Isabelle
50310
The problem is $$tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)$$
I asked a question like this previously, and understood that one once someone helped me, but now I am back to not understanding with this problem. Could someone show me the steps and help me understand?
Also, I know it's a part of memorizing the trig identities and becoming familiar with them, but if anyone has any tips or something that can help me that would be great too.
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited yesterday
Blue
47.7k870151
asked May 8 '14 at 22:27
Ila Isabelle
50310
edited yesterday
Blue
47.7k870151
asked May 8 '14 at 22:27
Ila Isabelle
50310
edited yesterday
Blue
47.7k870151
edited yesterday
Blue
47.7k870151
edited yesterday
Blue
47.7k870151
47.7k870151
asked May 8 '14 at 22:27
Ila Isabelle
50310
asked May 8 '14 at 22:27
Ila Isabelle
50310
asked May 8 '14 at 22:27
Ila Isabelle
50310
50310
Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
– David H
May 8 '14 at 22:33
tan^2(x)*sin^2(x)
– Ila Isabelle
May 8 '14 at 22:35
add a comment |
Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
– David H
May 8 '14 at 22:33
tan^2(x)*sin^2(x)
– Ila Isabelle
May 8 '14 at 22:35
Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
– David H
May 8 '14 at 22:33
Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
– David H
May 8 '14 at 22:33
tan^2(x)*sin^2(x)
– Ila Isabelle
May 8 '14 at 22:35
tan^2(x)*sin^2(x)
– Ila Isabelle
May 8 '14 at 22:35
add a comment |
4 Answers
4
active
oldest
votes
General hint: convert everything to sines and cosines, and then there are only a couple of identities to mess with. In this case
$$
tan^2 x - sin^2 x = sin^2 x( frac{1}{cos^2 x} - 1)
$$
Multiplying top and bottom by $cos^2 x$, you get
$$
sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
$$
answered May 8 '14 at 22:36
John Hughes
62.5k24090
add a comment |
$$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$
edited yesterday
KM101
5,4911423
answered yesterday
M. C.
213
New contributor
M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
$x^2-y^2$ is always $(x-y)(x+y)$, no matter how complicated $x$ and $y$ may be. So use that, then see what happens when you re-write $tan x$ in terms of $sin x$ and $cos x$. Factor out what you can. That should get you on the way.
answered May 8 '14 at 22:41
bob.sacamento
2,4141819
add a comment |
$tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)
$
Typically,
the only identity that you need is
$s^2+c^2 = 1$,
writing $s$ for $sin$
and $c$ for $cos$.
Writing
$t$ for $tan$,
the difference of the sides is
$begin{array}\
t^2-s^2-t^2s^2
&=dfrac{s^2}{c^2}-s^2-dfrac{s^2}{c^2}s^2\
&=dfrac{s^2-s^2c^2-s^4}{c^2}\
&=dfrac{s^2(1-c^2-s^2)}{c^2}\
&=dfrac{s^2(1-(c^2+s^2))}{c^2}\
&=0
qquadtext{since }c^2+s^2=1\
end{array}
$
The only place where
there might be a problem
is when
$c = 0$;
there
$t^2 = infty$
(abusing notation a little)
and $s^2=1$
so the identity still holds,
in a limitly way.
answered yesterday
marty cohen
72.7k549128
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
General hint: convert everything to sines and cosines, and then there are only a couple of identities to mess with. In this case
$$
tan^2 x - sin^2 x = sin^2 x( frac{1}{cos^2 x} - 1)
$$
Multiplying top and bottom by $cos^2 x$, you get
$$
sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
$$
answered May 8 '14 at 22:36
John Hughes
62.5k24090
add a comment |
General hint: convert everything to sines and cosines, and then there are only a couple of identities to mess with. In this case
$$
tan^2 x - sin^2 x = sin^2 x( frac{1}{cos^2 x} - 1)
$$
Multiplying top and bottom by $cos^2 x$, you get
$$
sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
$$
answered May 8 '14 at 22:36
John Hughes
62.5k24090
add a comment |
General hint: convert everything to sines and cosines, and then there are only a couple of identities to mess with. In this case
$$
tan^2 x - sin^2 x = sin^2 x( frac{1}{cos^2 x} - 1)
$$
Multiplying top and bottom by $cos^2 x$, you get
$$
sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
$$
answered May 8 '14 at 22:36
John Hughes
62.5k24090
General hint: convert everything to sines and cosines, and then there are only a couple of identities to mess with. In this case
$$
tan^2 x - sin^2 x = sin^2 x( frac{1}{cos^2 x} - 1)
$$
Multiplying top and bottom by $cos^2 x$, you get
$$
sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
$$
answered May 8 '14 at 22:36
John Hughes
62.5k24090
answered May 8 '14 at 22:36
John Hughes
62.5k24090
answered May 8 '14 at 22:36
John Hughes
62.5k24090
answered May 8 '14 at 22:36
John Hughes
62.5k24090
62.5k24090
add a comment |
add a comment |
$$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$
edited yesterday
KM101
5,4911423
answered yesterday
M. C.
213
New contributor
M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
$$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$
edited yesterday
KM101
5,4911423
answered yesterday
M. C.
213
New contributor
M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
$$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$
edited yesterday
KM101
5,4911423
answered yesterday
M. C.
213
New contributor
M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$
edited yesterday
KM101
5,4911423
answered yesterday
M. C.
213
New contributor
M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
KM101
5,4911423
edited yesterday
KM101
5,4911423
edited yesterday
KM101
5,4911423
5,4911423
answered yesterday
M. C.
213
New contributor
M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
M. C.
213
answered yesterday
M. C.
213
213
New contributor
M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
M. C. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$x^2-y^2$ is always $(x-y)(x+y)$, no matter how complicated $x$ and $y$ may be. So use that, then see what happens when you re-write $tan x$ in terms of $sin x$ and $cos x$. Factor out what you can. That should get you on the way.
answered May 8 '14 at 22:41
bob.sacamento
2,4141819
add a comment |
$x^2-y^2$ is always $(x-y)(x+y)$, no matter how complicated $x$ and $y$ may be. So use that, then see what happens when you re-write $tan x$ in terms of $sin x$ and $cos x$. Factor out what you can. That should get you on the way.
answered May 8 '14 at 22:41
bob.sacamento
2,4141819
add a comment |
$x^2-y^2$ is always $(x-y)(x+y)$, no matter how complicated $x$ and $y$ may be. So use that, then see what happens when you re-write $tan x$ in terms of $sin x$ and $cos x$. Factor out what you can. That should get you on the way.
answered May 8 '14 at 22:41
bob.sacamento
2,4141819
$x^2-y^2$ is always $(x-y)(x+y)$, no matter how complicated $x$ and $y$ may be. So use that, then see what happens when you re-write $tan x$ in terms of $sin x$ and $cos x$. Factor out what you can. That should get you on the way.
answered May 8 '14 at 22:41
bob.sacamento
2,4141819
answered May 8 '14 at 22:41
bob.sacamento
2,4141819
answered May 8 '14 at 22:41
bob.sacamento
2,4141819
answered May 8 '14 at 22:41
bob.sacamento
2,4141819
2,4141819
add a comment |
add a comment |
$tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)
$
Typically,
the only identity that you need is
$s^2+c^2 = 1$,
writing $s$ for $sin$
and $c$ for $cos$.
Writing
$t$ for $tan$,
the difference of the sides is
$begin{array}\
t^2-s^2-t^2s^2
&=dfrac{s^2}{c^2}-s^2-dfrac{s^2}{c^2}s^2\
&=dfrac{s^2-s^2c^2-s^4}{c^2}\
&=dfrac{s^2(1-c^2-s^2)}{c^2}\
&=dfrac{s^2(1-(c^2+s^2))}{c^2}\
&=0
qquadtext{since }c^2+s^2=1\
end{array}
$
The only place where
there might be a problem
is when
$c = 0$;
there
$t^2 = infty$
(abusing notation a little)
and $s^2=1$
so the identity still holds,
in a limitly way.
answered yesterday
marty cohen
72.7k549128
add a comment |
$tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)
$
Typically,
the only identity that you need is
$s^2+c^2 = 1$,
writing $s$ for $sin$
and $c$ for $cos$.
Writing
$t$ for $tan$,
the difference of the sides is
$begin{array}\
t^2-s^2-t^2s^2
&=dfrac{s^2}{c^2}-s^2-dfrac{s^2}{c^2}s^2\
&=dfrac{s^2-s^2c^2-s^4}{c^2}\
&=dfrac{s^2(1-c^2-s^2)}{c^2}\
&=dfrac{s^2(1-(c^2+s^2))}{c^2}\
&=0
qquadtext{since }c^2+s^2=1\
end{array}
$
The only place where
there might be a problem
is when
$c = 0$;
there
$t^2 = infty$
(abusing notation a little)
and $s^2=1$
so the identity still holds,
in a limitly way.
answered yesterday
marty cohen
72.7k549128
add a comment |
$tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)
$
Typically,
the only identity that you need is
$s^2+c^2 = 1$,
writing $s$ for $sin$
and $c$ for $cos$.
Writing
$t$ for $tan$,
the difference of the sides is
$begin{array}\
t^2-s^2-t^2s^2
&=dfrac{s^2}{c^2}-s^2-dfrac{s^2}{c^2}s^2\
&=dfrac{s^2-s^2c^2-s^4}{c^2}\
&=dfrac{s^2(1-c^2-s^2)}{c^2}\
&=dfrac{s^2(1-(c^2+s^2))}{c^2}\
&=0
qquadtext{since }c^2+s^2=1\
end{array}
$
The only place where
there might be a problem
is when
$c = 0$;
there
$t^2 = infty$
(abusing notation a little)
and $s^2=1$
so the identity still holds,
in a limitly way.
answered yesterday
marty cohen
72.7k549128
$tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)
$
Typically,
the only identity that you need is
$s^2+c^2 = 1$,
writing $s$ for $sin$
and $c$ for $cos$.
Writing
$t$ for $tan$,
the difference of the sides is
$begin{array}\
t^2-s^2-t^2s^2
&=dfrac{s^2}{c^2}-s^2-dfrac{s^2}{c^2}s^2\
&=dfrac{s^2-s^2c^2-s^4}{c^2}\
&=dfrac{s^2(1-c^2-s^2)}{c^2}\
&=dfrac{s^2(1-(c^2+s^2))}{c^2}\
&=0
qquadtext{since }c^2+s^2=1\
end{array}
$
The only place where
there might be a problem
is when
$c = 0$;
there
$t^2 = infty$
(abusing notation a little)
and $s^2=1$
so the identity still holds,
in a limitly way.
answered yesterday
marty cohen
72.7k549128
answered yesterday
marty cohen
72.7k549128
answered yesterday
marty cohen
72.7k549128
answered yesterday
marty cohen
72.7k549128
72.7k549128
add a comment |
add a comment |
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Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
– David H
May 8 '14 at 22:33
tan^2(x)*sin^2(x)
– Ila Isabelle
May 8 '14 at 22:35