Induction for Trigonometric Identities
De Moivre's theorem is one of mathematics' most elegant results — and it's proved entirely by induction. In this lesson you'll master how compound angle formulas power the inductive step for trig identities, proving results like $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ and $\cos(n\pi) = (-1)^n$ from scratch.
Expand $(\cos\theta + i\sin\theta)^2$ by hand. Use $i^2 = -1$ and the double angle formulas $\cos(2\theta) = \cos^2\theta - \sin^2\theta$ and $\sin(2\theta) = 2\sin\theta\cos\theta$. What do you notice about the result?
Trigonometric induction always comes down to two key moves: write the $(k+1)$ case as the $k$ case times one extra factor, then apply the compound angle formula to collapse it.
The compound angle formulas are the engine of every trig induction proof:
$\cos(A+B) = \cos A\cos B - \sin A\sin B$
$\sin(A+B) = \sin A\cos B + \cos A\sin B$
Set $A = k\theta$ and $B = \theta$ to connect the $k$ case to the $k+1$ case.
Key facts
- De Moivre's theorem: $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$
- Compound angle formulas for $\cos(A+B)$ and $\sin(A+B)$
- $\cos(n\pi) = (-1)^n$ for all $n \geq 0$
Concepts
- Why the inductive step for trig identities always uses a compound angle formula
- How multiplying by one extra factor connects $n=k$ to $n=k+1$
- The role of $i^2 = -1$ in complex number induction proofs
Skills
- Write a complete induction proof for De Moivre's theorem
- Prove simple trig identities (e.g. $\cos(n\pi)=(-1)^n$) by induction
- Identify and apply the correct compound angle formula in the inductive step
Trigonometric induction proofs often rely on compound angle formulas. When proving a statement for $n = k + 1$, you typically express the $(k+1)$ case in terms of the $k$ case plus one extra angle, then apply the compound angle formula.
The typical structure: set $A = k\theta$ (covered by the inductive hypothesis) and $B = \theta$ (the single extra step). The hypothesis gives you $\cos(k\theta)$ and $\sin(k\theta)$; the formula assembles $\cos((k+1)\theta)$ and $\sin((k+1)\theta)$.
Trigonometric induction proofs often rely on compound angle formulas . When proving a statement for $n = k + 1$, you typically express the $(k+1)$ case in terms of the $k$ case plus one extra angle, then apply the compound angle formula.
Pause — copy the compound angle identities for $\cos(A+B)$ and $\sin(A+B)$ that power every trig induction proof into your book.
Quick check: In the inductive step of a trig induction proof, you write $f(k+1)$ as $f(k)$ combined with one extra angle. Which tool do you then reach for to simplify?
We just saw that trig induction uses the compound angle formulas, writing $\cos((k+1)\theta)=\cos(k\theta+\theta)$ and expanding. That raises a question: exactly how does multiplying the hypothesis $(\cos k\theta+i\sin k\theta)$ by $(\cos\theta+i\sin\theta)$ produce $(\cos(k+1)\theta+i\sin(k+1)\theta)$? This card answers it → expanding and collecting real/imaginary parts recovers the compound angle identities directly.
Statement: Prove by induction that $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ for all positive integers $n$.
Step 1 — Base Case ($n = 1$):
LHS $= \cos\theta + i\sin\theta$; RHS $= \cos\theta + i\sin\theta$. LHS = RHS. True for $n = 1$.
Step 2 — Inductive Hypothesis:
Assume $(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)$
Step 3 — Inductive Step ($n = k + 1$):
Apply the inductive hypothesis:
Expand, using $i^2 = -1$:
Apply compound angle formulas:
This is the formula with $n = k + 1$. By the principle of mathematical induction, true for all $n \geq 1$. ∎
Statement: Prove by induction that $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ for all positive integers $n$.
Pause — copy the statement and complete induction proof of De Moivre's Theorem $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$ into your book.
Did you get this? True or false: when expanding $[\cos(k\theta)+i\sin(k\theta)][\cos\theta+i\sin\theta]$, the term $i\sin(k\theta)\cdot i\sin\theta$ equals $-\sin(k\theta)\sin\theta$.
Worked examples · 3 in a row, reveal as you go
Prove by induction that $\cos(n\pi) = (-1)^n$ for all integers $n \geq 0$.
Verify De Moivre's theorem for $n = 3$ directly (i.e. expand $(\cos\theta + i\sin\theta)^3$ and confirm it equals $\cos(3\theta) + i\sin(3\theta)$).
Prove by induction that $\sin(n\pi) = 0$ for all integers $n \geq 0$.
Fill the gap: In the De Moivre inductive step, after expanding the product we apply compound angle formulas to get the real part $\cos(k\theta)\cos\theta - \sin(k\theta)\sin\theta = \cos($$)$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\cos(A+B) = \cos A\cos B + \sin A\sin B$.
Activities · practice with the ideas
State the compound angle formula for $\sin(A+B)$ from memory, then verify it for $A = B = 45°$.
In the De Moivre inductive step, write out the product $[\cos(k\theta)+i\sin(k\theta)][\cos\theta+i\sin\theta]$ fully expanded, grouping real and imaginary parts.
Prove by induction that $\cos\left(\dfrac{n\pi}{2}\right)$ cycles through $1, 0, -1, 0, \ldots$ for $n = 0, 1, 2, 3, \ldots$. (Hint: use $\cos(A+\frac{\pi}{2}) = -\sin A$.)
Write the complete induction proof for De Moivre's theorem in your own words, in full exam style (base case, IH, step, conclusion).
A student claims "$\sin(k\pi)=0$ is just given; I don't need to use it." Explain why that claim is wrong in the context of proving $\cos(n\pi)=(-1)^n$.
Odd one out: Three of these statements about trigonometric induction proofs are true. Which one is FALSE?
Earlier you expanded $(\cos\theta + i\sin\theta)^2$. The result is $\cos(2\theta) + i\sin(2\theta)$ — exactly what De Moivre predicts.
The key insight is that $i^2 = -1$ turns the cross-term $i\sin\theta \cdot i\sin\theta$ into $-\sin^2\theta$, which, combined with $\cos^2\theta$, gives $\cos(2\theta)$. Every step in the general inductive proof uses exactly this same logic, scaled to $k\theta$.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. State the inductive hypothesis and the inductive step for De Moivre's theorem. You do not need to write the full proof. (2 marks)
Q2. Prove by induction that $\cos(n\pi) = (-1)^n$ for all integers $n \geq 0$. (3 marks)
Q3. Prove by induction that $(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ for all positive integers $n$. (4 marks)
Comprehensive answers (click to reveal)
Q1 (2 marks): IH: $(\cos\theta+i\sin\theta)^k = \cos(k\theta)+i\sin(k\theta)$ [1]. Step: multiply both sides by $(\cos\theta+i\sin\theta)$, expand using $i^2=-1$, apply compound angle formulas to reach $\cos((k+1)\theta)+i\sin((k+1)\theta)$ [1].
Q2 (3 marks): Base $n=0$: $\cos(0)=1=(-1)^0$ [1]. IH: $\cos(k\pi)=(-1)^k$. Step: $\cos((k+1)\pi)=\cos(k\pi+\pi)=\cos(k\pi)\cos\pi-\sin(k\pi)\sin\pi = (-1)^k\cdot(-1)-0\cdot0 = (-1)^{k+1}$ [1]. Conclusion [1].
Q3 (4 marks): Base $n=1$ [1]. IH [1]. Step: multiply IH by $(\cos\theta+i\sin\theta)$, expand, use $i^2=-1$ to get $[\cos(k\theta)\cos\theta-\sin(k\theta)\sin\theta]+i[\sin(k\theta)\cos\theta+\cos(k\theta)\sin\theta]=\cos((k+1)\theta)+i\sin((k+1)\theta)$ [1]. Conclusion [1].
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⚔ Enter the arenaClimb platforms by answering trig induction questions. Lighter alternative to the boss.
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