Mixing
Compare without a calculator: $a=cos20^o$, $b=sin20^o$, $c=tan20^0$
$begingroup$
$a=cos20^o$
$b=sin20^o$
$c=tan20^o$
Compare $a,b$ and $c$
I could figure out that $a=cos20^o=sin70^o$, then $b lt a$.
I can't put $tan20^0$ anywhere in the inequality. Usually, in these type of problems, $tan{x^o}$ would be greater than $45^o$ , so that I can compare it with the others, knowing that it would be greater than $1$.
How do I compare $a$ and $c$?
algebra-precalculus trigonometry
asked Jan 25 at 19:46
Eldar RahimliEldar Rahimli
36310
$endgroup$
add a comment |
$begingroup$
$a=cos20^o$
$b=sin20^o$
$c=tan20^o$
Compare $a,b$ and $c$
I could figure out that $a=cos20^o=sin70^o$, then $b lt a$.
I can't put $tan20^0$ anywhere in the inequality. Usually, in these type of problems, $tan{x^o}$ would be greater than $45^o$ , so that I can compare it with the others, knowing that it would be greater than $1$.
How do I compare $a$ and $c$?
algebra-precalculus trigonometry
asked Jan 25 at 19:46
Eldar RahimliEldar Rahimli
36310
$endgroup$
add a comment |
$begingroup$
$a=cos20^o$
$b=sin20^o$
$c=tan20^o$
Compare $a,b$ and $c$
I could figure out that $a=cos20^o=sin70^o$, then $b lt a$.
I can't put $tan20^0$ anywhere in the inequality. Usually, in these type of problems, $tan{x^o}$ would be greater than $45^o$ , so that I can compare it with the others, knowing that it would be greater than $1$.
How do I compare $a$ and $c$?
algebra-precalculus trigonometry
asked Jan 25 at 19:46
Eldar RahimliEldar Rahimli
36310
$endgroup$
$a=cos20^o$
$b=sin20^o$
$c=tan20^o$
Compare $a,b$ and $c$
I could figure out that $a=cos20^o=sin70^o$, then $b lt a$.
I can't put $tan20^0$ anywhere in the inequality. Usually, in these type of problems, $tan{x^o}$ would be greater than $45^o$ , so that I can compare it with the others, knowing that it would be greater than $1$.
How do I compare $a$ and $c$?
algebra-precalculus trigonometry
algebra-precalculus trigonometry
asked Jan 25 at 19:46
Eldar RahimliEldar Rahimli
36310
asked Jan 25 at 19:46
Eldar RahimliEldar Rahimli
36310
asked Jan 25 at 19:46
Eldar RahimliEldar Rahimli
36310
asked Jan 25 at 19:46
Eldar RahimliEldar Rahimli
36310
asked Jan 25 at 19:46
Eldar RahimliEldar Rahimli
36310
36310
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$cos 30^circ$ and $tan 30^circ$ can be memorized without using a calculator (see https://math.stackexchange.com/a/1554126/645472).
$cos 30^circ = frac{sqrt{3}}{2}$ is larger than $tan 30^circ = frac{1}{sqrt{3}}$ (you can see this by multiplying both with $sqrt{3}$). Moreover, $cos theta$ is decreasing for $theta$ going from $0^circ$ to $90^circ$, whereas $tan theta$ is increasing on that interval, so
$$
a = cos 20^circ > cos 30^circ > tan 30^circ > tan 20^circ = c
$$
Combining this with what you already found ($b<a$) and the answer of @Anadactothe ($c>b$), you get
$$
a > c > b
$$
answered Mar 2 at 0:32
Koen TielsKoen Tiels
1706
$endgroup$
add a comment |
$begingroup$
Use the appropriate half-angle identity to render $cos^2 20°=(1+cos 40°)/2=(1+sin 50°)/2$. Then since $1>sin 50°>sin 20°$, conclude that $cos^2 20° > sin 20°$ and therefore $cos 20° > sin 20°/cos 20°$.
answered Mar 2 at 1:50
Oscar LanziOscar Lanzi
13.2k12136
$endgroup$
add a comment |
$begingroup$
$c=frac{b}{a}>b$, while $b<sin 30^circ=frac{1}{2}implies frac{c}{a}=frac{b}{1-b^2}<frac{2}{3}$. This uses the fact that $frac{x}{1-x^2}=frac{1}{2}(frac{1}{1-x}-frac{1}{1+x})$ increases on $[0,,1]$.
answered Jan 25 at 19:55
J.G.J.G.
30.9k23149
$endgroup$
$begingroup$
Can you please elaborate on your answer? I didn't understand how $frac{b}{1-b^2} lt frac{2}{3}$
$endgroup$
– Eldar Rahimli
Jan 25 at 20:06
$begingroup$
@EldarRahimli See edit.
$endgroup$
– J.G.
Jan 25 at 20:09
add a comment |
$begingroup$
For this question, all you need is to know is the quotient identity of tangent, that is $tan(theta) =frac{sin(theta)}{cos(theta)}$
We know that $cos(0)$ is 1 and is decreasing on the interval $(0,pi)$, so $cos(20) < 1$
Considering that $tan(20) =frac {sin(20)}{cos(20)}$, and $cos(20) < 1$, it follows that $tan(20) > sin(20)$ because dividing by a number less than one increases the absolute value of an expression, and looking at the unit circle, sine, cosine, and tangent are positive in quadrant 1.
Therefore, cos(20) > tan(20) > sin(20)
edited Jan 25 at 20:11
H Huang
397111
answered Jan 25 at 19:56
AnadactotheAnadactothe
12411
$endgroup$
1
$begingroup$
How do you know whether $cos(20^circ)$ is greater or less than $tan(20^circ)$? You haven't mentioned this part. Further, this is precisely the part that the OP is asking about.
$endgroup$
– JMoravitz
Jan 25 at 19:59
$begingroup$
ah, it appears i can't read properly. let me edit my post.
$endgroup$
– Anadactothe
Jan 25 at 20:03
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
$begingroup$
$cos 30^circ$ and $tan 30^circ$ can be memorized without using a calculator (see https://math.stackexchange.com/a/1554126/645472).
$cos 30^circ = frac{sqrt{3}}{2}$ is larger than $tan 30^circ = frac{1}{sqrt{3}}$ (you can see this by multiplying both with $sqrt{3}$). Moreover, $cos theta$ is decreasing for $theta$ going from $0^circ$ to $90^circ$, whereas $tan theta$ is increasing on that interval, so
$$
a = cos 20^circ > cos 30^circ > tan 30^circ > tan 20^circ = c
$$
Combining this with what you already found ($b<a$) and the answer of @Anadactothe ($c>b$), you get
$$
a > c > b
$$
answered Mar 2 at 0:32
Koen TielsKoen Tiels
1706
$endgroup$
add a comment |
$begingroup$
$cos 30^circ$ and $tan 30^circ$ can be memorized without using a calculator (see https://math.stackexchange.com/a/1554126/645472).
$cos 30^circ = frac{sqrt{3}}{2}$ is larger than $tan 30^circ = frac{1}{sqrt{3}}$ (you can see this by multiplying both with $sqrt{3}$). Moreover, $cos theta$ is decreasing for $theta$ going from $0^circ$ to $90^circ$, whereas $tan theta$ is increasing on that interval, so
$$
a = cos 20^circ > cos 30^circ > tan 30^circ > tan 20^circ = c
$$
Combining this with what you already found ($b<a$) and the answer of @Anadactothe ($c>b$), you get
$$
a > c > b
$$
answered Mar 2 at 0:32
Koen TielsKoen Tiels
1706
$endgroup$
add a comment |
$begingroup$
$cos 30^circ$ and $tan 30^circ$ can be memorized without using a calculator (see https://math.stackexchange.com/a/1554126/645472).
$cos 30^circ = frac{sqrt{3}}{2}$ is larger than $tan 30^circ = frac{1}{sqrt{3}}$ (you can see this by multiplying both with $sqrt{3}$). Moreover, $cos theta$ is decreasing for $theta$ going from $0^circ$ to $90^circ$, whereas $tan theta$ is increasing on that interval, so
$$
a = cos 20^circ > cos 30^circ > tan 30^circ > tan 20^circ = c
$$
Combining this with what you already found ($b<a$) and the answer of @Anadactothe ($c>b$), you get
$$
a > c > b
$$
answered Mar 2 at 0:32
Koen TielsKoen Tiels
1706
$endgroup$
$cos 30^circ$ and $tan 30^circ$ can be memorized without using a calculator (see https://math.stackexchange.com/a/1554126/645472).
$cos 30^circ = frac{sqrt{3}}{2}$ is larger than $tan 30^circ = frac{1}{sqrt{3}}$ (you can see this by multiplying both with $sqrt{3}$). Moreover, $cos theta$ is decreasing for $theta$ going from $0^circ$ to $90^circ$, whereas $tan theta$ is increasing on that interval, so
$$
a = cos 20^circ > cos 30^circ > tan 30^circ > tan 20^circ = c
$$
Combining this with what you already found ($b<a$) and the answer of @Anadactothe ($c>b$), you get
$$
a > c > b
$$
answered Mar 2 at 0:32
Koen TielsKoen Tiels
1706
answered Mar 2 at 0:32
Koen TielsKoen Tiels
1706
answered Mar 2 at 0:32
Koen TielsKoen Tiels
1706
answered Mar 2 at 0:32
Koen TielsKoen Tiels
1706
1706
add a comment |
add a comment |
$begingroup$
Use the appropriate half-angle identity to render $cos^2 20°=(1+cos 40°)/2=(1+sin 50°)/2$. Then since $1>sin 50°>sin 20°$, conclude that $cos^2 20° > sin 20°$ and therefore $cos 20° > sin 20°/cos 20°$.
answered Mar 2 at 1:50
Oscar LanziOscar Lanzi
13.2k12136
$endgroup$
add a comment |
$begingroup$
Use the appropriate half-angle identity to render $cos^2 20°=(1+cos 40°)/2=(1+sin 50°)/2$. Then since $1>sin 50°>sin 20°$, conclude that $cos^2 20° > sin 20°$ and therefore $cos 20° > sin 20°/cos 20°$.
answered Mar 2 at 1:50
Oscar LanziOscar Lanzi
13.2k12136
$endgroup$
add a comment |
$begingroup$
Use the appropriate half-angle identity to render $cos^2 20°=(1+cos 40°)/2=(1+sin 50°)/2$. Then since $1>sin 50°>sin 20°$, conclude that $cos^2 20° > sin 20°$ and therefore $cos 20° > sin 20°/cos 20°$.
answered Mar 2 at 1:50
Oscar LanziOscar Lanzi
13.2k12136
$endgroup$
Use the appropriate half-angle identity to render $cos^2 20°=(1+cos 40°)/2=(1+sin 50°)/2$. Then since $1>sin 50°>sin 20°$, conclude that $cos^2 20° > sin 20°$ and therefore $cos 20° > sin 20°/cos 20°$.
answered Mar 2 at 1:50
Oscar LanziOscar Lanzi
13.2k12136
answered Mar 2 at 1:50
Oscar LanziOscar Lanzi
13.2k12136
answered Mar 2 at 1:50
Oscar LanziOscar Lanzi
13.2k12136
answered Mar 2 at 1:50
Oscar LanziOscar Lanzi
13.2k12136
13.2k12136
add a comment |
add a comment |
$begingroup$
$c=frac{b}{a}>b$, while $b<sin 30^circ=frac{1}{2}implies frac{c}{a}=frac{b}{1-b^2}<frac{2}{3}$. This uses the fact that $frac{x}{1-x^2}=frac{1}{2}(frac{1}{1-x}-frac{1}{1+x})$ increases on $[0,,1]$.
answered Jan 25 at 19:55
J.G.J.G.
30.9k23149
$endgroup$
$begingroup$
Can you please elaborate on your answer? I didn't understand how $frac{b}{1-b^2} lt frac{2}{3}$
$endgroup$
– Eldar Rahimli
Jan 25 at 20:06
$begingroup$
@EldarRahimli See edit.
$endgroup$
– J.G.
Jan 25 at 20:09
add a comment |
$begingroup$
$c=frac{b}{a}>b$, while $b<sin 30^circ=frac{1}{2}implies frac{c}{a}=frac{b}{1-b^2}<frac{2}{3}$. This uses the fact that $frac{x}{1-x^2}=frac{1}{2}(frac{1}{1-x}-frac{1}{1+x})$ increases on $[0,,1]$.
answered Jan 25 at 19:55
J.G.J.G.
30.9k23149
$endgroup$
$begingroup$
Can you please elaborate on your answer? I didn't understand how $frac{b}{1-b^2} lt frac{2}{3}$
$endgroup$
– Eldar Rahimli
Jan 25 at 20:06
$begingroup$
@EldarRahimli See edit.
$endgroup$
– J.G.
Jan 25 at 20:09
add a comment |
$begingroup$
$c=frac{b}{a}>b$, while $b<sin 30^circ=frac{1}{2}implies frac{c}{a}=frac{b}{1-b^2}<frac{2}{3}$. This uses the fact that $frac{x}{1-x^2}=frac{1}{2}(frac{1}{1-x}-frac{1}{1+x})$ increases on $[0,,1]$.
answered Jan 25 at 19:55
J.G.J.G.
30.9k23149
$endgroup$
$c=frac{b}{a}>b$, while $b<sin 30^circ=frac{1}{2}implies frac{c}{a}=frac{b}{1-b^2}<frac{2}{3}$. This uses the fact that $frac{x}{1-x^2}=frac{1}{2}(frac{1}{1-x}-frac{1}{1+x})$ increases on $[0,,1]$.
answered Jan 25 at 19:55
J.G.J.G.
30.9k23149
edited Jan 25 at 20:09
answered Jan 25 at 19:55
J.G.J.G.
30.9k23149
answered Jan 25 at 19:55
J.G.J.G.
30.9k23149
answered Jan 25 at 19:55
J.G.J.G.
30.9k23149
30.9k23149
$begingroup$
Can you please elaborate on your answer? I didn't understand how $frac{b}{1-b^2} lt frac{2}{3}$
$endgroup$
– Eldar Rahimli
Jan 25 at 20:06
$begingroup$
@EldarRahimli See edit.
$endgroup$
– J.G.
Jan 25 at 20:09
add a comment |
$begingroup$
Can you please elaborate on your answer? I didn't understand how $frac{b}{1-b^2} lt frac{2}{3}$
$endgroup$
– Eldar Rahimli
Jan 25 at 20:06
$begingroup$
@EldarRahimli See edit.
$endgroup$
– J.G.
Jan 25 at 20:09
$begingroup$
Can you please elaborate on your answer? I didn't understand how $frac{b}{1-b^2} lt frac{2}{3}$
$endgroup$
– Eldar Rahimli
Jan 25 at 20:06
$begingroup$
Can you please elaborate on your answer? I didn't understand how $frac{b}{1-b^2} lt frac{2}{3}$
$endgroup$
– Eldar Rahimli
Jan 25 at 20:06
$begingroup$
@EldarRahimli See edit.
$endgroup$
– J.G.
Jan 25 at 20:09
$begingroup$
@EldarRahimli See edit.
$endgroup$
– J.G.
Jan 25 at 20:09
add a comment |
$begingroup$
For this question, all you need is to know is the quotient identity of tangent, that is $tan(theta) =frac{sin(theta)}{cos(theta)}$
We know that $cos(0)$ is 1 and is decreasing on the interval $(0,pi)$, so $cos(20) < 1$
Considering that $tan(20) =frac {sin(20)}{cos(20)}$, and $cos(20) < 1$, it follows that $tan(20) > sin(20)$ because dividing by a number less than one increases the absolute value of an expression, and looking at the unit circle, sine, cosine, and tangent are positive in quadrant 1.
Therefore, cos(20) > tan(20) > sin(20)
edited Jan 25 at 20:11
H Huang
397111
answered Jan 25 at 19:56
AnadactotheAnadactothe
12411
$endgroup$
1
$begingroup$
How do you know whether $cos(20^circ)$ is greater or less than $tan(20^circ)$? You haven't mentioned this part. Further, this is precisely the part that the OP is asking about.
$endgroup$
– JMoravitz
Jan 25 at 19:59
$begingroup$
ah, it appears i can't read properly. let me edit my post.
$endgroup$
– Anadactothe
Jan 25 at 20:03
add a comment |
$begingroup$
For this question, all you need is to know is the quotient identity of tangent, that is $tan(theta) =frac{sin(theta)}{cos(theta)}$
We know that $cos(0)$ is 1 and is decreasing on the interval $(0,pi)$, so $cos(20) < 1$
Considering that $tan(20) =frac {sin(20)}{cos(20)}$, and $cos(20) < 1$, it follows that $tan(20) > sin(20)$ because dividing by a number less than one increases the absolute value of an expression, and looking at the unit circle, sine, cosine, and tangent are positive in quadrant 1.
Therefore, cos(20) > tan(20) > sin(20)
edited Jan 25 at 20:11
H Huang
397111
answered Jan 25 at 19:56
AnadactotheAnadactothe
12411
$endgroup$
1
$begingroup$
How do you know whether $cos(20^circ)$ is greater or less than $tan(20^circ)$? You haven't mentioned this part. Further, this is precisely the part that the OP is asking about.
$endgroup$
– JMoravitz
Jan 25 at 19:59
$begingroup$
ah, it appears i can't read properly. let me edit my post.
$endgroup$
– Anadactothe
Jan 25 at 20:03
add a comment |
$begingroup$
For this question, all you need is to know is the quotient identity of tangent, that is $tan(theta) =frac{sin(theta)}{cos(theta)}$
We know that $cos(0)$ is 1 and is decreasing on the interval $(0,pi)$, so $cos(20) < 1$
Considering that $tan(20) =frac {sin(20)}{cos(20)}$, and $cos(20) < 1$, it follows that $tan(20) > sin(20)$ because dividing by a number less than one increases the absolute value of an expression, and looking at the unit circle, sine, cosine, and tangent are positive in quadrant 1.
Therefore, cos(20) > tan(20) > sin(20)
edited Jan 25 at 20:11
H Huang
397111
answered Jan 25 at 19:56
AnadactotheAnadactothe
12411
$endgroup$
For this question, all you need is to know is the quotient identity of tangent, that is $tan(theta) =frac{sin(theta)}{cos(theta)}$
We know that $cos(0)$ is 1 and is decreasing on the interval $(0,pi)$, so $cos(20) < 1$
Considering that $tan(20) =frac {sin(20)}{cos(20)}$, and $cos(20) < 1$, it follows that $tan(20) > sin(20)$ because dividing by a number less than one increases the absolute value of an expression, and looking at the unit circle, sine, cosine, and tangent are positive in quadrant 1.
Therefore, cos(20) > tan(20) > sin(20)
edited Jan 25 at 20:11
H Huang
397111
answered Jan 25 at 19:56
AnadactotheAnadactothe
12411
edited Jan 25 at 20:11
H Huang
397111
edited Jan 25 at 20:11
H Huang
397111
edited Jan 25 at 20:11
H Huang
397111
397111
answered Jan 25 at 19:56
AnadactotheAnadactothe
12411
answered Jan 25 at 19:56
AnadactotheAnadactothe
12411
answered Jan 25 at 19:56
AnadactotheAnadactothe
12411
12411
1
$begingroup$
How do you know whether $cos(20^circ)$ is greater or less than $tan(20^circ)$? You haven't mentioned this part. Further, this is precisely the part that the OP is asking about.
$endgroup$
– JMoravitz
Jan 25 at 19:59
$begingroup$
ah, it appears i can't read properly. let me edit my post.
$endgroup$
– Anadactothe
Jan 25 at 20:03
add a comment |
1
$begingroup$
How do you know whether $cos(20^circ)$ is greater or less than $tan(20^circ)$? You haven't mentioned this part. Further, this is precisely the part that the OP is asking about.
$endgroup$
– JMoravitz
Jan 25 at 19:59
$begingroup$
ah, it appears i can't read properly. let me edit my post.
$endgroup$
– Anadactothe
Jan 25 at 20:03
1
1
$begingroup$
How do you know whether $cos(20^circ)$ is greater or less than $tan(20^circ)$? You haven't mentioned this part. Further, this is precisely the part that the OP is asking about.
$endgroup$
– JMoravitz
Jan 25 at 19:59
$begingroup$
How do you know whether $cos(20^circ)$ is greater or less than $tan(20^circ)$? You haven't mentioned this part. Further, this is precisely the part that the OP is asking about.
$endgroup$
– JMoravitz
Jan 25 at 19:59
$begingroup$
ah, it appears i can't read properly. let me edit my post.
$endgroup$
– Anadactothe
Jan 25 at 20:03
$begingroup$
ah, it appears i can't read properly. let me edit my post.
$endgroup$
– Anadactothe
Jan 25 at 20:03
add a comment |
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