Cosecant: build a reciprocal graph from sine

Explain it like I am five

Picture buying gum with 1 dollar. At 50 cents a piece you get 2 pieces; at 25 cents, 4; at 10 cents, 10. Pieces = 1 ÷ price in dollars, so cheaper gum means more pieces, and a price of 0 has no answer. That 1 ÷ number is the number's reciprocal, and cosecant is the reciprocal of sine: csc x = $1 over sin x$ .

Start from the sine wave and keep every input x. Change each height h to 1 ÷ h: height 1 stays 1, $1 over 2$ becomes 2, − $1 over 2$ becomes −2; the sign never changes. Where sine is 0, at x = nπ (0, π, 2π and so on), draw a dashed vertical asymptote, a line the graph approaches but never reaches, instead of a point.

One full turn: between 0 and π sine is a positive hump, so cosecant is a U opening upward with lowest point ( $π over 2$ , 1). Between π and 2π sine is a negative dip, so cosecant is an upside-down U with highest point ( $3 π over 2$ , −1). No point lands strictly between −1 and 1.

In plain words

Imagine a machine that keeps your position along a road but changes the height written on your sign. A height of one stays one. A height of one half becomes two. A negative height stays negative when the machine takes its reciprocal, meaning one divided by that height. Cosecant uses the sine wave as its starting picture. You keep every horizontal input and change its vertical output. Where sine has height zero, the machine cannot produce an answer. Those missing inputs separate the finished graph into curved pieces called branches. The pieces bend away from the road, which represents the x-axis. The branch's turning point nearest the x-axis is its vertex.

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Reminder

Coordinates. The first coordinate is input and the second is height: ( $π over 6$ , 2) means input $π over 6$ , output 2.
Radians. Graph inputs are radians: $π over 2$ = 90° and 2π = 360°.
Unit-circle sine. Sine is the vertical coordinate on the unit circle: sin( $π over 6$ ) = $1 over 2$ .
Reciprocals and fraction division. Divide by a fraction by multiplying by its reciprocal: 1 ÷ $1 over 2$ = 1 × $2 over 1$ = 2.
Negative division. A positive divided by a negative is negative: 1 ÷ (− $1 over 2$ ) = −2.
Rationalizing a denominator. Multiply top and bottom by the same nonzero root: $2 over square root of 2$ = $2 square root of 2 over 2$ = $square root of 2$ .
Zero denominator. A fraction cannot have zero in its denominator: sin π = 0 makes $1 over sin π$ undefined.
Integer multiples. An integer n includes negative integers, zero and positive integers: x = nπ includes −π, 0 and π.
Intervals and union. Open parentheses exclude endpoints and ∪ joins pieces: (0, π) ∪ (π, 2π) excludes 0, π and 2π.
Range. Range names usable output heights: csc x ≥ 1 or csc x ≤ −1 includes 1 and −1 but excludes 0.
Odd symmetry. Odd symmetry changes both signs: ( $π over 6$ , 2) gives (− $π over 6$ , −2).
Period. A period repeats outputs at shifted inputs: csc(x + 2π) = csc x wherever defined.
Vertical asymptote. Nearby outputs grow without bound at an asymptote: near x = 0, the small sine denominator makes cosecant's magnitude large.

Why it works. The reciprocal identity csc x = changes only the output, so an input cannot move sideways. Sine lies between −1 and 1. A positive sine height becomes at least 1 after reciprocation, and a negative height becomes at most −1. Heights 1 and −1 keep their values. Near a sine zero, division by a tiny nonzero height produces outputs of growing magnitude. This makes the vertical asymptotes and the curved branches follow directly from the sine guide.

RuleRule: csc x = , with x ≠ nπ for every integer n. Its vertical asymptotes are x = nπ, its range is (−∞, −1] ∪ [1, ∞), its period is 2π, and csc(−x) = −csc x.

The same idea, five ways

Say it

Say: Cosecant of x is one divided by sine of x.

Write it

Write: A reciprocal graph keeps each input and replaces its nonzero output with one divided by that output.

In math

(x, y) becomes (x, $1 over y$ ) when y ≠ 0
csc x = $1 over sin x$
Domain: {x | x is real and x ≠ nπ, for every integer n}
Range: y ≤ −1 or y ≥ 1
Range: (−∞, −1] ∪ [1, ∞)
Graph words: two curved branches per 2π, opening away from the x-axis

Like: a machine changes a sign's height but leaves its address alone.

See it

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

The same idea, other ways

As a picture

Start with the dashed sine wave. Every zero becomes a vertical boundary. The top at 1 and bottom at −1 become the nearest points of cosecant to the x-axis. All other nonzero heights move farther from the axis.

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

As a height-changing machine

Keep the input $π over 6$ on its road sign. Sine writes $1 over 2$ on that sign. The reciprocal machine changes the height to 2 because 1 ÷ $1 over 2$ = 2. The input is still $π over 6$ .

Only the height changes; the angle remains radians.

With shrinking heights

A sine height of $1 over 10$ would produce a cosecant height of 10. A height of $1 over 100$ would produce 100. The closer a nonzero sine height gets to zero, the farther its reciprocal moves from the x-axis.

1 ÷ = 10

1 ÷ = 100

sin x = 0: csc x is undefined

Small nonzero denominator heights produce large reciprocal heights, while zero produces no value.

Why the nearest heights stay fixed

One divided by 1 is 1, and one divided by −1 is −1. These are the shared guide-wave points. No sine height has magnitude larger than 1, so no cosecant height can have magnitude smaller than 1.

= 1

= −1

csc x ≥ 1 or csc x ≤ −1

The shared heights ±1 mark the closed edges of the cosecant range.

.1Reciprocal graph: keep x and replace the height

Think of the input as an address and the output as a measurement delivered there. Taking a reciprocal changes the measurement, not the address. At one address sine might deliver one half; cosecant delivers two at that same address. At another address sine might deliver negative one half; cosecant delivers negative two. An address where sine delivers zero cannot receive a cosecant measurement, because dividing one by zero is undefined.

Rule: a nonzero guide point (x, y) becomes (x, $1 over y$ ).
Rule: guide height 0 produces an excluded input, not a point.
Rule: reciprocal heights keep their sign; heights ±1 stay fixed.

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Reminder

Reciprocal signs. 1 ÷ $1 over 2$ = 2, and 1 ÷ (− $1 over 2$ ) = −2; the sign stays the same.

The same idea, five ways

Say it

Say: Keep the angle and take one divided by the height.

Write it

Write: Each nonzero sine output is replaced by its reciprocal at the same input.

In math

(x, sin x) becomes (x, csc x)
csc x = $1 over sin x$
sin x × csc x = 1 when sin x ≠ 0

Like: changing the measurement delivered to an address while keeping the address.

See it

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Worked exampleA reciprocal point rather than a reflected point

You need the cosecant point at the same input as a known sine point. Convert ( $π over 6$ , $1 over 2$ ) from the sine guide to cosecant.

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Keep x = $π over 6$ .A reciprocal graph changes the height and does not change the input.
Calculate 1 ÷ $1 over 2$ = 1 × $2 over 1$ = 2.Dividing by a fraction multiplies by its reciprocal.
Plot ( $π over 6$ , 2).The cosecant point has the original input and the new height.

Answer

(, 2).

Check Multiplication check: × 2 = 1. A reflected height of − would instead give a product of −, so it fails the reciprocal check.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Change the input to 2 when taking a reciprocal.

The angle is an address, not the height being divided into 1.

✓ Instead: Keep the input and change its sine height to cosecant height 2.

Tips and tricks

Write the original input again before computing the new height.

.2The positive cosecant branch on (0, π)

Picture a curved bowl above a road. Its lowest point is one unit above the road. The sides climb steeply as they approach two boundary lines. Sine is positive between zero and π, so cosecant stays positive throughout that interval. Sine reaches its highest value at the middle, where its reciprocal reaches the positive branch's lowest value. The middle point is called the vertex, meaning the turning point of this branch.

Rule: on 0 < x < π, csc x ≥ 1.
Rule: the vertex is ( $π over 2$ , 1), shared with sine.
Rule: the branch approaches x = 0 and x = π without reaching either line.

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Reminder

Reciprocal signs. 1 ÷ $1 over 2$ = 2, and 1 ÷ (− $1 over 2$ ) = −2; the sign stays the same.

The same idea, five ways

Say it

Say: Between zero and pi, cosecant is one or greater.

Write it

Write: The positive branch has its lowest point at pi over two.

In math

0 < x < π
x ∈ (0, π)
csc x ≥ 1
Vertex: ( $π over 2$ , 1)
Graph words: upward-opening curved branch

Like: a bowl above a road, with its bottom one unit above the road.

See it

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Worked exampleThree points on the positive branch

You need the middle and two supporting points for the positive piece. Sketch the cosecant branch on 0 < x < π.

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Draw boundary lines x = 0 and x = π, after checking sin 0 = sin π = 0.These zeros would make the denominator zero, so they mark excluded inputs and asymptotes.
Mark ( $π over 2$ , 1).The unit-circle sine value is sin( $π over 2$ ) = 1, whose reciprocal is 1.
Mark ( $π over 6$ , 2) and ( $5 π over 6$ , 2).Both unit-circle sine values are $1 over 2$ , so both reciprocal heights are 2. In particular, $5 π over 6$ = π − $π over 6$ , and vertical-axis reflection preserves sine.
Draw a smooth upward-opening curve through these points, approaching each boundary line.Sine stays positive and approaches zero at both ends, so cosecant grows positive without bound there.

Answer

Interval: (0, π).
Vertex: ( $π over 2$ , 1).
Supporting points: ( $π over 6$ , 2) and ( $5 π over 6$ , 2).
Shape: one upward-opening positive branch.

Check Independent range check: all three marked heights satisfy y ≥ 1. Multiplication check at : × 2 = 1.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: The positive cosecant branch has its lowest output 0.

That would require a sine reciprocal of 0, and no nonzero finite sine has that reciprocal.

✓ Instead: Its lowest output is 1 at ; it stays in y ≥ 1.

Tips and tricks

The largest positive sine height gives the smallest positive cosecant height.

.3The negative cosecant branch on (π, 2π)

Now picture an upside-down bowl below the road. Its highest point is one unit below the road. The sides fall as they approach the boundary lines. Sine is negative from π to two π, so cosecant is negative there too. The sine wave's bottom is negative one, which becomes negative one again after taking the reciprocal. This is the vertex of the negative branch. Its sides head downward, away from the x-axis, rather than upward through it.

Rule: on π < x < 2π, csc x ≤ −1.
Rule: the vertex is ( $3 π over 2$ , −1), shared with sine.
Rule: the branch approaches x = π and x = 2π, opening downward away from the x-axis.

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Reminder

Reciprocal signs. 1 ÷ $1 over 2$ = 2, and 1 ÷ (− $1 over 2$ ) = −2; the sign stays the same.

The same idea, five ways

Say it

Say: From pi to two pi, cosecant is negative one or less.

Write it

Write: The negative branch has its highest point at three pi over two.

In math

π < x < 2π
x ∈ (π, 2π)
csc x ≤ −1
Vertex: ( $3 π over 2$ , −1)
Graph words: downward-opening curved branch

Like: an upside-down bowl hanging below a road.

See it

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Worked exampleThree points on the negative branch

You need the turning point and two supporting points for the negative piece. Sketch the cosecant branch on π < x < 2π.

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Draw boundary lines x = π and x = 2π, checking sin π = sin 2π = 0.This confirms that the two boundary inputs make the denominator zero and must be excluded.
Mark ( $3 π over 2$ , −1).Sine is −1 at that input, and $1 over −1$ = −1.
Mark ( $7 π over 6$ , −2) and ( $11 π over 6$ , −2).Both sine values are − $1 over 2$ , so their reciprocal heights are −2. $7 π over 6$ = π + $π over 6$ reverses sine after a half-turn; $11 π over 6$ = − $π over 6$ + 2π keeps the reflected negative sine after a full turn.
Draw the downward-opening curved branch through the three points.Its outputs remain negative and grow in magnitude near both zero-denominator boundaries.

Answer

Interval: (π, 2π).
Vertex: ( $3 π over 2$ , −1).
Supporting points: ( $7 π over 6$ , −2) and ( $11 π over 6$ , −2).
Shape: one downward-opening negative branch.

Check Independent period and symmetry check: (, 2) gives (−, −2) by odd symmetry; adding 2π to that input gives (, −2), the plotted right-side point.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: The negative cosecant branch opens upward through the x-axis.

A negative denominator gives a negative reciprocal of magnitude at least 1; crossing the axis would enter the forbidden band.

✓ Instead: It opens downward, with highest output −1 at .

Tips and tricks

On a negative branch, closest to the x-axis means highest, because −1 is above −2.

Strategy: step by step

1. Lightly sketch y = sin x as a guide wave. Its horizontal inputs stay fixed throughout the construction.
2. Draw a vertical asymptote at every x-intercept of that wave. Set sin x = 0 to find inputs that would make the denominator zero, then check them in sine before excluding them.
3. Keep every guide-wave point at height 1 or −1. Their reciprocals are the same heights, so the two curves share these points.
4. Between consecutive asymptotes draw a curved U shaped branch opening away from the x-axis. Above the axis it opens upward; below the axis it opens downward. U shaped describes the bend; it does not give a new equation for the curve. Keep using the reciprocal formula to find its heights.
5. Copy the two-branch pattern left and right by 2π. Each branch stays in y ≥ 1 or y ≤ −1, and no point is drawn on an asymptote.

Strategy

Strategy: build cosecant from a sine guide

Is the sine height zero?

YesSet sin x = 0 to locate a forbidden denominator input, plug the input back into sine to verify zero, then draw an asymptote rather than a point.

NoKeep x and compute one divided by the sine height.

↓

Is the nonzero sine height positive?

YesPlace cosecant at height 1 or above and draw an upward-opening branch.

NoPlace cosecant at height −1 or below and draw a downward-opening branch.

↓

Is the sine height 1 or −1?

YesKeep that point unchanged as the branch's nearest point to the x-axis.

NoMove the point to its reciprocal height.

1. Find sine's zeros to locate excluded inputs and asymptotes.
2. Mark the shared points where sine is 1 or −1.
3. Use one reciprocal point on each side of a shared point to guide the curve.
4. Keep the sine sign, bend away from the x-axis, and repeat every 2π.

Worked exampleA complete cosecant period

You need the graph pieces across one full turn, including the locations where no point exists. Sketch y = csc x on 0 ≤ x ≤ 2π using sine as the guide.

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Read each output directly below its input; the inputs are angles in radians and sin x is the output height.

Read each output directly below its input; the inputs are angles in radians and csc x is the output height.

Draw the sine guide from x = 0 to x = 2π, using the sine fntable visual.The sine row gives the starting height under each angle, and the reciprocal construction keeps that angle fixed.
Set sin x = 0 to find denominator zeros: x = 0, π, 2π in this interval. Plug them back in: sin 0 = 0, sin π = 0, sin 2π = 0.This finds the inputs that cannot be used in $1 over sin x$ , so each gets an asymptote and no cosecant point.
Keep ( $π over 2$ , 1) and ( $3 π over 2$ , −1).The sine guide has height 1 and −1 there, and $1 over 1$ = 1 while $1 over −1$ = −1.
Read the columns under $π over 6$ and $5 π over 6$ in the sine table. Replace their $1 over 2$ heights by 1 ÷ $1 over 2$ = 2.These two reciprocal points guide the positive branch on (0, π). Rebuild the farther input: $5 π over 6$ = π − $π over 6$ , so reflection across the vertical axis keeps sine $1 over 2$ .
Read the columns under $7 π over 6$ and $11 π over 6$ . Replace their − $1 over 2$ heights by 1 ÷ (− $1 over 2$ ) = −2.A positive numerator divided by a negative denominator gives a negative output, guiding the negative branch on (π, 2π). Here $7 π over 6$ = π + $π over 6$ , so a half-turn reverses sine. Also $11 π over 6$ = 2π − $π over 6$ , so reflecting to − $π over 6$ and adding a full turn gives the same negative sine.
Draw an upward-opening curved branch through ( $π over 2$ , 1), and a downward-opening curved branch through ( $3 π over 2$ , −1), approaching their boundary lines.As sine approaches zero from the branch's sign, the reciprocal magnitude grows without bound. Neither branch enters −1 < y < 1.

Answer

Asymptotes in the interval: x = 0, π, 2π.
Positive branch: 0 < x < π, vertex ( $π over 2$ , 1).
Negative branch: π < x < 2π, vertex ( $3 π over 2$ , −1).
Repeat the pattern every 2π.

Check Independent symmetry check: the point (, 2) turns about the origin into (−, −2). Adding the period 2π gives (, −2), which matches the negative branch. Both vertices have magnitude 1, so the range check also agrees.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: a shared height stays put

You need the output at the positive sine peak. Evaluate csc( $π over 2$ ) and locate its point.

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Use sin( $π over 2$ ) = 1.The unit-circle point at 90° has vertical coordinate 1.
Take the reciprocal: csc( $π over 2$ ) = $1 over 1$ = 1.Cosecant is one divided by sine.
Plot ( $π over 2$ , 1).The input stays $π over 2$ and the output is 1.

Answer

csc( $π over 2$ ) = 1.
Point: ( $π over 2$ , 1).

Check Multiplication check: sin() × 1 = 1 × 1 = 1, so the two heights are reciprocals.

Rung 2Rung 2: a positive fraction moves outward

You need the output at a sine height of one half. Evaluate csc( $π over 6$ ) and locate its point.

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Use sin( $π over 6$ ) = $1 over 2$ .The 30° unit-circle vertical coordinate is one half.
Compute 1 ÷ $1 over 2$ = 1 × $2 over 1$ = 2.Dividing by a fraction means multiplying by its reciprocal.
Plot ( $π over 6$ , 2), above the vertex's height of 1.The positive reciprocal height stays positive and moves farther from the x-axis.

Answer

csc( $π over 6$ ) = 2.
Point: ( $π over 6$ , 2).

Check Multiplication check: × 2 = 1. The height 2 also fits the positive range y ≥ 1.

Rung 3Rung 3: a negative radical height

You need a negative cosecant output in exact form. Evaluate csc( $7 π over 4$ ) and locate its point.

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Use sin( $7 π over 4$ ) = − $square root of 2 over 2$ . $7 π over 4$ = − $π over 4$ + 2π. A full turn keeps the sine, and reflecting $π over 4$ across the horizontal axis gives sine − $square root of 2 over 2$ .
Write csc( $7 π over 4$ ) = 1 ÷ (− $square root of 2 over 2$ ) = − $2 over square root of 2$ .Taking the reciprocal exchanges the numerator and denominator of the sine height and keeps its negative sign.
Multiply numerator and denominator by $square root of 2$ : − $2 square root of 2 over 2$ = − $square root of 2$ .Since $square root of 2$ × $square root of 2$ = 2, this removes the root from the denominator without changing the fraction.
Plot ( $7 π over 4$ , − $square root of 2$ ).The input stays fixed and the output belongs on the negative branch.

Answer

csc( $7 π over 4$ ) = − $square root of 2$ .
Point: ( $7 π over 4$ , − $square root of 2$ ).

Check Multiplication check: (−) × (−) = = 1. The answer is below −1, as the negative branch requires.

Rung 4Rung 4: rebuild a full period across zero

You need both graph pieces across a full turn that includes negative inputs. Sketch y = csc x on −π ≤ x ≤ π and state the usable inputs within that interval.

The horizontal axis uses radians; the dashed sine wave supplies the heights, while cosecant bends away from the x-axis.

Lightly sketch sine on the interval, then solve sin x = 0 to locate forbidden inputs: −π, 0, π. Plug them back in: sin(−π) = sin 0 = sin π = 0.This identifies every zero denominator in the interval so those inputs can be excluded and used as asymptote locations.
Keep (− $π over 2$ , −1) and ( $π over 2$ , 1) from the guide wave.Their sine heights are −1 and 1, whose reciprocals are unchanged.
Add (− $π over 6$ , −2) and ( $π over 6$ , 2).The corresponding sine heights are − $1 over 2$ and $1 over 2$ , giving reciprocal heights −2 and 2.
Draw a negative downward-opening branch on (−π, 0) and a positive upward-opening branch on (0, π).Each branch keeps the sine sign and heads away from the x-axis as the sine denominator approaches zero.
Write the usable inputs as (−π, 0) ∪ (0, π).The three zero-denominator inputs are excluded, and the union joins the two remaining open intervals.

Answer

Asymptotes: x = −π, 0, π.
Vertices: (− $π over 2$ , −1) and ( $π over 2$ , 1).
Usable inputs in the requested interval: (−π, 0) ∪ (0, π).

Check Independent odd-symmetry check: turning (, 1) and (, 2) about the origin gives both listed negative points. The interval length is 2π, so the two pieces form a full period.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample, wrong: sine height becomes − on cosecant.

Why it fails: that is a reflection of the height. A reciprocal must multiply with the original nonzero height to make 1, but × (−) = −.

✓ Instead: Right: sine height becomes cosecant height 2 at the same input, because × 2 = 1.

✗ Not this: Counterexample, wrong: draw a cosecant point (0, 0) where sine crosses the axis.

Why it fails: sin 0 = 0 would make csc 0 = , which is undefined. A zero denominator cannot become an output of zero.

✓ Instead: Right: draw the asymptote x = 0 and leave that input without a cosecant point.

✗ Not this: Counterexample: a U shaped cosecant branch may enter −1 < y < 1.

Cosecant is a reciprocal sine curve. Its reciprocal heights satisfy y ≥ 1 or y ≤ −1, including the vertices ±1.

✓ Instead: Right: draw curved branches approaching the asymptotes, with vertices at ±1 and all other points farther from the x-axis.

Tips and tricks

Tip: write 'keep x, take 1 ÷ height' beside the sine guide before changing any point.
Tip: sine's zeros give cosecant's asymptotes; sine's heights ±1 give cosecant's vertices.
Tip: 'same sign, farther from zero' describes every nonzero sine height with magnitude below 1.
Tip: one positive and one negative branch make the 2π period. Cosecant is odd, so opposite inputs give opposite heights.

Trap. Trap: taking a reciprocal is not reflecting the sine wave. The input stays fixed, the sign stays fixed, and the nonzero height becomes one divided by that height.

Keep in mind

A reciprocal is not a reflection: height $1 over 2$ becomes 2, not − $1 over 2$ .
Heights smaller than 1 in size move away from the axis: 0.04 becomes 25 and −0.02 becomes −50.
The asymptotes of csc x sit at sine's zeros, x = nπ, and its turning points sit at sine's peaks and valleys, where csc x = 1 or −1.

Memory hookCosecant sits on sine: its U shapes rest on the tops and bottoms of the sine wave, and its walls stand where sine crosses zero.

Flash cards: say the answer out loud, then flip

What is a reciprocal?

1 divided by the number: the reciprocal of 0.5 is 2.

Where are the vertical asymptotes of csc x?

x = nπ, where sin x = 0.

What is the vertex of a cosecant branch?

Its turning point nearest the x-axis, at height 1 or −1.

sin x = 0.1. What is csc x?

10, because 1 ÷ 0.1 = 10.

sin x = −. What is csc x?

− = −2.5.

Where sine is 0, is the cosecant point (x, 0)?

No: 1 ÷ 0 is undefined, so there is no point. Draw an asymptote there.