B. Trigonometric Functions · Trigonometric Identities and Equations

Compound Angle Formulas for Tangent

B3.2About 65 minutes

Learning goals

I can follow the development of tan(a ± b) from the sine and cosine formulas, and state its restrictions.
I can use the tangent formulas to find exact values such as tan(π/12) = 2 − √3.
I can find a compound angle when the tangents of its parts are known, using the quadrant to choose the angle.
I can use the tangent formula to find the angle between two lines from their slopes.

Words to know

tangent compound angle formula
tan(a + b) = (tan a + tan b)/(1 − tan a tan b) and tan(a − b) = (tan a − tan b)/(1 + tan a tan b).

restriction
A value of the variable that must be excluded because an expression would be undefined, such as a denominator of 0.

slope
The steepness of a line, rise ÷ run. A line through the origin at angle a to the positive x-axis has slope tan a.

rationalize the denominator
Rewrite a fraction so the denominator has no radical; for (√3 + 1)/(√3 − 1) multiply the top and bottom by the conjugate √3 + 1.

The third member of the family

You have formulas for sin(a ± b) and cos(a ± b). Since tan θ = sin θ / cos θ, the tangent formula is only a division away:

tan(a + b) = sin(a + b) / cos(a + b) = (sin a cos b + cos a sin b) / (cos a cos b − sin a sin b)

Now divide every term, top and bottom, by cos a cos b. Each sin/cos pair becomes a tangent:

top: sin a cos b/(cos a cos b) + cos a sin b/(cos a cos b) = tan a + tan b
bottom: cos a cos b/(cos a cos b) − sin a sin b/(cos a cos b) = 1 − tan a tan b

So tan(a + b) = (tan a + tan b) / (1 − tan a tan b). Replacing b by −b (tangent is odd) gives tan(a − b) = (tan a − tan b) / (1 + tan a tan b).

Restrictions: tan a and tan b must be defined (a, b ≠ π/2 + kπ), and the denominator cannot be 0. When tan a tan b = 1, the angle a + b is an odd multiple of π/2, where tan(a + b) itself is undefined.

How the compound angle formulas are connected. You followed the first four; the double angle formulas in the next lesson come from setting b = a.

	sum	difference
sine	sin a cos b + cos a sin b	sin a cos b − cos a sin b
cosine	cos a cos b − sin a sin b	cos a cos b + sin a sin b
tangent	(tan a + tan b)/(1 − tan a tan b)	(tan a − tan b)/(1 + tan a tan b)

All six compound angle formulas. Notice the signs: sine keeps the sign, cosine flips it, and tangent keeps it on top and flips it on the bottom.

Worked example

Example 1: Exact value of tan(π/12)

Find the exact value of tan(π/12) in simplest form.

π/12 = π/3 − π/4, with tan(π/3) = √3 and tan(π/4) = 1.
tan(π/3 − π/4) = (√3 − 1)/(1 + √3 · 1) = (√3 − 1)/(√3 + 1).
Rationalize: multiply top and bottom by (√3 − 1): (√3 − 1)²/((√3)² − 1²) = (3 − 2√3 + 1)/2 = (4 − 2√3)/2.
= 2 − √3. Calculator check: 2 − √3 ≈ 0.2679 and tan(π/12) ≈ 0.2679 ✓.

tan(π/12) = 2 − √3

Worked example

Example 2: Exact value of tan(7π/12)

Find the exact value of tan(7π/12).

7π/12 = π/3 + π/4.
tan(π/3 + π/4) = (√3 + 1)/(1 − √3).
Rationalize with (1 + √3): (√3 + 1)²/(1 − 3) = (4 + 2√3)/(−2) = −2 − √3.
Sense check: 7π/12 is in quadrant II, where tangent is negative ✓.

tan(7π/12) = −2 − √3 (about −3.732)

Worked example

Example 3: Recognize the pattern

Find the exact value of (tan(5π/18) + tan(π/18)) / (1 − tan(5π/18) tan(π/18)).

This is the pattern of tan(a + b) with a = 5π/18 and b = π/18.
a + b = 6π/18 = π/3.

tan(π/3) = √3

Worked example

Example 4: Find the angle

a and b are acute angles with tan a = 2 and tan b = 3. Find the exact value of a + b.

tan(a + b) = (2 + 3)/(1 − 2 · 3) = 5/(−5) = −1.
Both angles are between 0 and π/2, so 0 < a + b < π.
In that interval, tangent is −1 only in quadrant II, at 3π/4.

a + b = 3π/4

Worked example

Example 5: Angle between two lines

Find the acute angle between the lines y = 3x and y = x/2.

Let the lines make angles a and b with the positive x-axis, so tan a = 3 and tan b = 1/2 (the slopes).
The angle between the lines is θ = a − b.
tan θ = (3 − 1/2)/(1 + 3 · 1/2) = (5/2)/(5/2) = 1.
θ is acute, so θ = π/4.

The lines meet at an angle of π/4 (45°).

The lines y = 3x and y = x/2 make angles a and b with the x-axis. Their slopes are tan a and tan b, so tan(a − b) gives the angle θ between them.

Just for fun

A road sign showing a 10% grade means the road rises 10 m for every 100 m across, so its slope (a tangent!) is 0.10. The angle is only tan⁻¹(0.10) ≈ 0.0997 rad, under 6°. Hills feel much steeper than their tangents, especially on a bike.

Hands-on activity

Measure the angle between two lines

You need: graph paper (or a sheet you grid with a ruler), a ruler, a protractor, a sharp pencil and a calculator.

Draw axes. Draw the line y = 3x through (0, 0) and (2, 6), and the line y = x/2 through (0, 0) and (8, 4).
Measure the angle between them with the protractor. Convert to radians (× π/180).
Compare with the formula result from Example 5: π/4 ≈ 0.785 rad (45°).
Now choose your own slopes m₁ and m₂ (both positive, with m₁ > m₂). Predict the angle with tan θ = (m₁ − m₂)/(1 + m₁m₂) and θ = tan⁻¹(...), then draw and measure.

What to look for: your measurement should be within about 2° of 45° for the first pair. For your own pair, measurement and prediction should agree within a couple of degrees. For example, slopes 2 and 1 give tan θ = 1/3, θ ≈ 0.322 rad (about 18.4°).

Your turn

Find the exact value of tan(5π/12) using 5π/12 = π/4 + π/6.

Simplify (tan 3x − tan x)/(1 + tan 3x tan x).

Use a compound angle formula to show that tan(π − x) = −tan x.

a and b are acute with tan a = 1/2 and tan b = 1/3. Find a + b.

Write tan(x − π/4) in terms of tan x.

sin a = 3/5 with a in quadrant I, and cos b = −5/13 with b in quadrant II. Find tan(a + b).

More practice: warm-up

State the formula for tan(a − b).

Use the formula to show tan(x + π) = tan x.

Try to use the formula for tan(π/4 + π/4). What happens, and why?

Evaluate tan(π/3 − π/6) with the formula.

What is the slope of a line that makes an angle of π/3 with the positive x-axis?

More practice: on your own

Find the exact value of tan(11π/12) using 11π/12 = 2π/3 + π/4.

Find the acute angle between the lines y = 2x and y = −x, to three decimal places.

Prove that tan(x + π/4) tan(x − π/4) = −1 wherever both sides are defined.

tan a = −2 with π/2 < a < π. Find tan(a + π/4).

Why can you not use the tangent formula to find tan(π/2 + π/4)? Find the value another way.

Challenge

Show that tan⁻¹1 + tan⁻¹2 + tan⁻¹3 = π.

Prove that (tan(a + b) − tan a)/(1 + tan(a + b) tan a) = tan b.

You stand 10 m from a wall. The bottom of a sign on the wall is 10 m above your eye level and the top is 20 m above. Find the angle the sign takes up in your view.

Check your understanding

Quiz

Which formula is correct?

What is the exact value of tan(π/12)?

Simplify (tan 4x − tan x)/(1 + tan 4x tan x).

Why can't the tangent formula be used to expand tan(π/2 + x)?

a and b are acute, tan a = 2 and tan b = 3. What is a + b?

What is the acute angle between the lines y = 3x and y = x/2?

For parents and teachers

Big idea (B3.2): tan(a + b) = (tan a + tan b)/(1 − tan a tan b) and tan(a − b) = (tan a − tan b)/(1 + tan a tan b). They come from dividing the sine formula by the cosine formula. They give exact values (tan(π/12) = 2 − √3) and solve geometry problems such as the angle between two lines.

How to help: Ask your teen to show you where each term comes from ("divide everything by cos a cos b"). When an answer involves a radical in the denominator, ask whether it is simplified; rationalizing with the conjugate is a common step.

Common difficulties:

Sign errors in the denominator (1 − for a sum, 1 + for a difference).
Using the formula when one angle is π/2 (tan π/2 is undefined).
Choosing the wrong angle from a tangent value; always decide the interval first (for two acute angles, 0 < a + b < π).
Rationalizing: (√3 − 1)/(√3 + 1) needs the conjugate √3 − 1 on top and bottom.