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Precalculus Flashcards: Proving The Pythagorean Identity

Study Proving The Pythagorean Identity in Precalculus with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

Precalculus

Proving The Pythagorean Identity

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QUESTION
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What identity relates sin⁡2(θ)\sin^2(\theta) and cos⁡2(θ)\cos^2(\theta) for any angle θ\theta?

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ANSWER

sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1. Fundamental trig identity that holds for all angles.

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What this deck covers

This deck focuses on Proving The Pythagorean Identity, giving you a quick way to review the definitions, rules, and examples that matter most for Precalculus.

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Flashcard 1: What identity relates sin⁡2(θ)\sin^2(\theta) and cos⁡2(θ)\cos^2(\theta) for any angle θ\theta?

Answer: sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1. Fundamental trig identity that holds for all angles.

Flashcard 2: Find cos⁡(θ)\cos(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant I.

Answer: cos⁡(θ)=45\cos(\theta)=\frac{4}{5}. Use cos⁡2=1−sin⁡2=1−925=1625\cos^2=1-\sin^2=1-\frac{9}{25}=\frac{16}{25}; QI means positive.

Flashcard 3: What is sin⁡(θ)\sin(\theta) if cos⁡(θ)=513\cos(\theta)=\frac{5}{13} and θ\theta is in Quadrant I?

Answer: 1213\frac{12}{13}. Use sin⁡2(θ)=1−cos⁡2(θ)=1−25169=144169\sin^2(\theta)=1-\cos^2(\theta)=1-\frac{25}{169}=\frac{144}{169}; positive in Q1.

Flashcard 4: What is tan⁡(θ)\tan(\theta) if cos⁡(θ)=−817\cos(\theta)=-\frac{8}{17} and θ\theta is in Quadrant II?

Answer: −158-\frac{15}{8}. Calculate tan⁡(θ)=sin⁡(θ)cos⁡(θ)=15/17−8/17=−158\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{15/17}{-8/17}=-\frac{15}{8}.

Flashcard 5: What is the Pythagorean identity relating sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta)?

Answer: sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1. Fundamental identity derived from the unit circle equation x2+y2=1x^2+y^2=1.

Flashcard 6: What is the sign of tan⁡(θ)\tan(\theta) in Quadrant III?

Answer: tan⁡(θ)>0\tan(\theta)>0. Since both sin⁡<0\sin<0 and cos⁡<0\cos<0 in QIII, their ratio is positive.

Flashcard 7: What are the signs of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta) in Quadrant II?

Answer: sin⁡(θ)>0, cos⁡(θ)<0\sin(\theta)>0,\ \cos(\theta)<0. In Q2, yy-values are positive and xx-values are negative.

Flashcard 8: What is the sign of cos⁡(θ)\cos(\theta) in Quadrant II?

Answer: cos⁡(θ)<0\cos(\theta)<0. In QII, xx-values are negative on the unit circle.

Flashcard 9: What is sin⁡(θ)\sin(\theta) if tan⁡(θ)=34\tan(\theta)=\frac{3}{4} and θ\theta is in Quadrant III?

Answer: −35-\frac{3}{5}. From tan⁡(θ)=34\tan(\theta)=\frac{3}{4} and Q3 signs, sin⁡(θ)=−35\sin(\theta)=-\frac{3}{5}.

Flashcard 10: What are the signs of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta) in Quadrant I?

Answer: sin⁡(θ)>0, cos⁡(θ)>0\sin(\theta)>0,\ \cos(\theta)>0. Both coordinates are positive in the first quadrant.

Flashcard 11: What is cos⁡(θ)\cos(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant I?

Answer: 45\frac{4}{5}. Use cos⁡2(θ)=1−sin⁡2(θ)=1−925=1625\cos^2(\theta)=1-\sin^2(\theta)=1-\frac{9}{25}=\frac{16}{25}; positive in Q1.

Flashcard 12: What is cos⁡(θ)\cos(\theta) if tan⁡(θ)=34\tan(\theta)=\frac{3}{4} and θ\theta is in Quadrant I?

Answer: 45\frac{4}{5}. From tan⁡(θ)=34\tan(\theta)=\frac{3}{4} and sin⁡2+cos⁡2=1\sin^2+\cos^2=1 with Q1 signs.

Flashcard 13: Find sin⁡(θ)\sin(\theta) if cos⁡(θ)=−1213\cos(\theta)=-\frac{12}{13} and θ\theta is in Quadrant II.

Answer: sin⁡(θ)=513\sin(\theta)=\frac{5}{13}. Use sin⁡2=1−cos⁡2=1−144169=25169\sin^2=1-\cos^2=1-\frac{144}{169}=\frac{25}{169}; QII means positive.

Flashcard 14: What is tan⁡(θ)\tan(\theta) if sin⁡(θ)=−1213\sin(\theta)=-\frac{12}{13} and θ\theta is in Quadrant III?

Answer: 125\frac{12}{5}. Calculate tan⁡(θ)=sin⁡(θ)cos⁡(θ)=−12/13−5/13=125\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{-12/13}{-5/13}=\frac{12}{5}.

Flashcard 15: What is the sign of tan⁡(θ)\tan(\theta) in Quadrant II?

Answer: tan⁡(θ)<0\tan(\theta)<0. Since sin⁡>0\sin>0 and cos⁡<0\cos<0 in QII, their ratio is negative.

Flashcard 16: What are the signs of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta) in Quadrant III?

Answer: sin⁡(θ)<0, cos⁡(θ)<0\sin(\theta)<0,\ \cos(\theta)<0. Both coordinates are negative in the third quadrant.

Flashcard 17: Find cos⁡(θ)\cos(\theta) if tan⁡(θ)=−34\tan(\theta)=-\frac{3}{4} and θ\theta is in Quadrant II.

Answer: cos⁡(θ)=−45\cos(\theta)=-\frac{4}{5}. From tan⁡=−34\tan=-\frac{3}{4} and sin⁡2+cos⁡2=1\sin^2+\cos^2=1, solve; QII cos negative.

Flashcard 18: What is sin⁡(θ)\sin(\theta) if cos⁡(θ)=−817\cos(\theta)=-\frac{8}{17} and θ\theta is in Quadrant II?

Answer: 1517\frac{15}{17}. Use sin⁡2(θ)=1−cos⁡2(θ)=1−64289=225289\sin^2(\theta)=1-\cos^2(\theta)=1-\frac{64}{289}=\frac{225}{289}; positive in Q2.

Flashcard 19: What is the sign of sin⁡(θ)\sin(\theta) in Quadrant III?

Answer: sin⁡(θ)<0\sin(\theta)<0. In QIII, yy-values are negative on the unit circle.

Flashcard 20: What is the sign of sin⁡(θ)\sin(\theta) in Quadrant II?

Answer: sin⁡(θ)>0\sin(\theta)>0. In QII, yy-values are positive on the unit circle.

Flashcard 21: What is the unit-circle justification for sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1 using a point (x,y)(x,y)?

Answer: On x2+y2=1x^2+y^2=1, x=cos⁡(θ)x=\cos(\theta), y=sin⁡(θ)y=\sin(\theta). Points on unit circle satisfy x2+y2=1x^2+y^2=1, giving the identity.

Flashcard 22: What is cos⁡2(θ)\cos^2(\theta) rewritten using sin⁡(θ)\sin(\theta) from sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1?

Answer: cos⁡2(θ)=1−sin⁡2(θ)\cos^2(\theta)=1-\sin^2(\theta). Rearrange the Pythagorean identity by subtracting sin⁡2(θ)\sin^2(\theta) from both sides.

Flashcard 23: What are the signs of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta) in Quadrant IV?

Answer: sin⁡(θ)<0, cos⁡(θ)>0\sin(\theta)<0,\ \cos(\theta)>0. In Q4, xx-values are positive and yy-values are negative.

Flashcard 24: What is sin⁡(θ)\sin(\theta) if cos⁡(θ)=513\cos(\theta)=\frac{5}{13} and θ\theta is in Quadrant IV?

Answer: −1213-\frac{12}{13}. Use sin⁡2(θ)=1−cos⁡2(θ)=1−25169=144169\sin^2(\theta)=1-\cos^2(\theta)=1-\frac{25}{169}=\frac{144}{169}; negative in Q4.

Flashcard 25: Find sin⁡(θ)\sin(\theta) if cos⁡(θ)=1213\cos(\theta)=\frac{12}{13} and θ\theta is in Quadrant IV.

Answer: sin⁡(θ)=−513\sin(\theta)=-\frac{5}{13}. Use sin⁡2=1−cos⁡2=1−144169=25169\sin^2=1-\cos^2=1-\frac{144}{169}=\frac{25}{169}; QIV means negative.

Flashcard 26: Find cos⁡(θ)\cos(\theta) if tan⁡(θ)=34\tan(\theta)=\frac{3}{4} and θ\theta is in Quadrant I.

Answer: cos⁡(θ)=45\cos(\theta)=\frac{4}{5}. From tan⁡=34\tan=\frac{3}{4} and sin⁡2+cos⁡2=1\sin^2+\cos^2=1, solve system; QI positive.

Flashcard 27: What is tan⁡(θ)\tan(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant II?

Answer: −34-\frac{3}{4}. Find cos⁡(θ)=−45\cos(\theta)=-\frac{4}{5} in Q2, then tan⁡(θ)=3/5−4/5=−34\tan(\theta)=\frac{3/5}{-4/5}=-\frac{3}{4}.

Flashcard 28: What is tan⁡(θ)\tan(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant I?

Answer: 34\frac{3}{4}. Find cos⁡(θ)=45\cos(\theta)=\frac{4}{5} in Q1, then tan⁡(θ)=3/54/5=34\tan(\theta)=\frac{3/5}{4/5}=\frac{3}{4}.

Flashcard 29: Find sin⁡(θ)\sin(\theta) if tan⁡(θ)=34\tan(\theta)=\frac{3}{4} and θ\theta is in Quadrant I.

Answer: sin⁡(θ)=35\sin(\theta)=\frac{3}{5}. From tan⁡=34\tan=\frac{3}{4} and sin⁡2+cos⁡2=1\sin^2+\cos^2=1, solve system; QI positive.

Flashcard 30: State the Pythagorean identity solved for cos⁡(θ)\cos(\theta) in terms of sin⁡(θ)\sin(\theta).

Answer: cos⁡(θ)=±1−sin⁡2(θ)\cos(\theta)=\pm\sqrt{1-\sin^2(\theta)}. Rearrange Pythagorean identity; sign depends on quadrant.

Flashcard 31: Find tan⁡(θ)\tan(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and cos⁡(θ)=−45\cos(\theta)=-\frac{4}{5}.

Answer: tan⁡(θ)=−34\tan(\theta)=-\frac{3}{4}. Direct division: tan⁡=sin⁡cos⁡=3/5−4/5=−34\tan=\frac{\sin}{\cos}=\frac{3/5}{-4/5}=-\frac{3}{4}.

Flashcard 32: Find cos⁡(θ)\cos(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant II.

Answer: cos⁡(θ)=−45\cos(\theta)=-\frac{4}{5}. Use cos⁡2=1−sin⁡2=1−925=1625\cos^2=1-\sin^2=1-\frac{9}{25}=\frac{16}{25}; QII means negative.

Flashcard 33: What is the sign of cos⁡(θ)\cos(\theta) in Quadrant III?

Answer: cos⁡(θ)<0\cos(\theta)<0. In QIII, xx-values are negative on the unit circle.

Flashcard 34: What is the definition of tan⁡(θ)\tan(\theta) in terms of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta)?

Answer: tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}. Ratio of opposite to adjacent sides in a right triangle.

Flashcard 35: Find sin⁡(θ)\sin(\theta) if tan⁡(θ)=−34\tan(\theta)=-\frac{3}{4} and θ\theta is in Quadrant II.

Answer: sin⁡(θ)=35\sin(\theta)=\frac{3}{5}. From tan⁡=−34\tan=-\frac{3}{4} and sin⁡2+cos⁡2=1\sin^2+\cos^2=1, solve; QII sin positive.

Flashcard 36: Find tan⁡(θ)\tan(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and cos⁡(θ)=45\cos(\theta)=\frac{4}{5}.

Answer: tan⁡(θ)=34\tan(\theta)=\frac{3}{4}. Direct division: tan⁡=sin⁡cos⁡=3/54/5=34\tan=\frac{\sin}{\cos}=\frac{3/5}{4/5}=\frac{3}{4}.

Flashcard 37: What is cos⁡(θ)\cos(\theta) if sin⁡(θ)=−1213\sin(\theta)=-\frac{12}{13} and θ\theta is in Quadrant III?

Answer: −513-\frac{5}{13}. Use cos⁡2(θ)=1−sin⁡2(θ)=1−144169=25169\cos^2(\theta)=1-\sin^2(\theta)=1-\frac{144}{169}=\frac{25}{169}; negative in Q3.

Flashcard 38: What is sin⁡2(θ)\sin^2(\theta) rewritten using cos⁡(θ)\cos(\theta) from sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1?

Answer: sin⁡2(θ)=1−cos⁡2(θ)\sin^2(\theta)=1-\cos^2(\theta). Rearrange the Pythagorean identity by subtracting cos⁡2(θ)\cos^2(\theta) from both sides.

Flashcard 39: State the Pythagorean identity solved for sin⁡(θ)\sin(\theta) in terms of cos⁡(θ)\cos(\theta).

Answer: sin⁡(θ)=±1−cos⁡2(θ)\sin(\theta)=\pm\sqrt{1-\cos^2(\theta)}. Rearrange Pythagorean identity; sign depends on quadrant.

Flashcard 40: What is cos⁡(θ)\cos(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant II?

Answer: −45-\frac{4}{5}. Use cos⁡2(θ)=1−sin⁡2(θ)=1−925=1625\cos^2(\theta)=1-\sin^2(\theta)=1-\frac{9}{25}=\frac{16}{25}; negative in Q2.