Complementary angle relationships — sin(90−θ) = cos θ etc.: Exercises

This page is a self-test. Each problem is stated cleanly first — try it on paper, then open the collapsible solution. Problems climb from recognising the pattern to building whole arguments with it.

Everything here rests on the six identities proven in the parent note. The one fact you must carry into every problem:

If any symbol here is unfamiliar, pause and revisit Right Triangle Trigonometry (where "opposite over hypotenuse" is born) and Trigonometric Identities first.

Level 1 — Recognition

Goal: spot the " $90° - θ$ " shape and read off the partner. No algebra yet.

Exercise 1.1

Rewrite each as a single trig function of the acute angle inside, using a cofunction identity: (a) $sin (90° - 40°)$ (b) $cos (90° - 12°)$ (c) $tan (90° - 70°)$

Recall Solution

What we do: match each to the identity with the same function on the left.

(a) $sin (90° - 40°) = cos (40°)$ . Why: $sin$ of a complement becomes $cos$ of the inside angle.
(b) $cos (90° - 12°) = sin (12°)$ . Why: $cos$ of a complement becomes $sin$ .
(c) $tan (90° - 70°) = cot (70°)$ . Why: $tan$ of a complement becomes $cot$ .

Notice we did not compute $90 - 40 = 50$ and swap — the identity keeps the inside angle ( $40°$ ), only the function name changes.

Exercise 1.2

Which of these angle pairs are complementary (add to $90°$ )? For each complementary pair, name one true cofunction statement it produces. (a) $35°$ and $55°$ (b) $20°$ and $70°$ (c) $60°$ and $40°$

Recall Solution

(a) $35 + 55 = 90$ ✓ complementary. So $sin 35° = cos 55°$ .
(b) $20 + 70 = 90$ ✓ complementary. So $tan 20° = cot 70°$ .
(c) $60 + 40 = 100 \neq = 90$ ✗ not complementary — no cofunction relation.

What to see: the whole machinery only fires when the sum is exactly $90°$ .

Level 2 — Application

Goal: use an identity to get a value or a matching form you actually need.

Exercise 2.1

Given $cos 25° \approx 0.9063$ , find $sin 65°$ without a calculator.

Recall Solution

What we do: notice $65° = 90° - 25°$ . $sin 65° = sin (90° - 25°) = cos 25° \approx 0.9063$ Why this tool: the cofunction identity converts a sine we don't know into a cosine we were handed — no new computation needed.

Exercise 2.2

Simplify $\frac{cos ( 90° - θ )}{sin θ}$ to a single number.

Recall Solution

Step 1: $cos (90° - θ) = sin θ$ (cofunction). Why: cosine of a complement is sine. Step 2: Substitute: $\frac{c o s ( 90° - θ )}{s i n θ} = \frac{s i n θ}{s i n θ} = 1$ valid whenever $sin θ \neq = 0$ (i.e. $θ \neq = 0°, 180°, \dots$ ), so the fraction is defined.

Exercise 2.3

Prove $tan (90° - θ) tan θ = 1$ for $0° < θ < 90°$ .

Recall Solution

Step 1: $tan (90° - θ) = cot θ$ (cofunction). Step 2: $cot θ = \frac{1}{tan θ}$ (reciprocal — see Trigonometric Identities). Step 3: Multiply: $tan (90° - θ) tan θ = \frac{1}{t a n θ} \cdot tan θ = 1$ Why it holds on that range: for $0° < θ < 90°$ both $tan θ$ and $cot θ$ are finite and non-zero, so the reciprocal step is legal. $■$

Level 3 — Analysis

Goal: reason about which angles satisfy a condition; watch the geometry.

The next problems lean on the picture below — the same right triangle seen from both acute angles.

Figure — Complementary angle relationships — sin(90−θ) = cos θ etc.

Exercise 3.1

If $sin A = cos B$ and both $A, B$ are acute, what is $A + B$ ? Justify.

Recall Solution

Step 1: Rewrite the right side as a sine: $cos B = sin (90° - B)$ . Step 2: So $sin A = sin (90° - B)$ . Step 3: Both $A$ and $90° - B$ lie in $(0°, 90°)$ where $sin$ is one-to-one (each value hit once — look at the rising sine on the figure). Equal sines force equal angles: $A = 90° - B \Rightarrow A + B = 90°$ The takeaway: " $sin A = cos B$ with both acute" means $A$ and $B$ are complementary. This is the workhorse for the next exercises.

Exercise 3.2

Find the acute angle $x$ with $cos (4 x) = sin (x + 10°)$ .

Recall Solution

Step 1: Turn cosine into sine: $cos (4 x) = sin (90° - 4 x)$ . Step 2: Equation becomes $sin (90° - 4 x) = sin (x + 10°)$ . Step 3: For acute arguments, equal sines ⇒ equal angles: $90° - 4 x = x + 10°$ Step 4: Solve: $90° - 10° = x + 4 x \Rightarrow 80° = 5 x \Rightarrow x = 16°$ . Check: $cos (64°) \approx 0.4384$ and $sin (26°) \approx 0.4384$ ✓. Both arguments $64°, 26°$ are acute, so the step-3 shortcut was valid.

Exercise 3.3

Explain, using the figure, why $sec (90° - θ) = csc θ$ — no algebra, just the triangle.

Recall Solution

What the figure shows: one triangle, hypotenuse $h$ . Side $a$ is opposite $θ$ and adjacent to $(90° - θ)$ ; side $b$ is adjacent to $θ$ and opposite to $(90° - θ)$ . Reason: $sec (90° - θ) = \frac{h}{adjacent to ( 90° - θ )} = \frac{h}{a}$ $csc θ = \frac{h}{opposite θ} = \frac{h}{a}$ Both equal $\frac{h}{a}$ because the side adjacent to $(90° - θ)$ is the side opposite $θ$ — the same segment, two viewpoints. Hence equal. $■$

Level 4 — Synthesis

Goal: combine cofunction identities with other tools into one clean argument.

Exercise 4.1

Simplify $\frac{sin ( 90° - θ )}{cos ( 90° - θ )} + \frac{cos θ}{sin θ}$ to a single expression.

Recall Solution

Step 1: $\frac{sin ( 90° - θ )}{cos ( 90° - θ )} = \frac{cos θ}{sin θ} = cot θ$ . Why: numerator → $cos θ$ , denominator → $sin θ$ . Step 2: The second term is already $\frac{cos θ}{sin θ} = cot θ$ . Step 3: Add: $cot θ + cot θ = 2 cot θ$ valid for $sin θ \neq = 0$ .

Exercise 4.2

A right triangle has acute angles $θ$ and $90° - θ$ . Show that $sin^{2} θ + sin^{2} (90° - θ) = 1.$

Recall Solution

Step 1: $sin (90° - θ) = cos θ$ , so $sin^{2} (90° - θ) = cos^{2} θ$ . Step 2: Substitute: $sin^{2} θ + sin^{2} (90° - θ) = sin^{2} θ + cos^{2} θ .$ Step 3: By the Pythagorean identity (from Trigonometric Identities), $sin^{2} θ + cos^{2} θ = 1$ . What it means geometrically: the two acute angles' sines are the triangle's two legs over the hypotenuse; squaring and adding is Pythagoras. $∴ sin^{2} θ + sin^{2} (90° - θ) = 1 ■$

Exercise 4.3

Evaluate exactly, using cofunction pairing: $cos^{2} 1° + cos^{2} 2° + cos^{2} 3° + \dots + cos^{2} 89°.$

Recall Solution

Step 1 — the pairing idea: $cos (90° - k °) = sin k °$ , so pair term $k$ with term $90 - k$ : $cos^{2} k ° + cos^{2} (90 - k) ° = cos^{2} k ° + sin^{2} k ° = 1.$ Step 2 — count the pairs: terms run $k = 1$ to $89$ . Pair $1$ with $89$ , $2$ with $88$ , …, $44$ with $46$ . That is $44$ pairs, each summing to $1$ , leaving the middle term $k = 45$ unpaired. Step 3 — the leftover: $cos^{2} 45° = (\frac{2}{2})^{2} = \frac{1}{2}$ . Step 4 — total: $44 \times 1 + \frac{1}{2} = 44.5$ Answer: $44.5$ .

Level 5 — Mastery

Goal: prove or evaluate where the cofunction step is one move inside a larger structure.

Exercise 5.1

Prove for all $θ$ where every term is defined: $tan θ \cdot tan (90° - θ) + cos^{2} θ + cos^{2} (90° - θ) = 2.$

Recall Solution

Piece 1: $tan θ \cdot tan (90° - θ) = tan θ \cdot cot θ = 1$ (Ex 2.3 result). Piece 2: $cos^{2} (90° - θ) = sin^{2} θ$ , so $cos^{2} θ + cos^{2} (90° - θ) = cos^{2} θ + sin^{2} θ = 1.$ Combine: $1 + 1 = 2$ . $■$ Domain: need $sin θ \neq = 0$ and $cos θ \neq = 0$ so both $tan$ and $cot$ exist, i.e. $θ$ not a multiple of $90°$ .

Exercise 5.2

Solve for all acute $x$ : $sin (2 x + 20°) = cos (3 x - 10°)$ .

Recall Solution

Step 1: Convert the cosine: $cos (3 x - 10°) = sin (90° - (3 x - 10°)) = sin (100° - 3 x)$ . Step 2: Equation: $sin (2 x + 20°) = sin (100° - 3 x)$ . Step 3 (acute case): set arguments equal: $2 x + 20° = 100° - 3 x \Rightarrow 5 x = 80° \Rightarrow x = 16°.$ Step 4 — check the alternative branch: $sin α = sin β$ also allows $α = 180° - β$ : $2 x + 20° = 180° - (100° - 3 x) = 80° + 3 x \Rightarrow - x = 60° \Rightarrow x = - 60°,$ which is not acute — discard. Step 5 — verify $x = 16°$ : arguments become $2 (16) + 20 = 52°$ and $3 (16) - 10 = 38°$ ; these are complementary ( $52 + 38 = 90$ ) so $sin 52° = cos 38°$ ✓, and both are acute. Answer: $x = 16°$ .

Exercise 5.3

The graph of $y = cos x$ is the graph of $y = sin x$ shifted. Use a cofunction identity to state the shift precisely, and its direction.

Recall Solution

Step 1: $cos x = sin (90° - x)$ . In radians, $cos x = sin (\frac{π}{2} - x)$ . Step 2 — read the transformation (see Graph Transformations): $sin (\frac{π}{2} - x) = sin (- (x - \frac{π}{2}))$ . Since sine is odd — $sin (- u) = - sin u$ , a fact from Even and Odd Functions — this is a reflection and a shift, but the net visible effect is simplest read the other way: $cos x = sin (x + \frac{π}{2}) .$ Step 3 — the shift: replacing $x$ by $x + \frac{π}{2}$ moves the sine graph left by $\frac{π}{2}$ (i.e. $90°$ ). So $cos$ is $sin$ shifted left by a quarter turn. Check a point: at $x = 0$ , $cos 0 = 1$ ; the sine curve reaches $1$ at $x = \frac{π}{2}$ , which after a left-shift of $\frac{π}{2}$ lands at $x = 0$ ✓.

Recall One-line self-check ::: If you see

$sin A = cos B$ with $A, B$ acute, what do you instantly know? The two angles are complementary: $A + B = 90°$ .