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Shift Identities


How to use
  1. The three tabs across the top switch between sin⁡\sin, cos⁡\cos and tan⁡\tan of the shifted angle. The identity bar underneath writes out the current identity. Learn more about switching functions
  2. The Shift buttons choose between +π+\pi, a half turn, and +π2+\frac{\pi}{2}, a quarter turn. Together with the tabs they select one of six identities. Learn more about choosing the shift
  3. Drag the θ slider through the full turn from 0°0° to 360°360°. The point P and its rotated copy P′ move together, and every value on the page recomputes. Learn more about the angle
  4. The figure shows P on the unit circle, the green arrow turning it to P′, and the legs of both right triangles coloured by which function they measure. The banner at the bottom checks the identity numerically. Learn more about the figure
  5. The panel on the right gives the coordinates of P and P′ and highlights the one consequence that proves the current identity. Learn more about the rotation panel
  6. The two cards under the figure evaluate the left and right sides of the identity at θ. They agree at every angle. Learn more about the numerical check
  7. The identity table under the tool lists all six identities with their effect and value. Click a row to open that identity. Learn more about the identity table

sin(θ + π/2) = cos θ
Shift
θ35°
−sin θcos θP′cos θsin θP+ π/2Oθ = 35°sin(θ + π/2) = cos θ · 0.819 = 0.819

sin(θ + π/2)

0.819

cos θ

0.819

Quarter turn
P = (cos θ, sin θ) = (0.819, 0.574)
P′ = (−sin θ, cos θ) = (-0.574, 0.819)

A quarter turn moves P to P′. The coordinates swap, and the new x picks up a minus sign: P′’s x-leg has the length of P’s y-leg, and P′’s y-leg the length of P’s x-leg.

sin(θ + π/2) is the y of P′, so it equals cos θ.
cos(θ + π/2) is the x of P′, so it equals −sin θ.
tan(θ + π/2) = y / x = cos θ / (−sin θ) = −cot θ.

Drag θ through the full turn: P′ stays a quarter turn ahead of P.

Function
Identity
Effect
Value





Switching Between Functions

The tab strip across the top of the tool has one tab for each of sin⁡\sin, cos⁡\cos and tan⁡\tan, each labelled with the current shift — for example sin⁡(θ+π2)\sin(\theta + \frac{\pi}{2}). The active tab is filled indigo.

Under the tabs, the identity bar writes out the identity being shown, with the shift in red and the right-hand side in the colour of the function it turns into: indigo for cosine, amber for sine. A minus sign, when there is one, sits outside the coloured part, so the change of sign is read separately from the change of function.

Switching tabs keeps both the shift and the angle, so the three identities for one shift can be compared at the same point. The figure changes only in which leg of P′ is being read and in what the banner checks.

The current function and shift are written into the page address as `?shiftFn=` and `?shift=`, so a link can open the tool on a particular identity. The identity table is a second way to switch.

Choosing the Shift

DemoQuarter turn and half turn
Step 0 of 5
Next to the angle slider, two Shift buttons choose how far the angle is moved: +π+\pi or +π2+\frac{\pi}{2}. The panel on the right is titled by the choice — Half turn or Quarter turn.

The shift is the whole geometry of the tool. Adding π\pi to an angle rotates its point on the unit circle by half a turn, to the diametrically opposite point. Adding π2\frac{\pi}{2} rotates it by a quarter turn, counterclockwise. The green arrow in the figure draws that rotation from P to P′, labelled with the shift.

The two shifts produce different kinds of identity. A half turn keeps every function the same and only changes signs — sin⁡\sin stays sin⁡\sin, up to a minus. A quarter turn swaps sine and cosine — sin⁡\sin becomes cos⁡\cos. The identity table shows this directly in its Effect column: sign flips or unchanged for the half turn, sin → cos-type changes for the quarter turn.

Adjusting the Angle

The θ slider runs through a full turn, from 0°0° to 360°360°, in whole degrees. The small red arc at the centre of the figure marks θ, and the red label in the top-left corner prints it.

As θ moves, P travels around the circle and P′ travels with it, always a fixed rotation ahead: half a turn for +π+\pi, a quarter turn for +π2+\frac{\pi}{2}. The panel's final line says exactly this, and it is the reason the identities hold for every angle rather than just the one on screen.

Dragging through all four quadrants is the best test of the signs. In the first quadrant every coordinate of P is positive and the minus signs of the identities are easy to see. In the other three, the coordinates of P are already negative, and the identities still hold — cos⁡(θ+π)=−cos⁡θ\cos(\theta + \pi) = -\cos\theta is positive whenever cos⁡θ\cos\theta is negative. The numerical check confirms it at every stop.

For tangent, some angles make one side infinite. The cards print ∞\infty there, on both sides at once.

Reading the Rotation Figure

DemoReading P and P′
Step 0 of 5
The figure is a unit circle with two points on it. P is the point at angle θ, with coordinates (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta). P′ is P rotated by the shift, so its coordinates are the cosine and sine of θ+π\theta + \pi or θ+π2\theta + \frac{\pi}{2}.

Each point has its own right triangle, dropped from the point to the xx-axis, with the legs drawn thick and labelled. On P, the horizontal leg is indigo and labelled cos⁡θ\cos\theta; the vertical leg is amber and labelled sin⁡θ\sin\theta. On P′, the legs are coloured by where they came from, not by their direction. After a quarter turn, P′'s vertical leg is indigo, because it is as long as P's cosine leg; its horizontal leg is amber, because it is as long as P's sine leg. The labels give the signed values, such as −sin⁡θ-\sin\theta.

The green arrow turns P into P′ and carries the shift as its label. The banner along the bottom states the identity and evaluates both sides at the current θ, so the figure checks itself.

The Rotation Panel

The panel to the right of the figure turns the picture into the proof.

At the top, in a white box, it gives the coordinates of both points as expressions and as numbers — for a quarter turn, P=(cos⁡θ,sin⁡θ)P = (\cos\theta, \sin\theta) and P′=(−sin⁡θ,cos⁡θ)P' = (-\sin\theta, \cos\theta) — each coordinate in the colour of its leg in the figure.

Below that, one sentence states what the rotation does: a quarter turn swaps the coordinates and puts a minus on the new xx; a half turn flips the sign of both.

Then come three consequence lines, one per function, and the line for the active tab is outlined in indigo. Each one reads the identity off the coordinates of P′: sine is its yy, cosine is its xx, tangent is yy over xx. For tangent after a half turn, the line shows the two minus signs cancelling.

The last lines point to this page — to the frozen state of the current identity and to the section explaining its rotation.

Checking Both Sides

Under the figure, two cards evaluate the identity at the current angle. The left card is labelled with the left-hand side, such as cos⁡(θ+π2)\cos(\theta + \frac{\pi}{2}), and the right card with the right-hand side, −sin⁡θ-\sin\theta.

The two values agree to three decimal places at every angle. The banner inside the figure makes the same comparison, so both the picture and the cards confirm each other.

Agreement at a single angle is not a proof, but agreement while the slider is being dragged through a full turn is strong evidence, and it catches sign mistakes immediately. If cos⁡(θ+π2)\cos(\theta + \frac{\pi}{2}) were +sin⁡θ+\sin\theta, the two cards would show equal magnitudes with opposite signs everywhere except where the sine is zero.

The proof itself is the geometry of the rotation figure: the coordinates of P′ are the functions of the shifted angle by definition, and the rotation fixes what those coordinates are.

Reading the Identity Table

DemoThe identity table
Step 0 of 5
Under the tool, a table lists all six shift identities: the three functions shifted by π\pi, then the three shifted by π2\frac{\pi}{2}. Its columns are the shifted function, the identity, the effect, and the value at the current θ.

The Effect column summarises what the shift does in words. Shifting by π\pi gives sign flips for sine and cosine and unchanged for tangent, the last one in indigo to set it apart. Shifting by π2\frac{\pi}{2} gives sin → cos, cos → −sin and tan → −cot.

The active identity is marked with an indigo bar on the left. Clicking any row switches the tool to that function and shift in one step, which makes the table the quickest way to move between the two halves.

Read down the value column with the slider fixed and relationships appear: the π\pi-shifted sine and cosine are the negatives of the ordinary values, and the π2\frac{\pi}{2}-shifted sine equals the ordinary cosine.

Why a Half Turn Flips Both Signs

On the unit circle, the point at angle θ is (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta), and the point at angle θ+π\theta + \pi is the same point turned half-way round the centre. Turning a point half-way round the origin sends (x,y)(x, y) to (−x,−y)(-x, -y) — it is a reflection through the centre. So

cos⁡(θ+π)=−cos⁡θsin⁡(θ+π)=−sin⁡θ\cos(\theta + \pi) = -\cos\theta \qquad \sin(\theta + \pi) = -\sin\theta


and both coordinates change sign while keeping their size. Tangent is the ratio of the two, and the two minus signs cancel:

tan⁡(θ+π)=−sin⁡θ−cos⁡θ=tan⁡θ\tan(\theta + \pi) = \frac{-\sin\theta}{-\cos\theta} = \tan\theta


That last identity is the statement that tangent has period π\pi, half the period of sine and cosine. It is also the reason the graph of tangent repeats after every half turn, with one branch per interval of length π\pi. The periodicity identities on the lesson page state the full-turn version for every function.

Why a Quarter Turn Swaps the Coordinates

Turning a point a quarter turn counterclockwise about the origin sends (x,y)(x, y) to (−y,x)(-y, x): what was the horizontal distance becomes the vertical one, and what was the vertical distance becomes the horizontal one, now pointing the other way. Applied to P=(cos⁡θ,sin⁡θ)P = (\cos\theta, \sin\theta) it gives

P′=(−sin⁡θ,cos⁡θ)P' = (-\sin\theta, \cos\theta)


and since P′P' is the point at angle θ+π2\theta + \frac{\pi}{2}, reading its coordinates gives

cos⁡(θ+π2)=−sin⁡θsin⁡(θ+π2)=cos⁡θ\cos\left(\theta + \tfrac{\pi}{2}\right) = -\sin\theta \qquad \sin\left(\theta + \tfrac{\pi}{2}\right) = \cos\theta


The two right triangles in the figure make this visible: they are congruent, turned a quarter turn relative to each other, which is why the tool colours P′'s legs by where they came from. Tangent follows as their ratio, cos⁡θ−sin⁡θ=−cot⁡θ\frac{\cos\theta}{-\sin\theta} = -\cot\theta.

The same swap underlies the co-function identities, which use π2−θ\frac{\pi}{2} - \theta instead of θ+π2\theta + \frac{\pi}{2}; the extra reflection there removes the minus sign.

Shifts and the Sine Wave

Every identity in this tool is also a statement about graphs, because adding a constant to the input shifts a graph sideways.

sin⁡(θ+π2)=cos⁡θ\sin(\theta + \frac{\pi}{2}) = \cos\theta says that the cosine curve is the sine curve moved π2\frac{\pi}{2} to the left — the two functions are the same wave, a quarter period apart. sin⁡(θ+π)=−sin⁡θ\sin(\theta + \pi) = -\sin\theta says that moving the sine curve half a period gives its reflection in the axis. tan⁡(θ+π)=tan⁡θ\tan(\theta + \pi) = \tan\theta says that the tangent curve is unchanged by a shift of π\pi, which is the definition of having period π\pi.

These are the phase shifts of the general form y=Asin⁡(Bx−C)+Dy = A\sin(Bx - C) + D with C=−π2C = -\frac{\pi}{2} or C=−πC = -\pi. The shift identities section of the lesson lists them alongside the rest of the family, and the phase shift section of the graphs lesson treats the shift of a curve in general.

Sine Shifted by Pi

The identity sin⁡(θ+π)=−sin⁡θ\sin(\theta + \pi) = -\sin\theta is the half turn read vertically.
−cos θ−sin θP′cos θsin θP+ πOθ = 35°sin(θ + π) = −sin θ · -0.574 = -0.574
sin(θ + π) = −sin θ, frozen at θ = 35°

The half turn carries P to the opposite point P′. Its height is the same length as P’s, pointing down: the sine flips sign.

P′ sits diametrically opposite P, so its height is P's height pointing the other way. The amber leg of P′ has exactly the length of P's amber leg, and the label −sin⁡θ-\sin\theta records the direction. The argument for both coordinates is set out in why a half turn flips both signs.

Cosine Shifted by Pi

The identity cos⁡(θ+π)=−cos⁡θ\cos(\theta + \pi) = -\cos\theta is the same half turn read horizontally.
−cos θ−sin θP′cos θsin θP+ πOθ = 35°cos(θ + π) = −cos θ · -0.819 = -0.819
cos(θ + π) = −cos θ, frozen at θ = 35°

The same half turn read horizontally: P′’s x-leg has the length of P’s, pointing left, so the cosine flips sign.

P′'s indigo leg has the length of P's indigo leg and points the opposite way along the xx-axis. Together with the sine identity, this is the statement that a half turn is a reflection through the centre of the circle, (x,y)↦(−x,−y)(x, y) \mapsto (-x, -y).

Tangent Shifted by Pi

The identity tan⁡(θ+π)=tan⁡θ\tan(\theta + \pi) = \tan\theta is the one half-turn identity with no minus sign.
−cos θ−sin θP′cos θsin θP+ πOθ = 35°tan(θ + π) = tan θ · 0.700 = 0.700
tan(θ + π) = tan θ, frozen at θ = 35°

Both coordinates of P′ are the negatives of P’s, so their ratio is unchanged. This is why tangent repeats every π.

Both coordinates of P′ are the negatives of P's, so their ratio — the slope of the radius — is unchanged: P and P′ lie on the same line through the centre. This is why tangent has period π\pi, and why the table's Effect column marks it unchanged.

Sine Shifted by Half Pi

The identity sin⁡(θ+π2)=cos⁡θ\sin(\theta + \frac{\pi}{2}) = \cos\theta is the quarter turn read vertically.
−sin θcos θP′cos θsin θP+ π/2Oθ = 35°sin(θ + π/2) = cos θ · 0.819 = 0.819
sin(θ + π/2) = cos θ, frozen at θ = 35°

The quarter turn swaps the legs: P′’s height, drawn in the cosine colour, is exactly as long as P’s x-leg.

P′'s vertical leg is drawn in the cosine colour because it is as long as P's horizontal leg, and it points up, so no sign changes. Graphically, this says the sine wave shifted a quarter period to the left is the cosine wave. The general argument is in why a quarter turn swaps the coordinates.

Cosine Shifted by Half Pi

The identity cos⁡(θ+π2)=−sin⁡θ\cos(\theta + \frac{\pi}{2}) = -\sin\theta is the quarter turn read horizontally.
−sin θcos θP′cos θsin θP+ π/2Oθ = 35°cos(θ + π/2) = −sin θ · -0.574 = -0.574
cos(θ + π/2) = −sin θ, frozen at θ = 35°

P′’s x-leg, in the sine colour, is as long as P’s height and points left: the cosine becomes minus the sine.

P′'s horizontal leg has the length of P's vertical leg, drawn in the sine colour, but it now points to the left, which is the minus sign. Of the quarter-turn identities, this is the one where the swap and the sign change happen together.

Tangent Shifted by Half Pi

The identity tan⁡(θ+π2)=−cot⁡θ\tan(\theta + \frac{\pi}{2}) = -\cot\theta follows from the other two quarter-turn identities.
−sin θcos θP′cos θsin θP+ π/2Oθ = 35°tan(θ + π/2) = −cot θ · -1.428 = -1.428
tan(θ + π/2) = −cot θ, frozen at θ = 35°

The ratio y/x of P′ is cos θ over −sin θ, which is minus the cotangent.

Dividing the yy of P′ by its xx gives cos⁡θ−sin⁡θ\frac{\cos\theta}{-\sin\theta}, which is −cot⁡θ-\cot\theta. Geometrically, the radius to P′ is perpendicular to the radius to P, and perpendicular slopes multiply to −1-1: tan⁡θ⋅tan⁡(θ+π2)=−1\tan\theta \cdot \tan(\theta + \frac{\pi}{2}) = -1.