RIGHT-ANGLED TRIANGLES
SOH CAH TOA
SJMATHTUBE — Tiny guide to choose
Pythagoras
or
SOH CAH TOA
Consider a right-angled triangle with angle \(x\) such that \(0^\circ < x < 90^\circ\). This \(x\) is the angle under consideration (the given angle in the question).
Naming the sides (with respect to \(x\)):
- Hypotenuse (H) – the longest side, always opposite the \(90^\circ\) right angle.
- Opposite (O) – the side directly across from the angle \(x\).
- Adjacent (A) – the side next to / nearby the angle \(x\). It is one of the legs and is not the hypotenuse.
Side key:
- O = Opposite
- A = Adjacent
- H = Hypotenuse
When you see a right-angled triangle, first look at what the question gives you:
- Only sides mentioned (no angle \(x\)) → use Pythagoras Theorem to find the missing side.
- Angles and sides mentioned → use SOH CAH TOA to find a missing side or a missing angle.
In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (Opposite and Adjacent).
\( H^{2} = a^{2} + b^{2} \)
Here \(H\) is the hypotenuse and \(a, b\) are the two legs (Opposite and Adjacent).
SOH CAH TOA is a short cut to remember the three basic trigonometric ratios in a right-angled triangle. Once the sides are named correctly with respect to \(x\), choose the ratio that uses the sides you know and solve for the unknown.
Short cut:
- SOH → \( \sin x = \frac{O}{H} \)
- CAH → \( \cos x = \frac{A}{H} \)
- TOA → \( \tan x = \frac{O}{A} \)
Think “Sine = Opposite over Hypotenuse,
Cosine = Adjacent over Hypotenuse,
Tangent = Opposite over Adjacent” while looking at the
triangle.
It is like converting a triangle into a secret code which only mathematicians understand.
TRIGONOMETRY FORMULAE
SJMATHTUBE — Exam-ready reference sheet
\( \sin^{2}\theta + \cos^{2}\theta = 1 \)
\( \sec^{2}\theta - \tan^{2}\theta = 1 \)
\( \csc^{2}\theta - \cot^{2}\theta = 1 \)
\( \sin(A+B) = \sin A \cos B + \cos A \sin B \)
\( \sin(A-B) = \sin A \cos B - \cos A \sin B \)
\( \cos(A+B) = \cos A \cos B - \sin A \sin B \)
\( \cos(A-B) = \cos A \cos B + \sin A \sin B \)
\( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)
\( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \)
\( \sin 2\theta = 2 \sin\theta \cos\theta \)
\( \cos 2\theta = \cos^{2}\theta - \sin^{2}\theta \)
\( \quad = 1 - 2\sin^{2}\theta \)
\( \quad = 2\cos^{2}\theta - 1 \)
\( \tan 2\theta = \frac{2 \tan\theta}{1 - \tan^{2}\theta} \)
\( \sin 3\theta = 3 \sin\theta - 4 \sin^{3}\theta \)
\( \cos 3\theta = 4 \cos^{3}\theta - 3 \cos\theta \)
\( \tan 3\theta = \frac{3 \tan\theta - \tan^{3}\theta}{1 - 3 \tan^{2}\theta} \)
In terms of \( \cos\theta \):
\( \sin^{2}\frac{\theta}{2} = \frac{1 - \cos\theta}{2} \)
\( \cos^{2}\frac{\theta}{2} = \frac{1 + \cos\theta}{2} \)
Also,
\( \tan\frac{\theta}{2} = \frac{\sin\theta}{1 + \cos\theta} = \frac{1 - \cos\theta}{\sin\theta} \)
\( \sin 2\theta = \frac{2 \tan\theta}{1 + \tan^{2}\theta} \)
\( \cos 2\theta = \frac{1 - \tan^{2}\theta}{1 + \tan^{2}\theta} \)
\( \tan 2\theta = \frac{2 \tan\theta}{1 - \tan^{2}\theta} \)
\( \sin C + \sin D = 2 \sin\frac{C + D}{2} \cos\frac{C - D}{2} \)
\( \sin C - \sin D = 2 \cos\frac{C + D}{2} \sin\frac{C - D}{2} \)
\( \cos C + \cos D = 2 \cos\frac{C + D}{2} \cos\frac{C - D}{2} \)
\( \cos C - \cos D = -2 \sin\frac{C + D}{2} \sin\frac{C - D}{2} \)
Sing this to remember the RHS:
sin cos
cos sin
cos cos
minus sin sin
\( \sin A \cos B = \frac{1}{2}\big[\sin(A+B) + \sin(A-B)\big] \)
\( \cos A \sin B = \frac{1}{2}\big[\sin(A+B) - \sin(A-B)\big] \)
\( \cos A \cos B = \frac{1}{2}\big[\cos(A+B) + \cos(A-B)\big] \)
\( \sin A \sin B = \frac{1}{2}\big[\cos(A-B) - \cos(A+B)\big] \)
\( 180^\circ = \pi \text{ radians} \)
\( \text{radians} = \text{degrees} \times \frac{\pi}{180^\circ} \)
\( \text{degrees} = \text{radians} \times \frac{180^\circ}{\pi} \)
Examples:
\( 60^\circ = 60 \times \frac{\pi}{180} = \frac{\pi}{3} \).
\( \frac{\pi}{4} = \frac{\pi}{4} \times \frac{180^\circ}{\pi} = 45^\circ \).
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