Mixing
Prove that the area of this triangle is given by $f(x)=sin(2x)$
$begingroup$
I've encountered this exercise in multiple textbooks now, and I've always been unable to solve it.
Let angle $DBC = x$. $C$ moves through the arc $BD$ such that $xin (0;frac{pi }{2})$. Prove that the area of triangle $[BCD]$ can be given by sin(2$x$).
geometry trigonometry circle
edited Jan 26 at 13:58
Wesley Strik
2,209424
asked Jan 26 at 12:51
Daniel OscarDaniel Oscar
35011
$endgroup$
add a comment |
$begingroup$
I've encountered this exercise in multiple textbooks now, and I've always been unable to solve it.
Let angle $DBC = x$. $C$ moves through the arc $BD$ such that $xin (0;frac{pi }{2})$. Prove that the area of triangle $[BCD]$ can be given by sin(2$x$).
geometry trigonometry circle
edited Jan 26 at 13:58
Wesley Strik
2,209424
asked Jan 26 at 12:51
Daniel OscarDaniel Oscar
35011
$endgroup$
add a comment |
$begingroup$
I've encountered this exercise in multiple textbooks now, and I've always been unable to solve it.
Let angle $DBC = x$. $C$ moves through the arc $BD$ such that $xin (0;frac{pi }{2})$. Prove that the area of triangle $[BCD]$ can be given by sin(2$x$).
geometry trigonometry circle
edited Jan 26 at 13:58
Wesley Strik
2,209424
asked Jan 26 at 12:51
Daniel OscarDaniel Oscar
35011
$endgroup$
I've encountered this exercise in multiple textbooks now, and I've always been unable to solve it.
Let angle $DBC = x$. $C$ moves through the arc $BD$ such that $xin (0;frac{pi }{2})$. Prove that the area of triangle $[BCD]$ can be given by sin(2$x$).
geometry trigonometry circle
geometry trigonometry circle
edited Jan 26 at 13:58
Wesley Strik
2,209424
asked Jan 26 at 12:51
Daniel OscarDaniel Oscar
35011
edited Jan 26 at 13:58
Wesley Strik
2,209424
asked Jan 26 at 12:51
Daniel OscarDaniel Oscar
35011
edited Jan 26 at 13:58
Wesley Strik
2,209424
edited Jan 26 at 13:58
Wesley Strik
2,209424
edited Jan 26 at 13:58
Wesley Strik
2,209424
2,209424
asked Jan 26 at 12:51
Daniel OscarDaniel Oscar
35011
asked Jan 26 at 12:51
Daniel OscarDaniel Oscar
35011
asked Jan 26 at 12:51
Daniel OscarDaniel Oscar
35011
35011
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$BC=2cos{x}$ and $CD=2sin{x}.$
Now, $$S_{Delta BCD}=frac{2sin{x}cdot2cos{x}}{2}.$$
Can you end it now?
answered Jan 26 at 13:03
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
Hint:
It is crucial to note that the angle $BCD$ is right. Then the area is
$$frac{BCcdot CD}2$$ and the sides are obtained by projecting the diameter.
answered Jan 26 at 13:08
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
@YvesDaoust Thales is what proves the angle BCD is right.
$endgroup$
– zdimension
Jan 26 at 13:21
add a comment |
$begingroup$
Observe that the $y$ coordinate of $C$ is $C_y = sin(angle DAC).$ Observe also that $angle DBC$ is inscribed in the circle shown in the figure and $angle DAC$ is the corresponding central angle, and therefore
$angle DAC = 2 angle DBC = 2x.$
So $C_y = sin(2x).$
You now have a triangle with base $BD = 2$ and height $C_y = sin(2x).$
Its area is $$frac12 bh = frac12 cdot 2 cdot sin(2x) = sin(2x).$$
answered Jan 26 at 13:28
David KDavid K
55.3k344120
$endgroup$
add a comment |
$begingroup$
If we first recognise that $BD$ is the diameter of a circle and $C$ is always on the circle, then by Thales's theorem we know that $BCD$ is right-angled. For right-angled triangles we know how to express its sides in terms of a given angle. We use the trigonometric relations in terms of hypothenuse, adjacent and opposite. The hypothenuse is always $2$ since the radius of the circle is $2$, if we let $x$ be our angle $DBC$ we can write:
$$ BC= 2cos(x)$$
$$ CD= 2sin(x)$$
Now we calculate the area by taking base times the height, so:
$$ frac{1}{2} cdot BC cdot CD = 2 sin(x) cos(x) = sin(2x)$$
Here we also used a trigonometric identity in the final step, a so-called double angle formula
answered Jan 26 at 13:20
Wesley StrikWesley Strik
2,209424
$endgroup$
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
$begingroup$
$BC=2cos{x}$ and $CD=2sin{x}.$
Now, $$S_{Delta BCD}=frac{2sin{x}cdot2cos{x}}{2}.$$
Can you end it now?
answered Jan 26 at 13:03
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
$BC=2cos{x}$ and $CD=2sin{x}.$
Now, $$S_{Delta BCD}=frac{2sin{x}cdot2cos{x}}{2}.$$
Can you end it now?
answered Jan 26 at 13:03
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
$begingroup$
$BC=2cos{x}$ and $CD=2sin{x}.$
Now, $$S_{Delta BCD}=frac{2sin{x}cdot2cos{x}}{2}.$$
Can you end it now?
answered Jan 26 at 13:03
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$BC=2cos{x}$ and $CD=2sin{x}.$
Now, $$S_{Delta BCD}=frac{2sin{x}cdot2cos{x}}{2}.$$
Can you end it now?
answered Jan 26 at 13:03
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 13:03
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 13:03
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 13:03
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
$begingroup$
Hint:
It is crucial to note that the angle $BCD$ is right. Then the area is
$$frac{BCcdot CD}2$$ and the sides are obtained by projecting the diameter.
answered Jan 26 at 13:08
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
@YvesDaoust Thales is what proves the angle BCD is right.
$endgroup$
– zdimension
Jan 26 at 13:21
add a comment |
$begingroup$
Hint:
It is crucial to note that the angle $BCD$ is right. Then the area is
$$frac{BCcdot CD}2$$ and the sides are obtained by projecting the diameter.
answered Jan 26 at 13:08
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
@YvesDaoust Thales is what proves the angle BCD is right.
$endgroup$
– zdimension
Jan 26 at 13:21
add a comment |
$begingroup$
Hint:
It is crucial to note that the angle $BCD$ is right. Then the area is
$$frac{BCcdot CD}2$$ and the sides are obtained by projecting the diameter.
answered Jan 26 at 13:08
Yves DaoustYves Daoust
131k676229
$endgroup$
Hint:
It is crucial to note that the angle $BCD$ is right. Then the area is
$$frac{BCcdot CD}2$$ and the sides are obtained by projecting the diameter.
answered Jan 26 at 13:08
Yves DaoustYves Daoust
131k676229
answered Jan 26 at 13:08
Yves DaoustYves Daoust
131k676229
answered Jan 26 at 13:08
Yves DaoustYves Daoust
131k676229
answered Jan 26 at 13:08
Yves DaoustYves Daoust
131k676229
131k676229
$begingroup$
@YvesDaoust Thales is what proves the angle BCD is right.
$endgroup$
– zdimension
Jan 26 at 13:21
add a comment |
$begingroup$
@YvesDaoust Thales is what proves the angle BCD is right.
$endgroup$
– zdimension
Jan 26 at 13:21
$begingroup$
@YvesDaoust Thales is what proves the angle BCD is right.
$endgroup$
– zdimension
Jan 26 at 13:21
$begingroup$
@YvesDaoust Thales is what proves the angle BCD is right.
$endgroup$
– zdimension
Jan 26 at 13:21
add a comment |
$begingroup$
Observe that the $y$ coordinate of $C$ is $C_y = sin(angle DAC).$ Observe also that $angle DBC$ is inscribed in the circle shown in the figure and $angle DAC$ is the corresponding central angle, and therefore
$angle DAC = 2 angle DBC = 2x.$
So $C_y = sin(2x).$
You now have a triangle with base $BD = 2$ and height $C_y = sin(2x).$
Its area is $$frac12 bh = frac12 cdot 2 cdot sin(2x) = sin(2x).$$
answered Jan 26 at 13:28
David KDavid K
55.3k344120
$endgroup$
add a comment |
$begingroup$
Observe that the $y$ coordinate of $C$ is $C_y = sin(angle DAC).$ Observe also that $angle DBC$ is inscribed in the circle shown in the figure and $angle DAC$ is the corresponding central angle, and therefore
$angle DAC = 2 angle DBC = 2x.$
So $C_y = sin(2x).$
You now have a triangle with base $BD = 2$ and height $C_y = sin(2x).$
Its area is $$frac12 bh = frac12 cdot 2 cdot sin(2x) = sin(2x).$$
answered Jan 26 at 13:28
David KDavid K
55.3k344120
$endgroup$
add a comment |
$begingroup$
Observe that the $y$ coordinate of $C$ is $C_y = sin(angle DAC).$ Observe also that $angle DBC$ is inscribed in the circle shown in the figure and $angle DAC$ is the corresponding central angle, and therefore
$angle DAC = 2 angle DBC = 2x.$
So $C_y = sin(2x).$
You now have a triangle with base $BD = 2$ and height $C_y = sin(2x).$
Its area is $$frac12 bh = frac12 cdot 2 cdot sin(2x) = sin(2x).$$
answered Jan 26 at 13:28
David KDavid K
55.3k344120
$endgroup$
Observe that the $y$ coordinate of $C$ is $C_y = sin(angle DAC).$ Observe also that $angle DBC$ is inscribed in the circle shown in the figure and $angle DAC$ is the corresponding central angle, and therefore
$angle DAC = 2 angle DBC = 2x.$
So $C_y = sin(2x).$
You now have a triangle with base $BD = 2$ and height $C_y = sin(2x).$
Its area is $$frac12 bh = frac12 cdot 2 cdot sin(2x) = sin(2x).$$
answered Jan 26 at 13:28
David KDavid K
55.3k344120
answered Jan 26 at 13:28
David KDavid K
55.3k344120
answered Jan 26 at 13:28
David KDavid K
55.3k344120
answered Jan 26 at 13:28
David KDavid K
55.3k344120
55.3k344120
add a comment |
add a comment |
$begingroup$
If we first recognise that $BD$ is the diameter of a circle and $C$ is always on the circle, then by Thales's theorem we know that $BCD$ is right-angled. For right-angled triangles we know how to express its sides in terms of a given angle. We use the trigonometric relations in terms of hypothenuse, adjacent and opposite. The hypothenuse is always $2$ since the radius of the circle is $2$, if we let $x$ be our angle $DBC$ we can write:
$$ BC= 2cos(x)$$
$$ CD= 2sin(x)$$
Now we calculate the area by taking base times the height, so:
$$ frac{1}{2} cdot BC cdot CD = 2 sin(x) cos(x) = sin(2x)$$
Here we also used a trigonometric identity in the final step, a so-called double angle formula
answered Jan 26 at 13:20
Wesley StrikWesley Strik
2,209424
$endgroup$
add a comment |
$begingroup$
If we first recognise that $BD$ is the diameter of a circle and $C$ is always on the circle, then by Thales's theorem we know that $BCD$ is right-angled. For right-angled triangles we know how to express its sides in terms of a given angle. We use the trigonometric relations in terms of hypothenuse, adjacent and opposite. The hypothenuse is always $2$ since the radius of the circle is $2$, if we let $x$ be our angle $DBC$ we can write:
$$ BC= 2cos(x)$$
$$ CD= 2sin(x)$$
Now we calculate the area by taking base times the height, so:
$$ frac{1}{2} cdot BC cdot CD = 2 sin(x) cos(x) = sin(2x)$$
Here we also used a trigonometric identity in the final step, a so-called double angle formula
answered Jan 26 at 13:20
Wesley StrikWesley Strik
2,209424
$endgroup$
add a comment |
$begingroup$
If we first recognise that $BD$ is the diameter of a circle and $C$ is always on the circle, then by Thales's theorem we know that $BCD$ is right-angled. For right-angled triangles we know how to express its sides in terms of a given angle. We use the trigonometric relations in terms of hypothenuse, adjacent and opposite. The hypothenuse is always $2$ since the radius of the circle is $2$, if we let $x$ be our angle $DBC$ we can write:
$$ BC= 2cos(x)$$
$$ CD= 2sin(x)$$
Now we calculate the area by taking base times the height, so:
$$ frac{1}{2} cdot BC cdot CD = 2 sin(x) cos(x) = sin(2x)$$
Here we also used a trigonometric identity in the final step, a so-called double angle formula
answered Jan 26 at 13:20
Wesley StrikWesley Strik
2,209424
$endgroup$
If we first recognise that $BD$ is the diameter of a circle and $C$ is always on the circle, then by Thales's theorem we know that $BCD$ is right-angled. For right-angled triangles we know how to express its sides in terms of a given angle. We use the trigonometric relations in terms of hypothenuse, adjacent and opposite. The hypothenuse is always $2$ since the radius of the circle is $2$, if we let $x$ be our angle $DBC$ we can write:
$$ BC= 2cos(x)$$
$$ CD= 2sin(x)$$
Now we calculate the area by taking base times the height, so:
$$ frac{1}{2} cdot BC cdot CD = 2 sin(x) cos(x) = sin(2x)$$
Here we also used a trigonometric identity in the final step, a so-called double angle formula
answered Jan 26 at 13:20
Wesley StrikWesley Strik
2,209424
answered Jan 26 at 13:20
Wesley StrikWesley Strik
2,209424
answered Jan 26 at 13:20
Wesley StrikWesley Strik
2,209424
answered Jan 26 at 13:20
Wesley StrikWesley Strik
2,209424
2,209424
add a comment |
add a comment |
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