B. Trigonometric Functions · Trigonometric Identities and EquationsEquivalent Expressions: Cofunction Identities
B3.1About 65 minutes
Learning goals- I can explain, using the two acute angles of a right triangle, why sin x and cos(π/2 − x) are equivalent.
- I can write the six cofunction identities and use them to rewrite trigonometric expressions.
- I can use cofunction identities to find a ratio of one angle from a ratio of its complement.
- I can check that two expressions are equivalent by testing values and by comparing graphs.
Words to know
equivalent expressions
Two expressions that give the same value for every value of the variable where both are defined, such as sin x and cos(π/2 − x).
complementary angles
Two angles whose sum is π/2 (90°). The complement of x is π/2 − x.
cofunction
The partner ratio of a trigonometric ratio: sine and cosine, tangent and cotangent, secant and cosecant are cofunction pairs.
cofunction identity
An identity saying that a ratio of x equals its cofunction of the complement π/2 − x, for example cos x = sin(π/2 − x).
radian
The angle unit used in this course: π radians = 180°, so π/2 = 90°, π/3 = 60°, π/4 = 45° and π/6 = 30°.
Two answers that match
Try this with exact values from the unit Radian Measure and Trigonometric Functions:
- sin(π/6) = 1/2 and cos(π/3) = 1/2
- sin(π/3) = √3/2 and cos(π/6) = √3/2
- sin(π/4) = √2/2 and cos(π/4) = √2/2
In each line, the two angles add up to π/2. That is not a coincidence. In this unit you will learn to see when two different-looking trigonometric expressions are really the same. This idea runs through everything that follows: compound angle formulas, proofs of identities and solving equations.
Remember from Grade 11: in Functions (MCR3U), the unit Trigonometric Ratios, Identities and Laws defined sine, cosine and tangent for any angle, introduced the reciprocal ratios csc, sec and cot, and proved identities with sin²x + cos²x = 1, tan x = sin x/cos x and the reciprocal identities. This year we work in radians, so put your calculator in radian mode (RAD) for this whole unit.
Why sin x = cos(π/2 − x)
Take any right triangle ABC with the right angle at C. The three angles add to π, so the two acute angles add to π/2. If one acute angle is x, the other is π/2 − x.
Now look at side a. It is opposite angle x, but it is adjacent to angle π/2 − x. So the same fraction a/c is the sine of one angle and the cosine of the other:
- sin x = a/c and cos(π/2 − x) = a/c, so sin x = cos(π/2 − x).
This is where the word cosine comes from: it is the complement's sine.
The six cofunction identities
The same reasoning works for every pair of partner ratios:
- sin x = cos(π/2 − x) and cos x = sin(π/2 − x)
- tan x = cot(π/2 − x) and cot x = tan(π/2 − x)
- sec x = csc(π/2 − x) and csc x = sec(π/2 − x)
We built these from a triangle, so the triangle only shows them for acute angles (0 < x < π/2). In the next lesson, and with the compound angle formulas in lesson 3, you will see that they are true for every real number x where both sides are defined. They are identities.
Seeing it with graphs
A graph gives a second way to check. Since cosine is an even function, cos(π/2 − x) = cos(−(x − π/2)) = cos(x − π/2). That is the graph of y = cos x shifted π/2 to the right. Shift the cosine curve right by π/2 and its peak moves from x = 0 to x = π/2, exactly where the sine curve peaks. The two curves land on top of each other everywhere.
Verify with technology
Graphing technology (Desmos, GeoGebra or a graphing calculator, set to radians) can check equivalence. Graph y = sin(x) and y = cos(π/2 − x) together. If they are equivalent, you will see only one curve, because the second sits exactly on top of the first. A table of values (x = 0.5, 1, 2, ...) should show identical outputs. Technology supports an equivalence; the triangle argument or an algebraic proof explains it.
Worked exampleExample 1: Rewrite with the cofunction
Write sin(π/5) as a cosine, and tan(π/12) as a cotangent.
- The complement of π/5 is π/2 − π/5 = 5π/10 − 2π/10 = 3π/10.
- So sin(π/5) = cos(3π/10).
- The complement of π/12 is π/2 − π/12 = 6π/12 − π/12 = 5π/12.
- So tan(π/12) = cot(5π/12).
sin(π/5) = cos(3π/10) and tan(π/12) = cot(5π/12). Check on a calculator in radian mode: sin(π/5) ≈ 0.5878 and cos(3π/10) ≈ 0.5878.
Worked exampleExample 2: A ratio from its complement
Given cos(π/8) ≈ 0.9239, find sin(3π/8) without a calculator.
- Check whether the angles are complements: π/8 + 3π/8 = 4π/8 = π/2. Yes.
- By the cofunction identity, sin(3π/8) = cos(π/2 − 3π/8) = cos(π/8).
Worked exampleExample 3: Find the missing angle
Find the angle a, with 0 < a < π/2, such that sec(2π/9) = csc a.
- The cofunction identity says sec x = csc(π/2 − x).
- So a = π/2 − 2π/9 = 9π/18 − 4π/18 = 5π/18.
a = 5π/18 (that is 50°, and 2π/9 is 40°: complements ✓).
Worked exampleExample 4: Simplify an expression
Simplify cos(π/2 − x) / sin(π/2 − x).
- Replace each part with its cofunction: cos(π/2 − x) = sin x and sin(π/2 − x) = cos x.
- The expression becomes sin x / cos x.
- By the quotient identity from Grade 11, sin x / cos x = tan x.
tan x (for values of x where cos x ≠ 0).
Just for funThe word cosine really does mean "complement's sine". Mathematicians in the 1600s shortened complementi sinus (Latin for "sine of the complement") to co. sinus, and the name stuck. Tangent and secant got the same treatment, which is why we have cotangent and cosecant.
Hands-on activityComplements with a paper triangle
You need: a sheet of paper, a ruler, a protractor, a pencil, scissors (use them carefully) and a calculator in radian mode.
- Draw a right triangle with legs 8 cm and 15 cm. Cut it out.
- Measure the hypotenuse. (It should be about 17 cm, since 8² + 15² = 289 = 17².)
- Label the angle opposite the 8 cm side as x and the other acute angle as y. Measure both with the protractor and convert to radians (multiply degrees by π/180). Check that x + y ≈ π/2 ≈ 1.571.
- Calculate sin x and cos y from the sides. Both should be 8/17 ≈ 0.47. Then calculate cos x and sin y. Both should be 15/17 ≈ 0.88.
- Turn the triangle so that y is at the bottom left. Notice how the side that was "opposite" is now "adjacent".
What to look for: x ≈ 0.49 rad (about 28°) and y ≈ 1.08 rad (about 62°), adding to about 1.57. Your sin x and cos y should match to two decimal places; small differences come from measuring.
Your turn
1Write cos(π/9) as a sine of another angle.
π/2 − π/9 = 9π/18 − 2π/18 = 7π/18, so cos(π/9) = sin(7π/18).
2Write cot(π/10) as a tangent of another angle.
π/2 − π/10 = 4π/10 = 2π/5, so cot(π/10) = tan(2π/5).
3Given sin(2π/7) ≈ 0.7818, find cos(3π/14).
π/2 − 2π/7 = 7π/14 − 4π/14 = 3π/14. So cos(3π/14) = sin(2π/7) ≈ 0.7818.
4Find a, with 0 < a < π/2, such that csc(π/10) = sec a.
csc x = sec(π/2 − x), so a = π/2 − π/10 = 2π/5.
5Simplify sin(π/2 − x) sec x.
sin(π/2 − x) = cos x, and sec x = 1/cos x. The product is 1 (where cos x ≠ 0).
6Check with exact values that sin(π/3) = cos(π/6).
sin(π/3) = √3/2 and cos(π/6) = √3/2. They are equal ✓, and π/3 + π/6 = π/2.
More practice: warm-up
1What is the complement of π/4?
π/2 − π/4 = π/4 (it is its own complement).
2What is the complement of π/6?
3What is the complement of π/12?
4What is the complement of π/5?
5What is the complement of 0.4 rad, to four decimal places?
π/2 − 0.4 ≈ 1.5708 − 0.4 = 1.1708 rad
More practice: on your own
1Show that cos(π/2 − x) tan(π/2 − x) = cos x.
cos(π/2 − x) = sin x and tan(π/2 − x) = cot x = cos x/sin x. The product is sin x × cos x/sin x = cos x (where sin x ≠ 0).
2In right △ABC with ∠C = π/2, sin A = 5/13. Find cos B and tan B.
A and B are complements, so cos B = sin A = 5/13. From sin A = 5/13, the sides are 5, 12, 13, so tan A = 5/12 and tan B = cot A = 12/5.
3x is acute and cos x = 0.3. Find sin(π/2 − x) and cos(π/2 − x).
sin(π/2 − x) = cos x = 0.3. cos(π/2 − x) = sin x = √(1 − 0.3²) = √0.91 ≈ 0.9539 (positive because x is acute).
4Find a, with 0 < a < π/2, such that tan(3π/8) = cot a.
tan x = cot(π/2 − x), so a = π/2 − 3π/8 = π/8.
5Simplify sin²(π/2 − x) + sin²x.
sin(π/2 − x) = cos x, so the expression is cos²x + sin²x = 1 (Pythagorean identity).
Challenge
1The triangle argument only works for acute x. Test the identity cos(π/2 − x) = sin x at x = 2π/3 using exact values. Then explain why the graphs show it works for every x.
cos(π/2 − 2π/3) = cos(−π/6) = cos(π/6) = √3/2, and sin(2π/3) = √3/2. Equal ✓. The graph of y = cos(π/2 − x) = cos(x − π/2) is the cosine curve shifted π/2 right, which lies exactly on the sine curve for all x, not just acute angles.
2Find the exact value of sin²(π/12) + sin²(5π/12).
π/12 and 5π/12 are complements, so sin(5π/12) = cos(π/12). The sum is sin²(π/12) + cos²(π/12) = 1.
3Find the exact value of sin²(π/10) + sin²(π/5) + sin²(3π/10) + sin²(2π/5).
Pair the complements: π/10 + 2π/5 = π/2 and π/5 + 3π/10 = π/2. So sin²(2π/5) = cos²(π/10) and sin²(3π/10) = cos²(π/5). The sum is (sin²(π/10) + cos²(π/10)) + (sin²(π/5) + cos²(π/5)) = 1 + 1 = 2.
Quiz
1. Which expression is equivalent to sin(π/8)?
2. What does the "co" in cosine, cotangent and cosecant refer to?
3. If 0 < a < π/2 and cos a = sin(π/5), what is a?
4. tan(π/2 − x) is equivalent to which expression?
5. Simplify cos(π/2 − x) / sin(π/2 − x).
6. You graph y = sin x and y = cos(π/2 − x) on the same screen in radian mode. What do you see?
Big idea (B3.1): two expressions are equivalent when they give the same value for every input. The first family your teen meets are the cofunction identities, such as sin x = cos(π/2 − x). They come from the simple fact that the two acute angles of a right triangle add to π/2 (90°).
How to help: Ask your teen to draw a right triangle, label one acute angle x and the other π/2 − x, and explain in their own words why the sine of one is the cosine of the other. Then have them check a pair on a calculator in radian mode, such as sin(0.3) and cos(1.5708 − 0.3).
Common difficulties:
- Leaving the calculator in degree mode. In this unit everything is in radians.
- Arithmetic with fractions of π: π/2 − π/5 needs a common denominator (5π/10 − 2π/10 = 3π/10).
- Pairing the wrong ratios. The pairs are sin/cos, tan/cot and sec/csc.
- Thinking the identity only works for acute angles. The triangle shows it for acute angles, but the graphs show it works for every x.