Proving the Pythagorean Identity - Precalculus

  1. What identity relates sin⁡2(θ)\sin^2(\theta) and cos⁡2(θ)\cos^2(\theta) for any angle θ\theta?
    sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1. Fundamental trig identity that holds for all angles.
  2. Find cos⁡(θ)\cos(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant I.
    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.
  3. What is sin⁡(θ)\sin(\theta) if cos⁡(θ)=513\cos(\theta)=\frac{5}{13} and θ\theta is in Quadrant I?
    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.
  4. What is tan⁡(θ)\tan(\theta) if cos⁡(θ)=−817\cos(\theta)=-\frac{8}{17} and θ\theta is in Quadrant II?
    −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}.
  5. What is the Pythagorean identity relating sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta)?
    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.
  6. What is the sign of tan⁡(θ)\tan(\theta) in Quadrant III?
    tan⁡(θ)>0\tan(\theta)>0. Since both sin⁡<0\sin<0 and cos⁡<0\cos<0 in QIII, their ratio is positive.
  7. What are the signs of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta) in Quadrant II?
    sin⁡(θ)>0, cos⁡(θ)<0\sin(\theta)>0,\ \cos(\theta)<0. In Q2, yy-values are positive and xx-values are negative.
  8. What is the sign of cos⁡(θ)\cos(\theta) in Quadrant II?
    cos⁡(θ)<0\cos(\theta)<0. In QII, xx-values are negative on the unit circle.
  9. What is sin⁡(θ)\sin(\theta) if tan⁡(θ)=34\tan(\theta)=\frac{3}{4} and θ\theta is in Quadrant III?
    −35-\frac{3}{5}. From tan⁡(θ)=34\tan(\theta)=\frac{3}{4} and Q3 signs, sin⁡(θ)=−35\sin(\theta)=-\frac{3}{5}.
  10. What are the signs of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta) in Quadrant I?
    sin⁡(θ)>0, cos⁡(θ)>0\sin(\theta)>0,\ \cos(\theta)>0. Both coordinates are positive in the first quadrant.
  11. What is cos⁡(θ)\cos(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant I?
    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.
  12. What is cos⁡(θ)\cos(\theta) if tan⁡(θ)=34\tan(\theta)=\frac{3}{4} and θ\theta is in Quadrant I?
    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.
  13. Find sin⁡(θ)\sin(\theta) if cos⁡(θ)=−1213\cos(\theta)=-\frac{12}{13} and θ\theta is in Quadrant II.
    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.
  14. What is tan⁡(θ)\tan(\theta) if sin⁡(θ)=−1213\sin(\theta)=-\frac{12}{13} and θ\theta is in Quadrant III?
    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}.
  15. What is the sign of tan⁡(θ)\tan(\theta) in Quadrant II?
    tan⁡(θ)<0\tan(\theta)<0. Since sin⁡>0\sin>0 and cos⁡<0\cos<0 in QII, their ratio is negative.
  16. What are the signs of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta) in Quadrant III?
    sin⁡(θ)<0, cos⁡(θ)<0\sin(\theta)<0,\ \cos(\theta)<0. Both coordinates are negative in the third quadrant.
  17. Find cos⁡(θ)\cos(\theta) if tan⁡(θ)=−34\tan(\theta)=-\frac{3}{4} and θ\theta is in Quadrant II.
    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.
  18. What is sin⁡(θ)\sin(\theta) if cos⁡(θ)=−817\cos(\theta)=-\frac{8}{17} and θ\theta is in Quadrant II?
    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.
  19. What is the sign of sin⁡(θ)\sin(\theta) in Quadrant III?
    sin⁡(θ)<0\sin(\theta)<0. In QIII, yy-values are negative on the unit circle.
  20. What is the sign of sin⁡(θ)\sin(\theta) in Quadrant II?
    sin⁡(θ)>0\sin(\theta)>0. In QII, yy-values are positive on the unit circle.
  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)?
    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.
  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?
    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.
  23. What are the signs of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta) in Quadrant IV?
    sin⁡(θ)<0, cos⁡(θ)>0\sin(\theta)<0,\ \cos(\theta)>0. In Q4, xx-values are positive and yy-values are negative.
  24. What is sin⁡(θ)\sin(\theta) if cos⁡(θ)=513\cos(\theta)=\frac{5}{13} and θ\theta is in Quadrant IV?
    −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.
  25. Find sin⁡(θ)\sin(\theta) if cos⁡(θ)=1213\cos(\theta)=\frac{12}{13} and θ\theta is in Quadrant IV.
    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.
  26. Find cos⁡(θ)\cos(\theta) if tan⁡(θ)=34\tan(\theta)=\frac{3}{4} and θ\theta is in Quadrant I.
    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.
  27. What is tan⁡(θ)\tan(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant II?
    −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}.
  28. What is tan⁡(θ)\tan(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant I?
    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}.
  29. Find sin⁡(θ)\sin(\theta) if tan⁡(θ)=34\tan(\theta)=\frac{3}{4} and θ\theta is in Quadrant I.
    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.
  30. State the Pythagorean identity solved for cos⁡(θ)\cos(\theta) in terms of sin⁡(θ)\sin(\theta).
    cos⁡(θ)=±1−sin⁡2(θ)\cos(\theta)=\pm\sqrt{1-\sin^2(\theta)}. Rearrange Pythagorean identity; sign depends on quadrant.
  31. Find tan⁡(θ)\tan(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and cos⁡(θ)=−45\cos(\theta)=-\frac{4}{5}.
    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}.
  32. Find cos⁡(θ)\cos(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant II.
    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.
  33. What is the sign of cos⁡(θ)\cos(\theta) in Quadrant III?
    cos⁡(θ)<0\cos(\theta)<0. In QIII, xx-values are negative on the unit circle.
  34. What is the definition of tan⁡(θ)\tan(\theta) in terms of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta)?
    tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}. Ratio of opposite to adjacent sides in a right triangle.
  35. Find sin⁡(θ)\sin(\theta) if tan⁡(θ)=−34\tan(\theta)=-\frac{3}{4} and θ\theta is in Quadrant II.
    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.
  36. Find tan⁡(θ)\tan(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and cos⁡(θ)=45\cos(\theta)=\frac{4}{5}.
    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}.
  37. What is cos⁡(θ)\cos(\theta) if sin⁡(θ)=−1213\sin(\theta)=-\frac{12}{13} and θ\theta is in Quadrant III?
    −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.
  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?
    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.
  39. State the Pythagorean identity solved for sin⁡(θ)\sin(\theta) in terms of cos⁡(θ)\cos(\theta).
    sin⁡(θ)=±1−cos⁡2(θ)\sin(\theta)=\pm\sqrt{1-\cos^2(\theta)}. Rearrange Pythagorean identity; sign depends on quadrant.
  40. What is cos⁡(θ)\cos(\theta) if sin⁡(θ)=35\sin(\theta)=\frac{3}{5} and θ\theta is in Quadrant II?
    −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.

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