Proving $tan^2(x)-sin^2(x) = tan^2(x)sin^2(x)$ (and trig identities in general)
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
$begingroup$
Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
$endgroup$
– David H
May 8 '14 at 22:33
$begingroup$
tan^2(x)*sin^2(x)
$endgroup$
– Ila Isabelle
May 8 '14 at 22:35
add a comment |
$begingroup$
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
$endgroup$
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 Jan 4 at 22:51
Blue
49.1k870156
49.1k870156
asked May 8 '14 at 22:27
Ila IsabelleIla Isabelle
50311
50311
$begingroup$
Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
$endgroup$
– David H
May 8 '14 at 22:33
$begingroup$
tan^2(x)*sin^2(x)
$endgroup$
– Ila Isabelle
May 8 '14 at 22:35
add a comment |
$begingroup$
Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
$endgroup$
– David H
May 8 '14 at 22:33
$begingroup$
tan^2(x)*sin^2(x)
$endgroup$
– Ila Isabelle
May 8 '14 at 22:35
$begingroup$
Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
$endgroup$
– David H
May 8 '14 at 22:33
$begingroup$
Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
$endgroup$
– David H
May 8 '14 at 22:33
$begingroup$
tan^2(x)*sin^2(x)
$endgroup$
– Ila Isabelle
May 8 '14 at 22:35
$begingroup$
tan^2(x)*sin^2(x)
$endgroup$
– Ila Isabelle
May 8 '14 at 22:35
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
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
$$
sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
$$
$endgroup$
add a comment |
$begingroup$
$x^2-y^2$ 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.
$endgroup$
add a comment |
$begingroup$
$$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$
$endgroup$
add a comment |
$begingroup$
$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.
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$
sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
$$
$endgroup$
add a comment |
$begingroup$
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
$$
sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
$$
$endgroup$
add a comment |
$begingroup$
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
$$
sin^2 x( frac{1}{cos^2 x} - 1) frac{cos^2 x}{cos^2 x} = sin^2 x frac{1}{cos^2 x}(1 - cos^2 x)= tan^2 x sin^2 x.
$$
$endgroup$
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 HughesJohn Hughes
64.6k24191
64.6k24191
add a comment |
add a comment |
$begingroup$
$x^2-y^2$ 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.
$endgroup$
add a comment |
$begingroup$
$x^2-y^2$ 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.
$endgroup$
add a comment |
$begingroup$
$x^2-y^2$ 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.
$endgroup$
$x^2-y^2$ 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.sacamentobob.sacamento
2,4311919
2,4311919
add a comment |
add a comment |
$begingroup$
$$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$
$endgroup$
add a comment |
$begingroup$
$$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$
$endgroup$
add a comment |
$begingroup$
$$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$
$endgroup$
$$tan^2(x)-sin^2(x) = frac{sin^2(x)}{cos^2(x)} -frac{sin^2(x)cos^2(x)}{cos^2(x)} = frac{sin^2x(1-cos^2(x))}{cos^2(x)} = frac{sin^4(x)}{cos^2(x)}$$
edited Jan 5 at 7:29
KM101
6,0901525
6,0901525
answered Jan 4 at 22:47
M. C.M. C.
449
449
add a comment |
add a comment |
$begingroup$
$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.
$endgroup$
add a comment |
$begingroup$
$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.
$endgroup$
add a comment |
$begingroup$
$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.
$endgroup$
$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 Jan 5 at 7:25
marty cohenmarty cohen
74.4k549129
74.4k549129
add a comment |
add a comment |
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$begingroup$
Okay, so the expression $tan^2{x}-sin^2{x}$ is your starting point. What's the endpoint you're trying to get to?
$endgroup$
– David H
May 8 '14 at 22:33
$begingroup$
tan^2(x)*sin^2(x)
$endgroup$
– Ila Isabelle
May 8 '14 at 22:35