1. Fundamental Trigonometric Identities

Trigonometric identities are mathematical equations that are true for all defined values of the angle \(\theta\). They form the building blocks for solving complex trigonometric proofs.

A. Pythagorean Identities

Derived directly from the Pythagorean theorem (\(h^2 = p^2 + b^2\)) applied to a right-angled triangle in a unit circle:

1. \(\sin^2 \theta + \cos^2 \theta = 1\)    \(\Rightarrow\)    \(\sin^2 \theta = 1 - \cos^2 \theta\)    \(\Rightarrow\)    \(\cos^2 \theta = 1 - \sin^2 \theta\)

2. \(\sec^2 \theta - \tan^2 \theta = 1\)    \(\Rightarrow\)    \(\sec^2 \theta = 1 + \tan^2 \theta\)    \(\Rightarrow\)    \(\tan^2 \theta = \sec^2 \theta - 1\)

3. \(\csc^2 \theta - \cot^2 \theta = 1\)    \(\Rightarrow\)    \(\csc^2 \theta = 1 + \cot^2 \theta\)    \(\Rightarrow\)    \(\cot^2 \theta = \csc^2 \theta - 1\)

B. Reciprocal and Quotient Identities

Reciprocal Identities:
\(\sin \theta \cdot \csc \theta = 1 \Rightarrow \csc \theta = \frac{1}{\sin \theta}\)
\(\cos \theta \cdot \sec \theta = 1 \Rightarrow \sec \theta = \frac{1}{\cos \theta}\)
\(\tan \theta \cdot \cot \theta = 1 \Rightarrow \cot \theta = \frac{1}{\tan \theta}\)
Quotient Identities:
\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)

\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)

2. The CAST Rule: Signs in the Four Quadrants

The coordinate plane is divided into four quadrants. The CAST rule helps you memorize which trigonometric ratios are positive in which quadrant. Moving counter-clockwise from the 4th quadrant, the acronym spells C-A-S-T.

Quadrant Angle Range Positive Ratios Negative Ratios
I (First) 0° to 90° All Ratios (\(\sin, \cos, \tan\)) None
II (Second) 90° to 180° \(\sin \theta\) and \(\csc \theta\) \(\cos, \sec, \tan, \cot\)
III (Third) 180° to 270° \(\tan \theta\) and \(\cot \theta\) \(\sin, \csc, \cos, \sec\)
IV (Fourth) 270° to 360° \(\cos \theta\) and \(\sec \theta\) \(\sin, \csc, \tan, \cot\)

3. Complete Conversion Formulas (All Degrees)

When calculating angles beyond 90°, you use reference angles. The rules are strict:

Quadrant I (90° - \(\theta\))

All Positive, Ratios Change

\(\sin(90^\circ - \theta) = \cos \theta\)
\(\cos(90^\circ - \theta) = \sin \theta\)
\(\tan(90^\circ - \theta) = \cot \theta\)
\(\csc(90^\circ - \theta) = \sec \theta\)
\(\sec(90^\circ - \theta) = \csc \theta\)
\(\cot(90^\circ - \theta) = \tan \theta\)

Quadrant II (90° + \(\theta\) / 180° - \(\theta\))

Only Sine/Cosecant Positive

Using 90° (Changes):
\(\sin(90^\circ + \theta) = \cos \theta\)
\(\cos(90^\circ + \theta) = -\sin \theta\)
\(\tan(90^\circ + \theta) = -\cot \theta\)

Using 180° (Stays):
\(\sin(180^\circ - \theta) = \sin \theta\)
\(\cos(180^\circ - \theta) = -\cos \theta\)
\(\tan(180^\circ - \theta) = -\tan \theta\)

Quadrant III (180° + \(\theta\) / 270° - \(\theta\))

Only Tangent/Cotangent Positive

Using 180° (Stays):
\(\sin(180^\circ + \theta) = -\sin \theta\)
\(\cos(180^\circ + \theta) = -\cos \theta\)
\(\tan(180^\circ + \theta) = \tan \theta\)

Using 270° (Changes):
\(\sin(270^\circ - \theta) = -\cos \theta\)
\(\cos(270^\circ - \theta) = -\sin \theta\)
\(\tan(270^\circ - \theta) = \cot \theta\)

Quadrant IV (270° + \(\theta\) / 360° - \(\theta\))

Only Cosine/Secant Positive

Using 270° (Changes):
\(\sin(270^\circ + \theta) = -\cos \theta\)
\(\cos(270^\circ + \theta) = \sin \theta\)
\(\tan(270^\circ + \theta) = -\cot \theta\)

Using 360° (Stays):
\(\sin(360^\circ - \theta) = -\sin \theta\)
\(\cos(360^\circ - \theta) = \cos \theta\)
\(\tan(360^\circ - \theta) = -\tan \theta\)

4. Strategies and Detailed Proofs

When asked to "Prove that LHS = RHS", follow these steps:

  1. Start with the most complex side (usually LHS) because it's easier to simplify down than to build up.
  2. Convert everything to \(\sin \theta\) and \(\cos \theta\) if you are stuck.
  3. Look for algebraic opportunities: factorizing, taking common denominators, or using \((a^2 - b^2)\).
  4. Always keep an eye on the RHS to guide your algebraic manipulation.

Example 1: Core Proof

Prove that: \(\frac{\sin \theta}{1 - \cos \theta} = \csc \theta + \cot \theta\)

Solution:
Taking the Left Hand Side (LHS):
LHS = \(\frac{\sin \theta}{1 - \cos \theta}\)

Step 1: Rationalize the denominator by multiplying numerator and denominator by the conjugate \((1 + \cos \theta)\).
= \(\frac{\sin \theta}{1 - \cos \theta} \times \frac{1 + \cos \theta}{1 + \cos \theta}\)

Step 2: Expand the denominator using \((a-b)(a+b) = a^2 - b^2\).
= \(\frac{\sin \theta(1 + \cos \theta)}{1 - \cos^2 \theta}\)

Step 3: Apply Pythagorean Identity (\(1 - \cos^2 \theta = \sin^2 \theta\)).
= \(\frac{\sin \theta(1 + \cos \theta)}{\sin^2 \theta}\)

Step 4: Cancel one \(\sin \theta\) from numerator and denominator.
= \(\frac{1 + \cos \theta}{\sin \theta}\)

Step 5: Separate the fraction.
= \(\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}\)

Step 6: Apply Reciprocal and Quotient Identities.
= \(\csc \theta + \cot \theta\)
= RHS. (Proved)

Example 2: CAST Rule Application

Find the exact value of \(\cos(210^\circ)\).

Solution:
Step 1: Express 210° using 180° or 270°. Let's use 180° so the ratio remains the same.
\(\cos(210^\circ) = \cos(180^\circ + 30^\circ)\)

Step 2: Identify the quadrant. 210° is between 180° and 270°, which is Quadrant III.
In Quadrant III, only Tangent and Cotangent are positive. Therefore, Cosine will be negative.

Step 3: Apply the conversion formula.
\(\cos(180^\circ + \theta) = -\cos \theta\)
\(\cos(180^\circ + 30^\circ) = -\cos(30^\circ)\)

Step 4: Substitute the exact standard value.
= \(-\frac{\sqrt{3}}{2}\)