Your weak spots

Insights load after your first practice round.

Module 7 · L12 of 20 ~40 min ⚡ +100 XP available

Proving Identities with Inverse Trig

Can you prove that $\tan^{-1}(1/2) + \tan^{-1}(1/3) = \pi/4$? It looks surprising — two arctan values summing to a "nice" angle. The proof uses the compound angle formula for tangent combined with careful range tracking. In this lesson you'll develop a toolkit of proof strategies for inverse trig identities, from the complementary pair to the arctan addition formula.

Today's hook — Is $\tan^{-1}(1/2) + \tan^{-1}(1/3)$ equal to $\pi/4$, less than $\pi/4$, or greater than $\pi/4$? Make a rough estimate by thinking about what $\tan^{-1}(1/2)$ and $\tan^{-1}(1/3)$ look like on the unit circle. Write your prediction before card 05.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

Before working through the proof: is $\tan^{-1}(1/2) + \tan^{-1}(1/3)$ equal to, less than, or greater than $\pi/4$? Explain your reasoning — think about approximate angle sizes. Note that $\tan(\pi/4) = 1$.

auto-saved

The two moves for inverse trig proofs

+5 XP to read

Every inverse trig identity proof uses one of two core moves: introduce a substitution (let $\alpha = \sin^{-1}x$) to convert inverse trig into an angle, or apply the compound angle formula after taking tan of both sides.

The key substitution strategy:

Let $\alpha = f^{-1}(a)$ and $\beta = f^{-1}(b)$ — converts to angles
Construct a right-angle triangle to find other trig ratios
Apply compound angle formulas to prove the result
Track ranges to confirm the result is in the correct principal range

$\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$

Substitution is key

Letting $\alpha = \tan^{-1}(a)$ converts the problem from "arctan arithmetic" into ordinary angle arithmetic. Then you can use triangle-based trig ratios freely.

Range check is the final step

After deriving $\tan(\alpha + \beta) = 1$, you still need to confirm that $\alpha + \beta$ lies in the correct principal range to conclude $\alpha + \beta = \pi/4$ (not $5\pi/4$).

Multiple identities from one framework

The same substitution approach proves: $\sin^{-1}x + \cos^{-1}x = \pi/2$; $\tan^{-1}x + \tan^{-1}(1/x) = \pi/2$ (for $x > 0$); and composite arctan additions.

What you'll master

Know

Key facts

$\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ for $x \in [-1,1]$
$\tan^{-1}x + \tan^{-1}\!\left(\dfrac{1}{x}\right) = \dfrac{\pi}{2}$ for $x > 0$
$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\!\left(\dfrac{a+b}{1-ab}\right)$ when $ab < 1$

Understand

Concepts

Why the substitution $\alpha = f^{-1}(x)$ converts inverse trig proofs into angle proofs
How right-triangle representations link inverse trig to ordinary trig ratios
Why a range check is essential at the end of every proof

Can do

Skills

Prove $\sin^{-1}x + \cos^{-1}x = \pi/2$ using the substitution method
Prove and apply the arctan addition formula
Verify identities like $\tan^{-1}(1/2) + \tan^{-1}(1/3) = \pi/4$

Key terms

Substitution methodSetting $\alpha = \sin^{-1}x$ (or $\cos^{-1}x$, $\tan^{-1}x$) to convert an inverse trig expression into a plain angle for manipulation.

Complementary identity$\sin^{-1}x + \cos^{-1}x = \pi/2$. Proved by showing $\cos(\pi/2 - \sin^{-1}x) = x$ and noting the range argument.

Arctan addition formula$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\!\left(\dfrac{a+b}{1-ab}\right)$ when $ab < 1$. Derived from the compound angle formula for $\tan$.

Range argumentThe final step in a proof: confirming that the derived value lies in the principal range of the relevant inverse function.

Right-triangle representationDrawing a right triangle where $\sin\alpha = x$ (opposite $= x$, hypotenuse $= 1$, adjacent $= \sqrt{1-x^2}$) to read off other trig ratios.

$ab < 1$ conditionThe condition for the standard arctan addition formula. If $ab > 1$, the result must be adjusted by $\pm\pi$.

The arctan addition formula — answering today's hook

core concept

Let $\alpha = \tan^{-1}(1/2)$ and $\beta = \tan^{-1}(1/3)$. Then $\tan\alpha = 1/2$ and $\tan\beta = 1/3$.

Apply the compound angle formula:

$$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{\tfrac{1}{2} + \tfrac{1}{3}}{1 - \tfrac{1}{2} \cdot \tfrac{1}{3}} = \frac{\tfrac{5}{6}}{\tfrac{5}{6}} = 1$$

So $\tan(\alpha + \beta) = 1$. Now the range argument: since $\alpha, \beta \in (0, \pi/2)$ (both $1/2$ and $1/3$ are positive), we have $\alpha + \beta \in (0, \pi)$. The unique angle in $(0, \pi)$ with $\tan = 1$ is $\pi/4$. Therefore:

$$\tan^{-1}\!\left(\frac{1}{2}\right) + \tan^{-1}\!\left(\frac{1}{3}\right) = \frac{\pi}{4}$$

So the hook answer was: exactly equal to $\pi/4$.

General formula. When $ab < 1$, we have $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\!\left(\dfrac{a+b}{1-ab}\right)$. Check: $a = 1/2$, $b = 1/3$, $ab = 1/6 < 1$ ✓. Formula gives $\tan^{-1}(5/6 \div 5/6) = \tan^{-1}(1) = \pi/4$ ✓.

Let $\alpha = \tan^{-1}(1/2)$ and $\beta = \tan^{-1}(1/3)$. Then $\tan\alpha = 1/2$ and $\tan\beta = 1/3$.

Pause — copy the arctan addition formula proof: $\tan^{-1}(1/2)+\tan^{-1}(1/3)=\pi/4$ via $\tan(\alpha+\beta)=\frac{p+q}{1-pq}=1$ into your book.

Quick check: Using the arctan addition formula with $a = 1$ and $b = 1$: what is $\tan^{-1}(1) + \tan^{-1}(1)$? (Note: $ab = 1$, so the standard formula does not apply directly — instead, evaluate each term directly.)

Proving $\sin^{-1}x + \cos^{-1}x = \pi/2$

core concept

We just saw the arctan addition formula proof: $\tan^{-1}(1/2)+\tan^{-1}(1/3)=\pi/4$, verified via $\tan(\alpha+\beta)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}}=\frac{5/6}{5/6}=1$, so $\alpha+\beta=\tan^{-1}(1)=\pi/4$. That raises a question: the identity $\sin^{-1}x+\cos^{-1}x=\pi/2$ is used constantly in derivative and integration problems — what is the cleanest proof of it? This card answers it → let $\theta=\sin^{-1}x$; then $\sin\theta=x$; $\cos(\pi/2-\theta)=\sin\theta=x$; since $\pi/2-\theta\in[0,\pi]$, we have $\cos^{-1}x=\pi/2-\theta$, so $\sin^{-1}x+\cos^{-1}x=\pi/2$.

This identity from Lesson 11 deserves a full proof. The substitution method is the cleanest approach.

Proof:

Let $\theta = \sin^{-1}x$, so $\sin\theta = x$ and $\theta \in [-\pi/2, \pi/2]$.

Then $\pi/2 - \theta \in [0, \pi]$ — this is the principal range of $\cos^{-1}$.

Now compute: $\cos(\pi/2 - \theta) = \sin\theta = x$.

Since $\pi/2 - \theta \in [0, \pi]$ and $\cos(\pi/2 - \theta) = x$, by definition of $\cos^{-1}$:

$$\cos^{-1}(x) = \dfrac{\pi}{2} - \theta = \dfrac{\pi}{2} - \sin^{-1}(x)$$

Rearranging: $\sin^{-1}x + \cos^{-1}x = \pi/2$. $\square$

Analogous identity. For $x > 0$: $\tan^{-1}x + \tan^{-1}(1/x) = \pi/2$. Proof: let $\alpha = \tan^{-1}x$, so $\tan\alpha = x$. Then $\tan(\pi/2 - \alpha) = \cot\alpha = 1/x$, and $\pi/2 - \alpha \in (0, \pi/2)$, so $\tan^{-1}(1/x) = \pi/2 - \alpha$.

This identity from Lesson 11 deserves a full proof. The substitution method is the cleanest approach.

Pause — copy the proof of $\sin^{-1}x+\cos^{-1}x=\pi/2$: let $\theta=\sin^{-1}x$; use $\cos(\pi/2-\theta)=\sin\theta=x$ and the range of $\cos^{-1}$ to conclude $\cos^{-1}x=\pi/2-\theta$ into your book.

Did you get this? True or false: for $x > 0$, the identity $\tan^{-1}x + \tan^{-1}\!\left(\dfrac{1}{x}\right) = \dfrac{\pi}{2}$ holds.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · PROVE AN IDENTITY

Prove that $\sin^{-1}x = \tan^{-1}\!\left(\dfrac{x}{\sqrt{1-x^2}}\right)$ for $x \in (-1, 1)$.

Let $\theta = \sin^{-1}x$, so $\sin\theta = x$ and $\theta \in (-\pi/2, \pi/2)$.

Substitution — converts the identity to a statement about the angle $\theta$.

PROBLEM 2 · EVALUATE COMPOSITE EXPRESSION

Find the exact value of $\sin\!\left(\cos^{-1}\!\left(\dfrac{3}{5}\right)\right)$.

Let $\theta = \cos^{-1}(3/5)$, so $\cos\theta = 3/5$ and $\theta \in [0,\pi]$.

Substitution. Note $\theta \in (0, \pi/2)$ since $3/5 > 0$, so $\sin\theta > 0$.

PROBLEM 3 · HARDER ARCTAN SUM

Prove that $\tan^{-1}(2) + \tan^{-1}(3) = \dfrac{3\pi}{4}$.

Let $\alpha = \tan^{-1}(2)$ and $\beta = \tan^{-1}(3)$. Check: $ab = 2 \times 3 = 6 > 1$, so the standard formula needs adjustment.

When $ab > 1$ and $a, b > 0$, the formula gives an angle outside $(-\pi/2, \pi/2)$; we must add $\pi$ to get the correct value.

Fill the gap: When proving $\sin(\cos^{-1}x) = \sqrt{1-x^2}$, the key step is: let $\theta = \cos^{-1}x$, then $\sin\theta =$ , using the Pythagorean identity and noting $\sin\theta \geq 0$ since $\theta \in [0,\pi]$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Omitting the range argument

After computing $\tan(\alpha + \beta) = -1$, many students write "so $\alpha + \beta = -\pi/4$" without checking the range. Always establish where $\alpha + \beta$ lies before selecting the correct inverse-tan value. This one step is what separates 4/4 from 3/4 in proofs.

Trap 02

Using the formula when $ab \geq 1$

$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\!\left(\dfrac{a+b}{1-ab}\right)$ only when $ab < 1$. When $ab > 1$ (and $a,b > 0$), you must add $\pi$. When $ab = 1$, the formula fails entirely (denominator is zero).

Trap 03

Wrong sign for $\cos\theta$ or $\sin\theta$

When evaluating $\cos(\sin^{-1}x)$ or $\sin(\cos^{-1}x)$, always check the sign using the range of the outer inverse function. $\cos(\sin^{-1}x) = +\sqrt{1-x^2}$ (positive, since $\theta \in [-\pi/2,\pi/2]$). $\sin(\cos^{-1}x) = +\sqrt{1-x^2}$ (positive, since $\theta \in [0,\pi]$).

Did you get this? True or false: $\cos\!\left(\sin^{-1}\!\left(\dfrac{5}{13}\right)\right) = \dfrac{12}{13}$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Evaluate $\cos\!\left(\sin^{-1}\!\left(-\dfrac{4}{5}\right)\right)$. Show your working.

Show that $\tan^{-1}(1/4) + \tan^{-1}(3/5) = \tan^{-1}(17/17) = \pi/4$. (Verify with the arctan addition formula.)

Prove that $\cos(\sin^{-1}x) = \sqrt{1-x^2}$ for $x \in (-1,1)$.

Find the exact value of $\sin\!\left(2\cos^{-1}\!\left(\dfrac{1}{3}\right)\right)$. (Hint: use the double angle formula $\sin 2\theta = 2\sin\theta\cos\theta$.)

Prove that for $x \in (-1,0)$, $\tan^{-1}x + \tan^{-1}(1/x) = -\pi/2$. (Compare with the $x > 0$ case.)

Odd one out: Three of these statements are correct. Which one is NOT?

Revisit your thinking

Earlier you estimated whether $\tan^{-1}(1/2) + \tan^{-1}(1/3)$ equalled $\pi/4$. The exact proof shows it does: the compound angle formula gives $\tan(\alpha + \beta) = 1$, and the range argument confirms $\alpha + \beta = \pi/4$ (not $5\pi/4$).

The proof template — substitute, compute, check range — applies to every inverse trig identity. The range check is not a formality; it is the step that makes the proof watertight.

auto-saved

Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Find the exact value of $\tan\!\left(\sin^{-1}\!\left(\dfrac{5}{13}\right)\right)$. (2 marks)

auto-saved

ApplyBand 43 marks

Q2. Prove that $\tan^{-1}(1/2) + \tan^{-1}(1/3) = \pi/4$, showing all steps including the range argument. (3 marks)

auto-saved

AnalyseBand 53 marks

Q3. Prove that $\sin^{-1}x = \cos^{-1}\!\left(\sqrt{1-x^2}\right)$ for $x \in [0, 1]$. (3 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity answers:

1. $\theta = \sin^{-1}(-4/5) \in [-\pi/2, \pi/2]$; $\sin\theta = -4/5$; $\cos\theta = +3/5$ (positive since $\theta$ is in $[-\pi/2,\pi/2]$ and cosine is positive there). Answer: $3/5$.

2. $a = 1/4$, $b = 3/5$, $ab = 3/20 < 1$. $(a+b)/(1-ab) = (5/20 + 12/20)/(1 - 3/20) = (17/20)/(17/20) = 1$. So the sum $= \tan^{-1}(1) = \pi/4$ ✓.

3. Let $\theta = \sin^{-1}x$, so $\sin\theta = x$ and $\theta \in (-\pi/2, \pi/2)$. Then $\cos^2\theta = 1 - \sin^2\theta = 1 - x^2$, and $\cos\theta = \sqrt{1-x^2}$ (positive since $\theta \in (-\pi/2,\pi/2)$). Hence $\cos(\sin^{-1}x) = \sqrt{1-x^2}$. $\square$

4. $\theta = \cos^{-1}(1/3)$; $\cos\theta = 1/3$; $\sin\theta = \sqrt{1 - 1/9} = \sqrt{8/9} = 2\sqrt{2}/3$. $\sin(2\theta) = 2\sin\theta\cos\theta = 2 \cdot \dfrac{2\sqrt{2}}{3} \cdot \dfrac{1}{3} = \dfrac{4\sqrt{2}}{9}$.

5. For $x \in (-1,0)$: let $\alpha = \tan^{-1}x \in (-\pi/2,0)$. Then $\tan\alpha = x < 0$. We want $\tan^{-1}(1/x)$: since $x < 0$, $1/x < 0$, so $\tan^{-1}(1/x) \in (-\pi/2,0)$. Now $\tan(-\pi/2 - \alpha) = \tan(\pi/2 + |\alpha|)$ — this diverges, so instead note $\alpha + \tan^{-1}(1/x) = -\pi/2$ by continuity and the analogous derivation: $\tan(-\pi/2 - \alpha) = \cot\alpha = 1/\tan\alpha = 1/x$, and $-\pi/2 - \alpha \in (-\pi/2, 0)$, so $\tan^{-1}(1/x) = -\pi/2 - \alpha$, giving $\alpha + \tan^{-1}(1/x) = -\pi/2$. $\square$

Q1 (2 marks): $\sin\theta = 5/13$; $\cos\theta = \sqrt{1 - 25/169} = 12/13$ [1]. $\tan\theta = 5/13 \div 12/13 = 5/12$ [1].

Q2 (3 marks): Let $\alpha = \tan^{-1}(1/2)$, $\beta = \tan^{-1}(1/3)$; $ab = 1/6 < 1$ ✓ [1]. $\tan(\alpha+\beta) = (1/2+1/3)/(1-1/6) = (5/6)/(5/6) = 1$ [1]. Range: $\alpha,\beta \in (0,\pi/4)$, so $\alpha+\beta \in (0,\pi/2)$; the unique angle with $\tan=1$ there is $\pi/4$ [1].

Q3 (3 marks): Let $\theta = \sin^{-1}x$; then $\sin\theta = x$ and $\theta \in [0,\pi/2]$ (since $x \in [0,1]$) [1]. $\cos\theta = \sqrt{1-x^2}$ (positive since $\theta \in [0,\pi/2]$) [1]. Since $\theta \in [0,\pi/2] \subset [0,\pi]$ and $\cos\theta = \sqrt{1-x^2}$, by definition $\theta = \cos^{-1}(\sqrt{1-x^2})$. But $\theta = \sin^{-1}x$, so $\sin^{-1}x = \cos^{-1}(\sqrt{1-x^2})$ [1]. $\square$

Boss battle · The Identity Prover

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering inverse trig identity questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 11 · Inverse Trigonometric Equations Lesson 13 · Graphical Solutions of Trig Equations →

Module overview · Extension 1 · Checkpoint 3