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Trigonometric Identities

Derive and apply the three fundamental Pythagorean identities.

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Deriving the Fundamental Identity

Show how dividing Pythagoras by AC^2 yields sin^2 + cos^2 = 1.

To find the ultimate connection between sine and cosine, we start with a familiar concept: the Pythagoras theorem.

Imagine a right triangle $ABC$ , right-angled at $B$ . The squares of the base and perpendicular add up to the square of the hypotenuse:

$AB^2 + BC^2 = AC^2$

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The Three Pythagorean Identities

Present the three core identities.

cos^{2} A + sin^{2} A = 1

Valid for:

0^{\circ} \leq A \leq 9 0^{\circ}

1 + tan^{2} A = sec^{2} A

Valid for:

0^{\circ} \leq A < 9 0^{\circ}

(tan and sec undefined at 90°)

cot^{2} A + 1 = cosec^{2} A

Valid for:

0^{\circ} < A \leq 9 0^{\circ}

(cot and cosec undefined at 0°)

Example 9: Expressing Ratios

Express cos, tan, sec in terms of sin.

Problem. Express the trigonometric ratios $\cos A$ , $\tan A$ , and $\sec A$ entirely in terms of $\sin A$ .

Strategy. We can use the fundamental trigonometric identity relating sine and cosine to find $\cos A$ , and then use basic ratio definitions for the rest.

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Example 12: Proving a Complex Identity

Prove a fractional trigonometric identity using sec^2 = 1 + tan^2.

Problem. Prove that $\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}$ using the identity $\sec^2 \theta = 1 + \tan^2 \theta$ .

Strategy. Look at the RHS. It only has $\sec \theta$ and $\tan \theta$ . This is our big clue to divide the LHS by $\cos \theta$ to convert everything into these terms first!

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Your Turn: Proving an Identity

Prove sec A(1 - sin A)(sec A + tan A) = 1.

0 of 2 blanks filled

Let's prove the identity: $\sec A (1 - \sin A)(\sec A + \tan A) = 1$ . It often helps to express all terms in sine and cosine. We know that $\sec A$ is the reciprocal of and $\tan A$ is the ratio of $\sin A$ over $\cos A$ . Substituting these gives: $(1/\cos A)(1 - \sin A)(1/\cos A + \sin A/\cos A)$ . Multiplying the terms inside the brackets in the numerator gives $(1 - \sin A)(1 + \sin A)$ , which simplifies to $1 - \sin^2 A$ . By rearranging the fundamental Pythagorean identity, we know $1 - \sin^2 A$ is equal to . Dividing this result by the denominator's $\cos^2 A$ yields $1$ , completing the proof.

Identity Multiple Choice

Evaluate 9 sec^2 A - 9 tan^2 A.

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Evaluate the expression: $9 \sec^2 A - 9 \tan^2 A$

← previousTrigonometric Ratios Of 0° And 90°next →Chapter Review

Deriving the Fundamental Identity

Show how dividing Pythagoras by AC^2 yields sin^2 + cos^2 = 1.

To find the ultimate connection between sine and cosine, we start with a familiar concept: the Pythagoras theorem.

Imagine a right triangle $ABC$ , right-angled at $B$ . The squares of the base and perpendicular add up to the square of the hypotenuse:

$AB^2 + BC^2 = AC^2$

1 / 4

Example 9: Expressing Ratios

Express cos, tan, sec in terms of sin.

Problem. Express the trigonometric ratios $\cos A$ , $\tan A$ , and $\sec A$ entirely in terms of $\sin A$ .

Strategy. We can use the fundamental trigonometric identity relating sine and cosine to find $\cos A$ , and then use basic ratio definitions for the rest.

1 / 4

Example 12: Proving a Complex Identity

Prove a fractional trigonometric identity using sec^2 = 1 + tan^2.

Problem. Prove that $\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}$ using the identity $\sec^2 \theta = 1 + \tan^2 \theta$ .

Strategy. Look at the RHS. It only has $\sec \theta$ and $\tan \theta$ . This is our big clue to divide the LHS by $\cos \theta$ to convert everything into these terms first!

1 / 5

Your Turn: Proving an Identity

Prove sec A(1 - sin A)(sec A + tan A) = 1.

0 of 2 blanks filled

Identity Multiple Choice

Evaluate 9 sec^2 A - 9 tan^2 A.

1 / 5

Evaluate the expression: $9 \sec^2 A - 9 \tan^2 A$