Trigonometric Identities Visual
Tying the identities back to the original right triangle and Pythagoras theorem.
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Derive and apply the three fundamental Pythagorean identities.
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Tying the identities back to the original right triangle and Pythagoras theorem.
bold editorial infographic, clean data-forward design, high-contrast color blocks, elegant typography hierarchy, geometr…
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 , right-angled at . The squares of the base and perpendicular add up to the square of the hypotenuse:
Present the three core identities.
Express cos, tan, sec in terms of sin.
Problem. Express the trigonometric ratios , , and entirely in terms of .
Strategy. We can use the fundamental trigonometric identity relating sine and cosine to find , and then use basic ratio definitions for the rest.
Prove a fractional trigonometric identity using sec^2 = 1 + tan^2.
Problem. Prove that using the identity .
Strategy. Look at the RHS. It only has and . This is our big clue to divide the LHS by to convert everything into these terms first!
Prove sec A(1 - sin A)(sec A + tan A) = 1.
Let's prove the identity: . It often helps to express all terms in sine and cosine. We know that is the reciprocal of and is the ratio of over . Substituting these gives: . Multiplying the terms inside the brackets in the numerator gives , which simplifies to . By rearranging the fundamental Pythagorean identity, we know is equal to . Dividing this result by the denominator's yields , completing the proof.
Evaluate 9 sec^2 A - 9 tan^2 A.
Evaluate the expression: