6.6, 6.7 - Trigonometry for any angle

6.6, 6.7 - Trigonometry for any angle

Extend sine, cosine and tangent beyond acute triangles using coordinates on a circle. You will use signs and reference angles, then sketch and read their graphs for angles of any size. All angles in this lesson are measured in degrees.

Trigonometric ratios from coordinates

Why does tangent repeat after half a turn, while sine and cosine take a full turn to repeat? A point moving round a circle makes the difference visible. Start at the positive horizontal axis and turn anticlockwise; after half a turn, both its coordinates change sign.

You already know

In a right-angled triangle, sin⁡θ=oppositehypotenuse\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}, cos⁡θ=adjacenthypotenuse\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} and tan⁡θ=oppositeadjacent\tan\theta=\frac{\text{opposite}}{\text{adjacent}}. On the formulae sheet.

For an acute angle, a point P(x,y)P(x,y) on a circle of radius rr, centred at the origin, makes a right triangle with horizontal side xx, vertical side yy and hypotenuse rr. The ratios are therefore y/ry/r, x/rx/r and y/xy/x. Keep these coordinate definitions as the point travels into the other quadrants: the coordinates now supply the signs that side lengths alone could not.

[DIAGRAM: asset_name: Coordinates define the ratios; asset_slug: l2fm-geometry-unit-circle; caption: Anticlockwise angle theta determines the signed coordinates of a point on the unit circle; description: A unit circle on x and y axes has point P at coordinates negative 3 fifths and 4 fifths in quadrant two. Radius OP is blue; a sage vertical drop and horizontal segment show the coordinate triangle. The anticlockwise angle theta is measured from the positive x axis, and acute reference angle alpha lies between OP and the negative x axis. A right-angle mark appears at the foot of the vertical drop.]
Anticlockwise angle theta determines the signed coordinates of a point on the unit circle

Using r=1r=1 makes the coordinates themselves equal to cosine and sine. The radius remains positive even when a coordinate is negative.

Coordinate definitions

For a point P(x,y)P(x,y) on a circle with centre O(0,0)O(0,0) and radius r>0r>0, reached by an anticlockwise angle θ\theta,

sin⁡θ=yr,cos⁡θ=xr,tan⁡θ=yx(x≠0)\sin\theta=\frac yr,\qquad\cos\theta=\frac xr,\qquad\tan\theta=\frac yx\quad(x\ne0)

On the unit circle, P=(cos⁡θ,sin⁡θ)P=(\cos\theta,\sin\theta). Tangent is undefined when x=0x=0, including 90∘90^\circ and 270∘270^\circ.
Not on the formulae sheet: learn it.

The definitions work on the axes too. At 90∘90^\circ, P=(0,1)P=(0,1): sine is 11, cosine is 00, and tangent has no value because division by zero is undefined.

Worked example 1: a point in the second quadrant

Question. A radius of length 55 ends at P(−3,4)P(-3,4). Its anticlockwise angle from the positive horizontal axis is θ\theta. Work out sin⁡θ\sin\theta, cos⁡θ\cos\theta and tan⁡θ\tan\theta.

Plan. Use signed coordinates in the three definitions.

sin⁡θ=45vertical coordinate over radiuscos⁡θ=−35horizontal coordinate over radiustan⁡θ=4−3=−43vertical over horizontal\begin{aligned} \sin\theta&=\frac45 && \text{vertical coordinate over radius} \\ \cos\theta&=-\frac35 && \text{horizontal coordinate over radius} \\ \tan\theta&=\frac4{-3}=-\frac43 && \text{vertical over horizontal} \end{aligned}

Why this step? Replacing signed coordinates by positive side lengths would lose the quadrant information.

Answer. 45\frac45, −35-\frac35, −43-\frac43, respectively.

Check. (−3)2+42=25=52(-3)^2+4^2=25=5^2, so the point and radius agree. Its positive height gives positive sine.

Worked example 2 reflects only the point in the horizontal axis. The radius stays 55, while the vertical coordinate changes sign.

Worked example 2: a point in the third quadrant

Question. A radius of length 55 ends at Q(−3,−4)Q(-3,-4), at anticlockwise angle ϕ\phi. Work out sin⁡ϕ\sin\phi, cos⁡ϕ\cos\phi and tan⁡ϕ\tan\phi.

Plan. Use the same definitions with the new signed coordinates.

sin⁡ϕ=−45negative vertical coordinatecos⁡ϕ=−35negative horizontal coordinatetan⁡ϕ=−4−3=43two negatives divide positively\begin{aligned} \sin\phi&=-\frac45 && \text{negative vertical coordinate} \\ \cos\phi&=-\frac35 && \text{negative horizontal coordinate} \\ \tan\phi&=\frac{-4}{-3}=\frac43 && \text{two negatives divide positively} \end{aligned}

Why this step? The two ratios to the positive radius are negative, but the ratio of the two negative coordinates is positive.

Answer. −45-\frac45, −35-\frac35, 43\frac43, respectively.

Check. Reflection in the horizontal axis reverses sine and tangent but leaves cosine unchanged, as these answers show.

On a calculator, use degree mode. Negative sine and cosine values are normal outside the first quadrant.

Watch out

Wrong: tan⁡90∘=0\tan90^\circ=0, because its horizontal coordinate is zero. Right: tangent divides by that coordinate, so it is undefined. Cosine divides the zero coordinate by the positive radius, giving zero.

Use a new point and radius without first calculating the angle.

Reference angles and signs

A reference angle supplies the magnitude of a ratio, while the quadrant supplies its sign.

Reference angle

For an angle not on an axis, its reference angle is the acute angle between its radius and the horizontal axis. For example, the reference angle of 142∘142^\circ is 180∘−142∘=38∘180^\circ-142^\circ=38^\circ. It is not the angle measured to the vertical axis.

Reflecting a point in an axis changes one coordinate sign and preserves its distance from each axis. Rotating by 180∘180^\circ changes both signs. The absolute coordinate ratios therefore match those of the acute reference triangle.

QuadrantAngle intervalSineCosineTangent
First0∘<θ<90∘0^\circ<\theta<90^\circpositivepositivepositive
Second90∘<θ<180∘90^\circ<\theta<180^\circpositivenegativenegative
Third180∘<θ<270∘180^\circ<\theta<270^\circnegativenegativepositive
Fourth270∘<θ<360∘270^\circ<\theta<360^\circnegativepositivenegative

The axes are boundaries, so use coordinates directly there. At 0∘,90∘,180∘,270∘,360∘0^\circ,90^\circ,180^\circ,270^\circ,360^\circ, the unit-circle points are (1,0),(0,1),(−1,0),(0,−1),(1,0)(1,0),(0,1),(-1,0),(0,-1),(1,0), respectively. The sign table and reference-angle method are not on the formulae sheet: learn them.

Worked example 3: an angle in the second quadrant

Question. You are given sin⁡38∘=a\sin38^\circ=a, cos⁡38∘=b\cos38^\circ=b and tan⁡38∘=c\tan38^\circ=c. Write down sin⁡142∘\sin142^\circ, cos⁡142∘\cos142^\circ and tan⁡142∘\tan142^\circ.

Plan. Use reference angle 38∘38^\circ and second-quadrant signs.

142∘=180∘−38∘reference angle 38∘sin⁡142∘=apositive vertical coordinatecos⁡142∘=−bnegative horizontal coordinatetan⁡142∘=−copposite coordinate signs\begin{aligned} 142^\circ&=180^\circ-38^\circ && \text{reference angle }38^\circ \\ \sin142^\circ&=a && \text{positive vertical coordinate} \\ \cos142^\circ&=-b && \text{negative horizontal coordinate} \\ \tan142^\circ&=-c && \text{opposite coordinate signs} \end{aligned}

Why this step? The magnitude comes from the reference triangle; the signs come from the actual point's position.

Answer. a,−b,−ca,-b,-c, respectively.

Check. Reflection of the 38∘38^\circ radius in the vertical axis preserves height and reverses horizontal position.

Worked example 4 changes only the quadrant, putting the radius 38∘38^\circ below the negative horizontal axis.

Worked example 4: the same reference angle in the third quadrant

Question. You are given sin⁡38∘=a\sin38^\circ=a, cos⁡38∘=b\cos38^\circ=b and tan⁡38∘=c\tan38^\circ=c. Write down sin⁡218∘\sin218^\circ, cos⁡218∘\cos218^\circ and tan⁡218∘\tan218^\circ.

Plan. Keep reference angle 38∘38^\circ and use third-quadrant signs.

218∘=180∘+38∘same acute reference anglesin⁡218∘=−abelow the horizontal axiscos⁡218∘=−bleft of the vertical axistan⁡218∘=ctwo negative coordinates\begin{aligned} 218^\circ&=180^\circ+38^\circ && \text{same acute reference angle} \\ \sin218^\circ&=-a && \text{below the horizontal axis} \\ \cos218^\circ&=-b && \text{left of the vertical axis} \\ \tan218^\circ&=c && \text{two negative coordinates} \end{aligned}

Why this step? Both coordinates are negative here, so their quotient is positive even though sine and cosine are negative.

Answer. −a,−b,c-a,-b,c, respectively.

Check. The 218∘218^\circ radius is a half-turn from the 38∘38^\circ radius, reversing both coordinates.

Choose a positive acute reference angle, then apply the signs of the actual quadrant.

Watch out

Wrong: the reference angle of 218∘218^\circ is 52∘52^\circ, obtained from 270∘−218∘270^\circ-218^\circ. That measures to the vertical axis. Right: use the horizontal axis, giving 218∘−180∘=38∘218^\circ-180^\circ=38^\circ.

Take the same reference triangle into the fourth quadrant.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account