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Why does $costheta=sin(fracpi2-theta)$ even when $theta>90^circ$?
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I get why this relationship holds true in $90$ degree triangles, but I don't get why this relationship holds true when $theta>90^circ$? I get that when $theta>90^circ$, you find the acute angle that describes $theta$, if you see what I mean. But there is no way of showing that in this function! So why would the fact that $costheta=sin(fracpi2-theta)$ still hold true here?
algebra-precalculus trigonometry
edited Jul 19 at 7:40
N. F. Taussig
38.2k93053
asked Jul 19 at 7:28
Ethan Chan
603322
 |Â
show 3 more comments
up vote
1
down vote
favorite
I get why this relationship holds true in $90$ degree triangles, but I don't get why this relationship holds true when $theta>90^circ$? I get that when $theta>90^circ$, you find the acute angle that describes $theta$, if you see what I mean. But there is no way of showing that in this function! So why would the fact that $costheta=sin(fracpi2-theta)$ still hold true here?
algebra-precalculus trigonometry
edited Jul 19 at 7:40
N. F. Taussig
38.2k93053
asked Jul 19 at 7:28
Ethan Chan
603322
Use the difference identity and see why.
– Sean Roberson
Jul 19 at 7:29
1
The definitions of trig functions for non-acute angles are set up so that all these identities continue to hold...
– Lord Shark the Unknown
Jul 19 at 7:30
@LordSharktheUnknown I know how they are defined for non acute angles, but can you explain to me how the continue to hold even then? As in how has the definition been designed to ensure that they continue to hold?
– Ethan Chan
Jul 19 at 7:31
3
You should be formatting your mathematical expressions with MathJax.
– N. F. Taussig
Jul 19 at 7:41
Do you know the definition of trigonometric functions for a general angle $theta$ which may be positive or negative and less than or greater than $pi/2$? Based on such a definition you can establish the addition formulas and get all such identities.
– Paramanand Singh
Jul 19 at 7:58
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I get why this relationship holds true in $90$ degree triangles, but I don't get why this relationship holds true when $theta>90^circ$? I get that when $theta>90^circ$, you find the acute angle that describes $theta$, if you see what I mean. But there is no way of showing that in this function! So why would the fact that $costheta=sin(fracpi2-theta)$ still hold true here?
algebra-precalculus trigonometry
edited Jul 19 at 7:40
N. F. Taussig
38.2k93053
asked Jul 19 at 7:28
Ethan Chan
603322
I get why this relationship holds true in $90$ degree triangles, but I don't get why this relationship holds true when $theta>90^circ$? I get that when $theta>90^circ$, you find the acute angle that describes $theta$, if you see what I mean. But there is no way of showing that in this function! So why would the fact that $costheta=sin(fracpi2-theta)$ still hold true here?
algebra-precalculus trigonometry
edited Jul 19 at 7:40
N. F. Taussig
38.2k93053
asked Jul 19 at 7:28
Ethan Chan
603322
edited Jul 19 at 7:40
N. F. Taussig
38.2k93053
edited Jul 19 at 7:40
N. F. Taussig
38.2k93053
edited Jul 19 at 7:40
N. F. Taussig
38.2k93053
38.2k93053
asked Jul 19 at 7:28
Ethan Chan
603322
asked Jul 19 at 7:28
Ethan Chan
603322
asked Jul 19 at 7:28
Ethan Chan
603322
603322
Use the difference identity and see why.
– Sean Roberson
Jul 19 at 7:29
1
The definitions of trig functions for non-acute angles are set up so that all these identities continue to hold...
– Lord Shark the Unknown
Jul 19 at 7:30
@LordSharktheUnknown I know how they are defined for non acute angles, but can you explain to me how the continue to hold even then? As in how has the definition been designed to ensure that they continue to hold?
– Ethan Chan
Jul 19 at 7:31
3
You should be formatting your mathematical expressions with MathJax.
– N. F. Taussig
Jul 19 at 7:41
Do you know the definition of trigonometric functions for a general angle $theta$ which may be positive or negative and less than or greater than $pi/2$? Based on such a definition you can establish the addition formulas and get all such identities.
– Paramanand Singh
Jul 19 at 7:58
 |Â
show 3 more comments
Use the difference identity and see why.
– Sean Roberson
Jul 19 at 7:29
1
The definitions of trig functions for non-acute angles are set up so that all these identities continue to hold...
– Lord Shark the Unknown
Jul 19 at 7:30
@LordSharktheUnknown I know how they are defined for non acute angles, but can you explain to me how the continue to hold even then? As in how has the definition been designed to ensure that they continue to hold?
– Ethan Chan
Jul 19 at 7:31
3
You should be formatting your mathematical expressions with MathJax.
– N. F. Taussig
Jul 19 at 7:41
Do you know the definition of trigonometric functions for a general angle $theta$ which may be positive or negative and less than or greater than $pi/2$? Based on such a definition you can establish the addition formulas and get all such identities.
– Paramanand Singh
Jul 19 at 7:58
Use the difference identity and see why.
– Sean Roberson
Jul 19 at 7:29
Use the difference identity and see why.
– Sean Roberson
Jul 19 at 7:29
1
1
The definitions of trig functions for non-acute angles are set up so that all these identities continue to hold...
– Lord Shark the Unknown
Jul 19 at 7:30
The definitions of trig functions for non-acute angles are set up so that all these identities continue to hold...
– Lord Shark the Unknown
Jul 19 at 7:30
@LordSharktheUnknown I know how they are defined for non acute angles, but can you explain to me how the continue to hold even then? As in how has the definition been designed to ensure that they continue to hold?
– Ethan Chan
Jul 19 at 7:31
@LordSharktheUnknown I know how they are defined for non acute angles, but can you explain to me how the continue to hold even then? As in how has the definition been designed to ensure that they continue to hold?
– Ethan Chan
Jul 19 at 7:31
3
3
You should be formatting your mathematical expressions with MathJax.
– N. F. Taussig
Jul 19 at 7:41
You should be formatting your mathematical expressions with MathJax.
– N. F. Taussig
Jul 19 at 7:41
Do you know the definition of trigonometric functions for a general angle $theta$ which may be positive or negative and less than or greater than $pi/2$? Based on such a definition you can establish the addition formulas and get all such identities.
– Paramanand Singh
Jul 19 at 7:58
Do you know the definition of trigonometric functions for a general angle $theta$ which may be positive or negative and less than or greater than $pi/2$? Based on such a definition you can establish the addition formulas and get all such identities.
– Paramanand Singh
Jul 19 at 7:58
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Draw a centre-$O$ circle of radius $1$ and add in the radius with an end at $(1,,0)$, and rotate that radius through an angle $theta$ anticlockwise so its end becomes $(costheta,,sintheta)$. You could have got the same result by rotating the radius at $(0,,1)$ clockwise through $fracpi2-theta$, so $(costheta,,sintheta)=(sin(fracpi2-theta),,cos(fracpi2-theta))$.
answered Jul 19 at 13:54
J.G.
13.2k11424
add a comment |Â
up vote
1
down vote
Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive $$cos x=sin(fracpi2-x)$$ for $0<x<fracpi2$.
Then one would learn the Maclaurin series of $sin$ and $cos$, and then one can show that they have an infinite radius of convergence.
Since $(0,fracpi2)subsetmathbb C$ and $cos x=sin(fracpi2-x)$ in this interval, by identity theorem this immediately implies the two functions($cos x$ and $sin(fracpi2-x)$) are equal for any $xin mathbb C$.
answered Jul 19 at 13:50
Szeto
4,1981521
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Draw a centre-$O$ circle of radius $1$ and add in the radius with an end at $(1,,0)$, and rotate that radius through an angle $theta$ anticlockwise so its end becomes $(costheta,,sintheta)$. You could have got the same result by rotating the radius at $(0,,1)$ clockwise through $fracpi2-theta$, so $(costheta,,sintheta)=(sin(fracpi2-theta),,cos(fracpi2-theta))$.
answered Jul 19 at 13:54
J.G.
13.2k11424
add a comment |Â
up vote
3
down vote
accepted
Draw a centre-$O$ circle of radius $1$ and add in the radius with an end at $(1,,0)$, and rotate that radius through an angle $theta$ anticlockwise so its end becomes $(costheta,,sintheta)$. You could have got the same result by rotating the radius at $(0,,1)$ clockwise through $fracpi2-theta$, so $(costheta,,sintheta)=(sin(fracpi2-theta),,cos(fracpi2-theta))$.
answered Jul 19 at 13:54
J.G.
13.2k11424
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Draw a centre-$O$ circle of radius $1$ and add in the radius with an end at $(1,,0)$, and rotate that radius through an angle $theta$ anticlockwise so its end becomes $(costheta,,sintheta)$. You could have got the same result by rotating the radius at $(0,,1)$ clockwise through $fracpi2-theta$, so $(costheta,,sintheta)=(sin(fracpi2-theta),,cos(fracpi2-theta))$.
answered Jul 19 at 13:54
J.G.
13.2k11424
Draw a centre-$O$ circle of radius $1$ and add in the radius with an end at $(1,,0)$, and rotate that radius through an angle $theta$ anticlockwise so its end becomes $(costheta,,sintheta)$. You could have got the same result by rotating the radius at $(0,,1)$ clockwise through $fracpi2-theta$, so $(costheta,,sintheta)=(sin(fracpi2-theta),,cos(fracpi2-theta))$.
answered Jul 19 at 13:54
J.G.
13.2k11424
answered Jul 19 at 13:54
J.G.
13.2k11424
answered Jul 19 at 13:54
J.G.
13.2k11424
answered Jul 19 at 13:54
J.G.
13.2k11424
13.2k11424
add a comment |Â
add a comment |Â
up vote
1
down vote
Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive $$cos x=sin(fracpi2-x)$$ for $0<x<fracpi2$.
Then one would learn the Maclaurin series of $sin$ and $cos$, and then one can show that they have an infinite radius of convergence.
Since $(0,fracpi2)subsetmathbb C$ and $cos x=sin(fracpi2-x)$ in this interval, by identity theorem this immediately implies the two functions($cos x$ and $sin(fracpi2-x)$) are equal for any $xin mathbb C$.
answered Jul 19 at 13:50
Szeto
4,1981521
add a comment |Â
up vote
1
down vote
Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive $$cos x=sin(fracpi2-x)$$ for $0<x<fracpi2$.
Then one would learn the Maclaurin series of $sin$ and $cos$, and then one can show that they have an infinite radius of convergence.
Since $(0,fracpi2)subsetmathbb C$ and $cos x=sin(fracpi2-x)$ in this interval, by identity theorem this immediately implies the two functions($cos x$ and $sin(fracpi2-x)$) are equal for any $xin mathbb C$.
answered Jul 19 at 13:50
Szeto
4,1981521
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive $$cos x=sin(fracpi2-x)$$ for $0<x<fracpi2$.
Then one would learn the Maclaurin series of $sin$ and $cos$, and then one can show that they have an infinite radius of convergence.
Since $(0,fracpi2)subsetmathbb C$ and $cos x=sin(fracpi2-x)$ in this interval, by identity theorem this immediately implies the two functions($cos x$ and $sin(fracpi2-x)$) are equal for any $xin mathbb C$.
answered Jul 19 at 13:50
Szeto
4,1981521
Even a student that has only learned the ‘geometric definition’ of trigonometric functions can derive $$cos x=sin(fracpi2-x)$$ for $0<x<fracpi2$.
Then one would learn the Maclaurin series of $sin$ and $cos$, and then one can show that they have an infinite radius of convergence.
Since $(0,fracpi2)subsetmathbb C$ and $cos x=sin(fracpi2-x)$ in this interval, by identity theorem this immediately implies the two functions($cos x$ and $sin(fracpi2-x)$) are equal for any $xin mathbb C$.
answered Jul 19 at 13:50
Szeto
4,1981521
answered Jul 19 at 13:50
Szeto
4,1981521
answered Jul 19 at 13:50
Szeto
4,1981521
answered Jul 19 at 13:50
Szeto
4,1981521
4,1981521
add a comment |Â
add a comment |Â
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Use the difference identity and see why.
– Sean Roberson
Jul 19 at 7:29
1
The definitions of trig functions for non-acute angles are set up so that all these identities continue to hold...
– Lord Shark the Unknown
Jul 19 at 7:30
@LordSharktheUnknown I know how they are defined for non acute angles, but can you explain to me how the continue to hold even then? As in how has the definition been designed to ensure that they continue to hold?
– Ethan Chan
Jul 19 at 7:31
3
You should be formatting your mathematical expressions with MathJax.
– N. F. Taussig
Jul 19 at 7:41
Do you know the definition of trigonometric functions for a general angle $theta$ which may be positive or negative and less than or greater than $pi/2$? Based on such a definition you can establish the addition formulas and get all such identities.
– Paramanand Singh
Jul 19 at 7:58