How should I go about finding angles at which two trigonometric ratios are equal?
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Let's say we have an equation, $ sin x = cos x $. We know that between $ 0^o text{and} 90^o $, $ sin $ and $ cos $ are equal for $ x = 45^o $. This is something we can easily figure because we memorize the trig ratios for angles between 0 and 90. How should I go about finding other angles for which an equality between two ratios hold?
I found by trial and error that this equation is true for $ x = 225^o $-- but I'd like to if there is a method to do it without trial and error and for all other trigonometric ratios as well (for example, say, $ tan x = cot x $).
trigonometry
edited Dec 29 '18 at 13:41
Bernard
123k741117
asked Dec 29 '18 at 13:33
WorldGovWorldGov
324211
$endgroup$
add a comment |
$begingroup$
Let's say we have an equation, $ sin x = cos x $. We know that between $ 0^o text{and} 90^o $, $ sin $ and $ cos $ are equal for $ x = 45^o $. This is something we can easily figure because we memorize the trig ratios for angles between 0 and 90. How should I go about finding other angles for which an equality between two ratios hold?
I found by trial and error that this equation is true for $ x = 225^o $-- but I'd like to if there is a method to do it without trial and error and for all other trigonometric ratios as well (for example, say, $ tan x = cot x $).
trigonometry
edited Dec 29 '18 at 13:41
Bernard
123k741117
asked Dec 29 '18 at 13:33
WorldGovWorldGov
324211
$endgroup$
add a comment |
$begingroup$
Let's say we have an equation, $ sin x = cos x $. We know that between $ 0^o text{and} 90^o $, $ sin $ and $ cos $ are equal for $ x = 45^o $. This is something we can easily figure because we memorize the trig ratios for angles between 0 and 90. How should I go about finding other angles for which an equality between two ratios hold?
I found by trial and error that this equation is true for $ x = 225^o $-- but I'd like to if there is a method to do it without trial and error and for all other trigonometric ratios as well (for example, say, $ tan x = cot x $).
trigonometry
edited Dec 29 '18 at 13:41
Bernard
123k741117
asked Dec 29 '18 at 13:33
WorldGovWorldGov
324211
$endgroup$
Let's say we have an equation, $ sin x = cos x $. We know that between $ 0^o text{and} 90^o $, $ sin $ and $ cos $ are equal for $ x = 45^o $. This is something we can easily figure because we memorize the trig ratios for angles between 0 and 90. How should I go about finding other angles for which an equality between two ratios hold?
I found by trial and error that this equation is true for $ x = 225^o $-- but I'd like to if there is a method to do it without trial and error and for all other trigonometric ratios as well (for example, say, $ tan x = cot x $).
trigonometry
trigonometry
edited Dec 29 '18 at 13:41
Bernard
123k741117
asked Dec 29 '18 at 13:33
WorldGovWorldGov
324211
edited Dec 29 '18 at 13:41
Bernard
123k741117
asked Dec 29 '18 at 13:33
WorldGovWorldGov
324211
edited Dec 29 '18 at 13:41
Bernard
123k741117
edited Dec 29 '18 at 13:41
Bernard
123k741117
edited Dec 29 '18 at 13:41
Bernard
123k741117
123k741117
asked Dec 29 '18 at 13:33
WorldGovWorldGov
324211
asked Dec 29 '18 at 13:33
WorldGovWorldGov
324211
asked Dec 29 '18 at 13:33
WorldGovWorldGov
324211
324211
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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Use that from $$sin(x)=cos(x)$$ we get $$tan(x)=1$$ if $$cos(x)neq 0$$
answered Dec 29 '18 at 13:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
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add a comment |
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Rewrite the equation to have only one trigonometric ratio rather than two
$$sin x = cos x iff 1 = frac{sin x}{cos x} iff tan x = 1$$
for $cos x neq 0$.
For $n in mathbb{Z}$, we have
$$tan x = 1 iff x = begin{cases} arctan 1 = frac{pi}{4}+2pi n \ pi+arctan 1 = frac{pi}{4}+pi+ 2pi n end{cases}$$
which means $x = frac{pi}{4}+pi n$, which is $x = 45°+180°n$ in degrees.
For the other example you gave, we have $tan x = cot x$, but $cot x = frac{1}{tan x}$, so rewrite the equation as
$$tan x = frac{1}{tan x}$$
$$tan^2 x = 1 iff tan x = pm 1$$
The two conditions give
$$tan x = 1 iff x = frac{pi}{4}+pi n$$
$$tan x = -1 iff x = frac{3pi}{4}+pi n$$
which can be combined to give $x = frac{pi}{4}+frac{pi n}{2}$, which is $x = 45°+90°n$ in degrees.
answered Dec 29 '18 at 13:37
KM101KM101
6,0901525
$endgroup$
$begingroup$
You have to prove first that sine and cosine can"t be simultaneously be $0$.
$endgroup$
– Bernard
Dec 29 '18 at 13:43
$begingroup$
I have included the condition $cos x neq 0$.
$endgroup$
– KM101
Dec 29 '18 at 13:46
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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votes
active
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votes
$begingroup$
Use that from $$sin(x)=cos(x)$$ we get $$tan(x)=1$$ if $$cos(x)neq 0$$
answered Dec 29 '18 at 13:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
add a comment |
$begingroup$
Use that from $$sin(x)=cos(x)$$ we get $$tan(x)=1$$ if $$cos(x)neq 0$$
answered Dec 29 '18 at 13:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
add a comment |
$begingroup$
Use that from $$sin(x)=cos(x)$$ we get $$tan(x)=1$$ if $$cos(x)neq 0$$
answered Dec 29 '18 at 13:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
$endgroup$
Use that from $$sin(x)=cos(x)$$ we get $$tan(x)=1$$ if $$cos(x)neq 0$$
answered Dec 29 '18 at 13:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Dec 29 '18 at 13:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Dec 29 '18 at 13:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
answered Dec 29 '18 at 13:35
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.9k42866
77.9k42866
add a comment |
add a comment |
$begingroup$
Rewrite the equation to have only one trigonometric ratio rather than two
$$sin x = cos x iff 1 = frac{sin x}{cos x} iff tan x = 1$$
for $cos x neq 0$.
For $n in mathbb{Z}$, we have
$$tan x = 1 iff x = begin{cases} arctan 1 = frac{pi}{4}+2pi n \ pi+arctan 1 = frac{pi}{4}+pi+ 2pi n end{cases}$$
which means $x = frac{pi}{4}+pi n$, which is $x = 45°+180°n$ in degrees.
For the other example you gave, we have $tan x = cot x$, but $cot x = frac{1}{tan x}$, so rewrite the equation as
$$tan x = frac{1}{tan x}$$
$$tan^2 x = 1 iff tan x = pm 1$$
The two conditions give
$$tan x = 1 iff x = frac{pi}{4}+pi n$$
$$tan x = -1 iff x = frac{3pi}{4}+pi n$$
which can be combined to give $x = frac{pi}{4}+frac{pi n}{2}$, which is $x = 45°+90°n$ in degrees.
answered Dec 29 '18 at 13:37
KM101KM101
6,0901525
$endgroup$
$begingroup$
You have to prove first that sine and cosine can"t be simultaneously be $0$.
$endgroup$
– Bernard
Dec 29 '18 at 13:43
$begingroup$
I have included the condition $cos x neq 0$.
$endgroup$
– KM101
Dec 29 '18 at 13:46
add a comment |
$begingroup$
Rewrite the equation to have only one trigonometric ratio rather than two
$$sin x = cos x iff 1 = frac{sin x}{cos x} iff tan x = 1$$
for $cos x neq 0$.
For $n in mathbb{Z}$, we have
$$tan x = 1 iff x = begin{cases} arctan 1 = frac{pi}{4}+2pi n \ pi+arctan 1 = frac{pi}{4}+pi+ 2pi n end{cases}$$
which means $x = frac{pi}{4}+pi n$, which is $x = 45°+180°n$ in degrees.
For the other example you gave, we have $tan x = cot x$, but $cot x = frac{1}{tan x}$, so rewrite the equation as
$$tan x = frac{1}{tan x}$$
$$tan^2 x = 1 iff tan x = pm 1$$
The two conditions give
$$tan x = 1 iff x = frac{pi}{4}+pi n$$
$$tan x = -1 iff x = frac{3pi}{4}+pi n$$
which can be combined to give $x = frac{pi}{4}+frac{pi n}{2}$, which is $x = 45°+90°n$ in degrees.
answered Dec 29 '18 at 13:37
KM101KM101
6,0901525
$endgroup$
$begingroup$
You have to prove first that sine and cosine can"t be simultaneously be $0$.
$endgroup$
– Bernard
Dec 29 '18 at 13:43
$begingroup$
I have included the condition $cos x neq 0$.
$endgroup$
– KM101
Dec 29 '18 at 13:46
add a comment |
$begingroup$
Rewrite the equation to have only one trigonometric ratio rather than two
$$sin x = cos x iff 1 = frac{sin x}{cos x} iff tan x = 1$$
for $cos x neq 0$.
For $n in mathbb{Z}$, we have
$$tan x = 1 iff x = begin{cases} arctan 1 = frac{pi}{4}+2pi n \ pi+arctan 1 = frac{pi}{4}+pi+ 2pi n end{cases}$$
which means $x = frac{pi}{4}+pi n$, which is $x = 45°+180°n$ in degrees.
For the other example you gave, we have $tan x = cot x$, but $cot x = frac{1}{tan x}$, so rewrite the equation as
$$tan x = frac{1}{tan x}$$
$$tan^2 x = 1 iff tan x = pm 1$$
The two conditions give
$$tan x = 1 iff x = frac{pi}{4}+pi n$$
$$tan x = -1 iff x = frac{3pi}{4}+pi n$$
which can be combined to give $x = frac{pi}{4}+frac{pi n}{2}$, which is $x = 45°+90°n$ in degrees.
answered Dec 29 '18 at 13:37
KM101KM101
6,0901525
$endgroup$
Rewrite the equation to have only one trigonometric ratio rather than two
$$sin x = cos x iff 1 = frac{sin x}{cos x} iff tan x = 1$$
for $cos x neq 0$.
For $n in mathbb{Z}$, we have
$$tan x = 1 iff x = begin{cases} arctan 1 = frac{pi}{4}+2pi n \ pi+arctan 1 = frac{pi}{4}+pi+ 2pi n end{cases}$$
which means $x = frac{pi}{4}+pi n$, which is $x = 45°+180°n$ in degrees.
For the other example you gave, we have $tan x = cot x$, but $cot x = frac{1}{tan x}$, so rewrite the equation as
$$tan x = frac{1}{tan x}$$
$$tan^2 x = 1 iff tan x = pm 1$$
The two conditions give
$$tan x = 1 iff x = frac{pi}{4}+pi n$$
$$tan x = -1 iff x = frac{3pi}{4}+pi n$$
which can be combined to give $x = frac{pi}{4}+frac{pi n}{2}$, which is $x = 45°+90°n$ in degrees.
answered Dec 29 '18 at 13:37
KM101KM101
6,0901525
edited Dec 29 '18 at 13:50
answered Dec 29 '18 at 13:37
KM101KM101
6,0901525
answered Dec 29 '18 at 13:37
KM101KM101
6,0901525
answered Dec 29 '18 at 13:37
KM101KM101
6,0901525
6,0901525
$begingroup$
You have to prove first that sine and cosine can"t be simultaneously be $0$.
$endgroup$
– Bernard
Dec 29 '18 at 13:43
$begingroup$
I have included the condition $cos x neq 0$.
$endgroup$
– KM101
Dec 29 '18 at 13:46
add a comment |
$begingroup$
You have to prove first that sine and cosine can"t be simultaneously be $0$.
$endgroup$
– Bernard
Dec 29 '18 at 13:43
$begingroup$
I have included the condition $cos x neq 0$.
$endgroup$
– KM101
Dec 29 '18 at 13:46
$begingroup$
You have to prove first that sine and cosine can"t be simultaneously be $0$.
$endgroup$
– Bernard
Dec 29 '18 at 13:43
$begingroup$
You have to prove first that sine and cosine can"t be simultaneously be $0$.
$endgroup$
– Bernard
Dec 29 '18 at 13:43
$begingroup$
I have included the condition $cos x neq 0$.
$endgroup$
– KM101
Dec 29 '18 at 13:46
$begingroup$
I have included the condition $cos x neq 0$.
$endgroup$
– KM101
Dec 29 '18 at 13:46
add a comment |
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