The Scenario
A student tries to prove the identity $\dfrac{1}{1-\sin\theta}\equiv\sec^2\theta(1+\sin\theta)$ and writes this "proof":
$$\frac{1}{1-\sin\theta}=\sec^2\theta(1+\sin\theta) \;\Rightarrow\; 1=\sec^2\theta(1-\sin\theta)(1+\sin\theta)=\sec^2\theta(1-\sin^2\theta)=\sec^2\theta\cdot\cos^2\theta=1.$$
Do you see the problem? The "proof" begins by assuming the very identity it is trying to prove. Circular reasoning proves nothing. Worse, a careless proof might divide both sides by $\cos\theta$ or $1-\sin\theta$ — quantities that can be zero for some angles.
Review · Activating Prior Knowledge
Quick check — pick the correct statement in each question before you start proving.
Dividing $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ gives $1+\tan^2\theta=\sec^2\theta$; dividing by $\sin^2\theta$ gives $1+\cot^2\theta=\cosec^2\theta$. These Pythagorean forms, plus compound-angle and double-angle formulae, are your proof toolkit.
You may add/subtract, multiply/divide by a non-zero quantity, factorise and combine fractions. But dividing by $\cos\theta$ or $\sin\theta$ is only legal when that factor is definitely non-zero — in identity proofs, that can silently destroy solutions, so avoid it.
The Question
Re-prove $\dfrac{1}{1-\sin\theta}\equiv\sec^2\theta(1+\sin\theta)$ the correct way: start from the left-hand side and transform it into the right-hand side, citing one identity at each step. Which identities do you use, and in what order?
💡 The Bridge (click to reveal)
Starting from the left: multiply numerator and denominator by $1+\sin\theta$:
$$\frac{1}{1-\sin\theta}=\frac{1+\sin\theta}{(1-\sin\theta)(1+\sin\theta)}=\frac{1+\sin\theta}{1-\sin^2\theta}=\frac{1+\sin\theta}{\cos^2\theta}=\sec^2\theta(1+\sin\theta).$$
This is the rigorous style of proof we practise today.
A trigonometric identity is a statement that two expressions are equal for all values of the variable. To prove an identity, you transform one side into the other using known formulae. The key is choosing the right formula for each step.
When proving $\text{LHS} \equiv \text{RHS}$, start from the side that looks more complex and simplify it toward the simpler side. This gives you a clear direction and avoids working "in both directions at once".
The double-angle formula for cosine comes in three equivalent forms. Each form is useful in different situations:
$$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$$
Choosing rule:
- If the expression has $1 + \cos 2A$, use $\cos 2A = 2\cos^2 A - 1$ — gives $2\cos^2 A$.
- If the expression has $1 - \cos 2A$, use $\cos 2A = 1 - 2\sin^2 A$ — gives $2\sin^2 A$.
- If the expression has $\cos^2 A - \sin^2 A$ directly, keep $\cos 2A = \cos^2 A - \sin^2 A$.
Setting $t = \tan\theta$ converts trigonometric expressions into algebraic ones. The key identities are:
$$\sin\theta = \dfrac{t}{\sqrt{1+t^2}}, \quad \cos\theta = \dfrac{1}{\sqrt{1+t^2}}, \quad \sin 2\theta = \dfrac{2t}{1+t^2}, \quad \cos 2\theta = \dfrac{1-t^2}{1+t^2}$$
This substitution is powerful for proving identities involving $\tan n\theta$, since $\tan 2\theta = \dfrac{2t}{1-t^2}$ and higher multiples follow by repeated application.
Together we prove $\dfrac{1-\cos 2\theta}{1+\cos 2\theta}\equiv\tan^2\theta$. At each step, tap the right form of $\cos 2\theta$ — three forms exist, but only one works each time.
LHS $=\dfrac{1-\cos 2\theta}{1+\cos 2\theta}$. We transform only the left-hand side; begin with the numerator $1-\cos 2\theta$.
Which form of $\cos 2\theta$ collapses $1-\cos 2\theta$ into a single squared term? Tap your answer.
Now the denominator $1+\cos 2\theta$. Which form makes it a single squared term? Tap your answer.
Put it together:
$$\text{LHS}=\frac{2\sin^2\theta}{2\cos^2\theta}=\frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta=\text{RHS}.\ \checkmark$$
Prove $\cosec 2\theta + \cot 2\theta \equiv \cot\theta$. Hint: combine into a single fraction, then apply the rule “$1+\cos 2\theta \to 2\cos^2\theta$”.
💬 Discuss with your class
When you saw $1-\cos 2\theta$, why did $1-2\sin^2\theta$ pop out at once while $2\cos^2\theta-1$ would make a mess? Generalise: which form of $\cos 2\theta$ do you reach for with $1-\cos 2A$, and which with $1+\cos 2A$?
You are the identity designer. Step 1: pick a RHS target. The tool then builds a more complex LHS for you and verifies it holds at every angle.
Choose the simpler side you want your identity to end at.
A valid proof starts from the LHS and simplifies to the RHS. This LHS was built by applying a known formula in reverse, then an algebraic move such as multiplying by $\dfrac{\cos\theta}{\cos\theta}$.
Drag the slider. Both sides must always agree — that is what makes an identity, not just an equation.
Swap your identity with a partner; each of you proves the other's. If you get stuck, the author gives one hint: which form of $\cos 2\theta$ did they use?
💬 Discuss with your class
Which of the four identities was hardest to prove, and why? Notice that every construction started from a known formula applied in reverse — how does reversing a formula create a more “complex-looking” identity?
Prove that
(i) $\dfrac{1 - \cos 2\theta}{1 + \cos 2\theta} \equiv \tan^2 \theta$
(ii) $\cosec 2\theta + \cot 2\theta \equiv \cot \theta$
(iii) $\tan 4\theta \equiv \dfrac{4t(1 - t^2)}{1 - 6t^2 + t^4}$ where $t = \tan \theta$
👁 Reveal Solution
Start from LHS (the more complex side). Choose forms of $\cos 2\theta$:
Numerator: $1 - \cos 2\theta = 1 - (1 - 2\sin^2\theta) = 2\sin^2\theta$
Denominator: $1 + \cos 2\theta = 1 + (2\cos^2\theta - 1) = 2\cos^2\theta$
LHS $= \dfrac{2\sin^2\theta}{2\cos^2\theta} = \dfrac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta =$ RHS. $\checkmark$
Start from LHS. Combine into a single fraction:
LHS $= \dfrac{1}{\sin 2\theta} + \dfrac{\cos 2\theta}{\sin 2\theta} = \dfrac{1 + \cos 2\theta}{\sin 2\theta}$
Use $1 + \cos 2\theta = 2\cos^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$:
LHS $= \dfrac{2\cos^2\theta}{2\sin\theta\cos\theta} = \dfrac{\cos\theta}{\sin\theta} = \cot\theta =$ RHS. $\checkmark$
Use $\tan 4\theta = \dfrac{2\tan 2\theta}{1 - \tan^2 2\theta}$. Substitute $\tan 2\theta = \dfrac{2t}{1-t^2}$:
$\tan^2 2\theta = \dfrac{4t^2}{(1-t^2)^2}$
$1 - \tan^2 2\theta = \dfrac{(1-t^2)^2 - 4t^2}{(1-t^2)^2} = \dfrac{1 - 2t^2 + t^4 - 4t^2}{(1-t^2)^2} = \dfrac{1 - 6t^2 + t^4}{(1-t^2)^2}$
$2\tan 2\theta = \dfrac{4t}{1-t^2}$
$\tan 4\theta = \dfrac{4t/(1-t^2)}{(1-6t^2+t^4)/(1-t^2)^2} = \dfrac{4t(1-t^2)}{1-6t^2+t^4} =$ RHS. $\checkmark$
Express
(i) $(\cos x - \sin x)^2$ in terms of $\sin 2x$
(ii) $\cos^4 x - \sin^4 x$ in terms of $\cos 2x$
(iii) $2\cos^2 x - 3\sin^2 x$ in terms of $\cos 2x$
👁 Reveal Solution
$(\cos x - \sin x)^2 = \cos^2 x - 2\sin x\cos x + \sin^2 x$
$= (\cos^2 x + \sin^2 x) - 2\sin x\cos x$
$= 1 - \sin 2x$
$\boxed{(\cos x - \sin x)^2 = 1 - \sin 2x}$
Factorise as difference of squares:
$\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)$
$= (\cos^2 x - \sin^2 x) \cdot 1 = \cos 2x$
$\boxed{\cos^4 x - \sin^4 x = \cos 2x}$
Use $\cos^2 x = \dfrac{1+\cos 2x}{2}$ and $\sin^2 x = \dfrac{1-\cos 2x}{2}$:
$2\cos^2 x - 3\sin^2 x = 2\cdot\dfrac{1+\cos 2x}{2} - 3\cdot\dfrac{1-\cos 2x}{2}$
$= (1 + \cos 2x) - \dfrac{3(1 - \cos 2x)}{2}$
$= \dfrac{2(1+\cos 2x) - 3(1-\cos 2x)}{2} = \dfrac{2 + 2\cos 2x - 3 + 3\cos 2x}{2}$
$= \dfrac{5\cos 2x - 1}{2}$
$\boxed{2\cos^2 x - 3\sin^2 x = \dfrac{5\cos 2x - 1}{2}}$
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
Simplify the following.
(i) $2\sin 3\theta \cos 3\theta$ (ii) $\cos^2 3\theta - \sin^2 3\theta$ (iii) $\cos^2 3\theta + \sin^2 3\theta$
(iv) $1 - 2\sin^2\!\left(\dfrac{\theta}{2}\right)$ (v) $\sin(\theta - \alpha)\cos\alpha + \cos(\theta - \alpha)\sin\alpha$
(vi) $3\sin\theta\cos\theta$ (vii) $\dfrac{\sin 2\theta}{2\sin\theta}$ (viii) $\cos 2\theta - 2\cos^2\theta$
💡 Hint
Use the double-angle formulae in reverse: $2\sin A\cos A = \sin 2A$, $\cos^2 A - \sin^2 A = \cos 2A$, $1 - 2\sin^2 A = \cos 2A$, $2\cos^2 A = 1 + \cos 2A$. For part (v), recognise the compound angle formula $\sin(A+B) = \sin A\cos B + \cos A\sin B$.
👁 Reveal Solution
Using $2\sin A\cos A = \sin 2A$ with $A = 3\theta$:
$\boxed{2\sin 3\theta\cos 3\theta = \sin 6\theta}$
Using $\cos^2 A - \sin^2 A = \cos 2A$ with $A = 3\theta$:
$\boxed{\cos^2 3\theta - \sin^2 3\theta = \cos 6\theta}$
Using the Pythagorean identity $\cos^2 A + \sin^2 A = 1$:
$\boxed{\cos^2 3\theta + \sin^2 3\theta = 1}$
Using $1 - 2\sin^2 A = \cos 2A$ with $A = \dfrac{\theta}{2}$:
$\cos 2\cdot\dfrac{\theta}{2} = \cos\theta$
$\boxed{1 - 2\sin^2\!\left(\dfrac{\theta}{2}\right) = \cos\theta}$
This matches $\sin A\cos B + \cos A\sin B = \sin(A+B)$ with $A = \theta-\alpha$, $B = \alpha$:
$\sin((\theta-\alpha)+\alpha) = \sin\theta$
$\boxed{\sin(\theta-\alpha)\cos\alpha + \cos(\theta-\alpha)\sin\alpha = \sin\theta}$
$3\sin\theta\cos\theta = \dfrac{3}{2}\cdot 2\sin\theta\cos\theta = \dfrac{3}{2}\sin 2\theta$
$\boxed{3\sin\theta\cos\theta = \dfrac{3}{2}\sin 2\theta}$
$\dfrac{\sin 2\theta}{2\sin\theta} = \dfrac{2\sin\theta\cos\theta}{2\sin\theta} = \cos\theta$
$\boxed{\dfrac{\sin 2\theta}{2\sin\theta} = \cos\theta}$
Using $2\cos^2\theta = 1 + \cos 2\theta$:
$\cos 2\theta - 2\cos^2\theta = \cos 2\theta - (1 + \cos 2\theta) = -1$
$\boxed{\cos 2\theta - 2\cos^2\theta = -1}$
(i) Show that $\cos(\theta - 60^{\circ}) + \cos(\theta + 60^{\circ}) \equiv \cos\theta$.
(ii) Given that $\dfrac{\cos(2x - 60^{\circ}) + \cos(2x + 60^{\circ})}{\cos(x - 60^{\circ}) + \cos(x + 60^{\circ})} = 3$, find the exact value of $\cos x$.
💡 Hint
Part (i): Expand $\cos(\theta - 60^{\circ})$ and $\cos(\theta + 60^{\circ})$ using the compound angle formula. The $\sin\theta$ terms cancel.
Part (ii): Apply the result of part (i) to both numerator and denominator. The numerator becomes $\cos 2x$ and the denominator becomes $\cos x$. Then use $\cos 2x = 2\cos^2 x - 1$.
👁 Reveal Solution
Expand using $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$:
$\cos(\theta-60^{\circ}) = \cos\theta\cos 60^{\circ} + \sin\theta\sin 60^{\circ} = \dfrac{1}{2}\cos\theta + \dfrac{\sqrt{3}}{2}\sin\theta$
$\cos(\theta+60^{\circ}) = \cos\theta\cos 60^{\circ} - \sin\theta\sin 60^{\circ} = \dfrac{1}{2}\cos\theta - \dfrac{\sqrt{3}}{2}\sin\theta$
Adding: $\cos(\theta-60^{\circ}) + \cos(\theta+60^{\circ}) = 2\cdot\dfrac{1}{2}\cos\theta = \cos\theta$. $\checkmark$
Using part (i): numerator $= \cos 2x$, denominator $= \cos x$.
So $\dfrac{\cos 2x}{\cos x} = 3$, giving $\cos 2x = 3\cos x$.
Using $\cos 2x = 2\cos^2 x - 1$:
$2\cos^2 x - 1 = 3\cos x$
$2\cos^2 x - 3\cos x - 1 = 0$
Let $u = \cos x$: $2u^2 - 3u - 1 = 0$
$u = \dfrac{3 \pm \sqrt{9 + 8}}{4} = \dfrac{3 \pm \sqrt{17}}{4}$
Check: $\dfrac{3+\sqrt{17}}{4} \approx 1.78$ (invalid, since $|\cos x| \leq 1$)
$\dfrac{3-\sqrt{17}}{4} \approx -0.281$ (valid)
$\boxed{\cos x = \dfrac{3 - \sqrt{17}}{4}}$
Solve $2\cos\!\left(\theta - \dfrac{\pi}{6}\right) = 1$ for $-\pi \leqslant \theta \leqslant \pi$
💡 Hint
Divide both sides by 2 to get $\cos\!\left(\theta - \dfrac{\pi}{6}\right) = \dfrac{1}{2}$. Let $\phi = \theta - \dfrac{\pi}{6}$. Solve $\cos\phi = \dfrac{1}{2}$, then convert back to $\theta$.
Remember: $\cos\phi = \dfrac{1}{2}$ gives $\phi = \pm\dfrac{\pi}{3} + 2n\pi$. Adjust the range for $\theta$ accordingly.
👁 Reveal Solution
$\cos\!\left(\theta - \dfrac{\pi}{6}\right) = \dfrac{1}{2}$
Let $\phi = \theta - \dfrac{\pi}{6}$. Then $\cos\phi = \dfrac{1}{2}$ gives $\phi = \pm\dfrac{\pi}{3} + 2n\pi$.
So $\theta - \dfrac{\pi}{6} = \pm\dfrac{\pi}{3} + 2n\pi$, giving $\theta = \dfrac{\pi}{6} \pm \dfrac{\pi}{3} + 2n\pi$.
$\theta = \dfrac{\pi}{6} + \dfrac{\pi}{3} = \dfrac{\pi}{2}$ (in range)
$\theta = \dfrac{\pi}{6} - \dfrac{\pi}{3} = -\dfrac{\pi}{6}$ (in range)
Other values of $n$ give solutions outside the range.
$\boxed{\theta = -\dfrac{\pi}{6}, \quad \theta = \dfrac{\pi}{2}}$
Solve $\cos(45^{\circ} - \theta) = 2\sin(30^{\circ} + \theta)$ for $-180^{\circ} \leqslant \theta \leqslant 180^{\circ}$
💡 Hint
Expand both sides using compound angle formulae:
$\cos(45^{\circ} - \theta) = \cos 45^{\circ}\cos\theta + \sin 45^{\circ}\sin\theta = \dfrac{1}{\sqrt{2}}(\cos\theta + \sin\theta)$
$2\sin(30^{\circ} + \theta) = 2(\sin 30^{\circ}\cos\theta + \cos 30^{\circ}\sin\theta) = \cos\theta + \sqrt{3}\sin\theta$
Set them equal and collect terms in $\cos\theta$ and $\sin\theta$, then find $\tan\theta$.
👁 Reveal Solution
$\cos(45^{\circ} - \theta) = \cos 45^{\circ}\cos\theta + \sin 45^{\circ}\sin\theta = \dfrac{1}{\sqrt{2}}(\cos\theta + \sin\theta)$
$2\sin(30^{\circ} + \theta) = 2\sin 30^{\circ}\cos\theta + 2\cos 30^{\circ}\sin\theta = \cos\theta + \sqrt{3}\sin\theta$
$\dfrac{1}{\sqrt{2}}(\cos\theta + \sin\theta) = \cos\theta + \sqrt{3}\sin\theta$
Multiply by $\sqrt{2}$: $\cos\theta + \sin\theta = \sqrt{2}\cos\theta + \sqrt{6}\sin\theta$
Rearrange: $(1 - \sqrt{2})\cos\theta = (\sqrt{6} - 1)\sin\theta$
$\tan\theta = \dfrac{1 - \sqrt{2}}{\sqrt{6} - 1}$
Multiply numerator and denominator by $(\sqrt{6}+1)$:
$\tan\theta = \dfrac{(1-\sqrt{2})(\sqrt{6}+1)}{(\sqrt{6}-1)(\sqrt{6}+1)} = \dfrac{(1-\sqrt{2})(\sqrt{6}+1)}{6-1} = \dfrac{(1-\sqrt{2})(\sqrt{6}+1)}{5}$
This is the exact form. (A common student error is to assume this simplifies to a rational fraction such as $-\tfrac{2}{7}$; however $7(1-\sqrt{2}) \neq -2(\sqrt{6}-1)$, since $7\sqrt{2}-2\sqrt{6} \approx 5.0005 \neq 5$. Keep the surd form.)
Numerically: $\tan\theta \approx \dfrac{(-0.4142)(3.4495)}{5} \approx -0.2858$.
$\theta = \arctan\!\left(\dfrac{(1-\sqrt{2})(\sqrt{6}+1)}{5}\right) \approx -15.9^{\circ}$
Since tan has period $180^{\circ}$, the second solution is $\theta = -15.9^{\circ} + 180^{\circ} = 164.1^{\circ}$.
Both values lie in $[-180^{\circ}, 180^{\circ}]$.
$\boxed{\theta \approx -15.9^{\circ}, \quad \theta \approx 164.1^{\circ}}$ (to 1 d.p.)
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.