$\sin\theta = \pm\sqrt{1-\cos^2\theta}$ · $\cos\theta = \pm\sqrt{1-\sin^2\theta}$
- Use $\sin^2\theta + \cos^2\theta \equiv 1$ to get the squared value you need.
- Square-root it — remembering that this gives $\pm$.
- Use the quadrant, or the information given, to choose the correct sign.
- If a tangent is wanted, divide sine by cosine.
$\theta$ is acute and $\cos\theta = \tfrac{5}{13}$. Find the exact values of $\sin\theta$ and $\tan\theta$.
$\theta$ is obtuse and $\sin\theta = \tfrac{7}{25}$. Find $\cos\theta$ and $\tan\theta$ exactly.
$\tan\theta = -\tfrac34$ and $270° < \theta < 360°$. Find $\sin\theta$ and $\cos\theta$.
Never move terms across the $\equiv$ — that assumes what you are proving.
- Start with the messier side — there is more to simplify.
- Write any $\tan$ as $\dfrac{\sin}{\cos}$.
- Combine fractions over a common denominator.
- Look for $\sin^2 + \cos^2$ to replace with $1$, or a $1 - \sin^2$ to replace with $\cos^2$.
- State clearly when you reach the required form.
Prove that $\sin\theta\,\tan\theta + \cos\theta \equiv \dfrac{1}{\cos\theta}$.
Prove that $\dfrac{1 - \cos^2\theta}{1 + \cos\theta} \equiv 1 - \cos\theta$.
Prove that $\dfrac{\tan\theta}{\sin\theta} \equiv \dfrac{1}{\cos\theta}$.
| If you see | Replace with |
|---|---|
| $\sin^2\theta + \cos^2\theta$ | $1$ |
| $1 - \sin^2\theta$ | $\cos^2\theta$ |
| $1 - \cos^2\theta$ | $\sin^2\theta$ |
| $\tan\theta\cos\theta$ | $\sin\theta$ |
| $\dfrac{\sin\theta}{\cos\theta}$ | $\tan\theta$ |
Simplify (a) $\dfrac{\sin^2\theta}{1-\sin^2\theta}$, (b) $(1+\sin\theta)(1-\sin\theta)$, (c) $\cos^2\theta\tan^2\theta + \cos^2\theta$.
PT3
$\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}$.
PT4
$\sin^2\theta + \cos^2\theta \equiv 1$.
Why PT4
Pythagoras on the unit circle.
Notation
$\sin^2\theta$ means $(\sin\theta)^2$.
Square-rooting
Gives $\pm$; use the quadrant to choose.
Triangle trick
Sketch the right-angled triangle from the given ratio.
Proving
Work on one side only.
First move
Write every $\tan$ as $\dfrac{\sin}{\cos}$.
Second move
Combine over a common denominator.
Watch for
$1 - \cos^2\theta$ as a difference of two squares.
$\theta$ is acute with $\sin\theta = \tfrac{4}{5}$. Find $\cos\theta$ and $\tan\theta$.
▶ Show solution
$\cos^2\theta = 1 - \tfrac{16}{25} = \tfrac{9}{25}$, so $\cos\theta = \tfrac{3}{5}$ (positive as $\theta$ is acute).
$\tan\theta = \dfrac{4/5}{3/5} = \dfrac{4}{3}$
Simplify $\dfrac{1-\cos^2\theta}{\sin\theta}$.
▶ Show solution
$1 - \cos^2\theta = \sin^2\theta$
$\dfrac{\sin^2\theta}{\sin\theta} = \sin\theta$
Simplify $\tan\theta \cos\theta$.
▶ Show solution
$\dfrac{\sin\theta}{\cos\theta} \times \cos\theta = \sin\theta$
$\theta$ is obtuse with $\cos\theta = -\tfrac{3}{5}$. Find $\sin\theta$ and $\tan\theta$.
▶ Show solution
$\sin^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}$, so $\sin\theta = \pm\tfrac45$.
Obtuse means the second quadrant, where sine is positive: $\sin\theta = \dfrac{4}{5}$.
$\tan\theta = \dfrac{4/5}{-3/5} = -\dfrac{4}{3}$
Prove that $(\sin\theta + \cos\theta)^2 \equiv 1 + 2\sin\theta\cos\theta$.
▶ Show solution
$\text{LHS} = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$
$= \left(\sin^2\theta + \cos^2\theta\right) + 2\sin\theta\cos\theta$
$= 1 + 2\sin\theta\cos\theta = \text{RHS}$ ✓
Simplify $\dfrac{\cos\theta}{\tan\theta}$.
▶ Show solution
$\cos\theta \div \dfrac{\sin\theta}{\cos\theta} = \cos\theta \times \dfrac{\cos\theta}{\sin\theta}$
$= \dfrac{\cos^2\theta}{\sin\theta}$
Show that $\sin^4\theta - \cos^4\theta \equiv \sin^2\theta - \cos^2\theta$.
▶ Show solution
$\text{LHS}$ is a difference of two squares:
$= \left(\sin^2\theta - \cos^2\theta\right)\left(\sin^2\theta + \cos^2\theta\right)$
The second bracket is $1$, so
$= \sin^2\theta - \cos^2\theta = \text{RHS}$ ✓
Given $\tan\theta = \tfrac{5}{12}$ and $\theta$ is acute, find $\sin\theta$ and $\cos\theta$.
▶ Show solution
Opposite $5$, adjacent $12$, so the hypotenuse is $\sqrt{25+144} = 13$.
$\sin\theta = \dfrac{5}{13}$, $\cos\theta = \dfrac{12}{13}$
Both positive, since $\theta$ is acute.
Prove that $\dfrac{1}{1+\tan^2\theta} \equiv \cos^2\theta$.
▶ Show solution
Work on the left. First, $\tan^2\theta = \dfrac{\sin^2\theta}{\cos^2\theta}$:
$1 + \tan^2\theta = 1 + \dfrac{\sin^2\theta}{\cos^2\theta} = \dfrac{\cos^2\theta + \sin^2\theta}{\cos^2\theta}$
The numerator is $1$ by PT4, so $1 + \tan^2\theta = \dfrac{1}{\cos^2\theta}$.
Therefore $\dfrac{1}{1+\tan^2\theta} = \dfrac{1}{1/\cos^2\theta} = \cos^2\theta = \text{RHS}$ ✓
(a) Show that the equation $2\cos^2\theta + 3\sin\theta = 3$ can be written as $2\sin^2\theta - 3\sin\theta + 1 = 0$. (b) Hence solve it for $0° \leqslant \theta \leqslant 360°$. (c) Explain why one of the factors gives only one solution in the interval while the other gives two.
▶ Show solution
(a) The equation mixes $\cos^2$ with $\sin$, so convert to a single ratio using PT4:
$\cos^2\theta = 1 - \sin^2\theta$
$2\left(1 - \sin^2\theta\right) + 3\sin\theta = 3$
$2 - 2\sin^2\theta + 3\sin\theta = 3$
$0 = 2\sin^2\theta - 3\sin\theta + 1$ as required.
(b) This is a quadratic in $\sin\theta$. Let $s = \sin\theta$:
$2s^2 - 3s + 1 = 0 \;\Rightarrow\; (2s-1)(s-1) = 0$
So $\sin\theta = \tfrac12$ or $\sin\theta = 1$.
$\sin\theta = \tfrac12$: $\theta = 30°$ or $180° - 30° = 150°$
$\sin\theta = 1$: $\theta = 90°$
$\theta = 30°,\; 90°,\; 150°$
(c) The value $\sin\theta = \tfrac12$ lies strictly between $-1$ and $1$, so the horizontal line $y = \tfrac12$ cuts the sine curve twice in one full period — once going up and once coming down.
But $\sin\theta = 1$ is the maximum value of the sine function. The line $y = 1$ only touches the curve at its peak, which happens once in the interval, at $\theta = 90°$.
Whenever a trigonometric equation produces $\pm 1$ for sine or cosine, expect one fewer solution than usual.