Precalculus - Trigonometric Relationships

40 flashcards · Mathematics · by luckydog789 (harrison holbrook)

1. tan⁡(180∘+θ)\tan(180^{\circ}+\theta)tan(180∘+θ)

   tan⁡(180∘+θ)=tan⁡(θ)\tan(180^{\circ}+\theta)=\tan(\theta)tan(180∘+θ)=tan(θ)

2. Reciprocal relationships

   Used in the following flashcards

3. csc⁡(θ)\csc({\theta})csc(θ)

   csc⁡(θ)=1sin⁡(θ)\csc({\theta})=\frac{1}{\sin({\theta})}csc(θ)=sin(θ)1​

4. sec⁡(θ)\sec({\theta})sec(θ)

   sec⁡(θ)=1cos⁡(θ)\sec({\theta})=\frac{1}{\cos({\theta})}sec(θ)=cos(θ)1​

5. cot⁡(θ)\cot({\theta})cot(θ)

   cot⁡(θ)=1tan⁡(θ)\cot({\theta})=\frac{1}{\tan({\theta})}cot(θ)=tan(θ)1​ =cos⁡(θ)sin⁡(θ)=\frac{\cos({\theta})}{\sin({\theta})}=sin(θ)cos(θ)​

6. Cofunction relationships

   Used in the following flashcards

7. sin⁡(θ) and cos⁡(θ)\sin({\theta}) \text{ and } \cos({\theta})sin(θ) and cos(θ)

   sin⁡(θ)=cos⁡(90∘−θ) andcos⁡(θ)=sin⁡(90∘−θ)\sin(\theta)=\cos(90^{\circ}-\theta) \text{ and} \cos(\theta)=\sin(90^{\circ}-\theta)sin(θ)=cos(90∘−θ) andcos(θ)=sin(90∘−θ)

8. tan⁡(θ) and cot⁡(θ)\tan({\theta}) \text{ and } \cot({\theta})tan(θ) and cot(θ)

   tan⁡(θ)=cot⁡(90∘−θ) andcot⁡(θ)=tan⁡(90∘−θ)\tan(\theta)=\cot(90^{\circ}-\theta) \text{ and} \cot(\theta)=\tan(90^{\circ}-\theta)tan(θ)=cot(90∘−θ) andcot(θ)=tan(90∘−θ)

9. sec⁡(θ) and csc⁡(θ)\sec({\theta}) \text{ and } \csc({\theta})sec(θ) and csc(θ)

   sec⁡(θ)=csc⁡(90∘−θ) andcsc⁡(θ)=sec⁡(90∘−θ)\sec(\theta)=\csc(90^{\circ}-\theta) \text{ and} \csc(\theta)=\sec(90^{\circ}-\theta)sec(θ)=csc(90∘−θ) andcsc(θ)=sec(90∘−θ)

10. Relationships with negatives

    Used in the following flashcards

11. sin⁡(−θ) andcos⁡(−θ)\sin(-\theta) \text { and} \cos(-\theta)sin(−θ) andcos(−θ)

    sin⁡(−θ)=−sin⁡(θ) andcos⁡(−θ)=cos⁡(θ)\sin(-\theta)=-\sin(\theta) \text { and} \cos(-\theta)=\cos(\theta)sin(−θ)=−sin(θ) andcos(−θ)=cos(θ)

12. csc⁡(−θ) andsec⁡(−θ)\csc(-\theta) \text { and} \sec(-\theta)csc(−θ) andsec(−θ)

    csc⁡(−θ)=−csc⁡(θ) andsec⁡(−θ)=sec⁡(θ)\csc(-\theta)=-\csc(\theta) \text { and} \sec(-\theta)=\sec(\theta)csc(−θ)=−csc(θ) andsec(−θ)=sec(θ)

13. tan⁡(−θ) andcot⁡(−θ)\tan(-\theta) \text { and} \cot(-\theta)tan(−θ) andcot(−θ)

    tan⁡(−θ)=−tan⁡(θ) andcot⁡(−θ)=−cot⁡(θ)\tan(-\theta)=-\tan(\theta) \text { and} \cot(-\theta)=-\cot(\theta)tan(−θ)=−tan(θ) andcot(−θ)=−cot(θ)

14. Pythagorean relationships

    Used in the following flashcards

15. sin⁡2(θ)+cos⁡2(θ)\sin^2(\theta)+\cos^2(\theta)sin2(θ)+cos2(θ)

    sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta)+\cos^2(\theta)=1sin2(θ)+cos2(θ)=1

16. sin⁡2(θ)\sin^2(\theta)sin2(θ)

    sin⁡2(θ)=1−cos⁡2(θ)\sin^2(\theta)=1-\cos^2(\theta)sin2(θ)=1−cos2(θ)

17. cos⁡2(θ)\cos^2(\theta)cos2(θ)

    cos⁡2(θ)=1−sin⁡2(θ)\cos^2(\theta)=1-\sin^2(\theta)cos2(θ)=1−sin2(θ)

18. sec⁡2(θ)−tan⁡2(θ)\sec^2(\theta)-\tan^2(\theta)sec2(θ)−tan2(θ)

    sec⁡2(θ)−tan⁡2(θ)=1\sec^2(\theta)-\tan^2(\theta)=1sec2(θ)−tan2(θ)=1

19. sec⁡2(θ)\sec^2(\theta)sec2(θ)

    sec⁡2(θ)=1+tan⁡2(θ)\sec^2(\theta)=1+\tan^2(\theta)sec2(θ)=1+tan2(θ)

20. tan⁡2(θ)\tan^2(\theta)tan2(θ)

    tan⁡2(θ)=sec⁡2(θ)−1\tan^2(\theta)=\sec^2(\theta)-1tan2(θ)=sec2(θ)−1

21. csc⁡2(θ)−cot⁡2(θ)\csc^2(\theta)-\cot^2(\theta)csc2(θ)−cot2(θ)

    csc⁡2(θ)−cot⁡2(θ)=1\csc^2(\theta)-\cot^2(\theta)=1csc2(θ)−cot2(θ)=1

22. csc⁡2(θ)\csc^2(\theta)csc2(θ)

    csc⁡2(θ)=1+cot⁡2(θ)\csc^2(\theta)=1+\cot^2(\theta)csc2(θ)=1+cot2(θ)

23. cot⁡2(θ)\cot^2(\theta)cot2(θ)

    cot⁡2(θ)=csc⁡2(θ)−1\cot^2(\theta)=\csc^2(\theta)-1cot2(θ)=csc2(θ)−1

24. Algebraic relationships

    Used in the following flashcards

25. sin⁡(α±β)\sin(\alpha \pm \beta)sin(α±β)

    sin⁡(α±β)=sin⁡(α)cos⁡(β)±cos⁡(α)sin⁡(β)\sin(\alpha \pm \beta)=\sin(\alpha)\cos(\beta)\pm\cos(\alpha)\sin(\beta)sin(α±β)=sin(α)cos(β)±cos(α)sin(β)

26. cos⁡(α±β)\cos(\alpha \pm \beta)cos(α±β)

    cos⁡(α±β)=cos⁡(α)cos⁡(β)∓sin⁡(α)cos⁡(β)\cos(\alpha \pm \beta)=\cos(\alpha)\cos(\beta)\mp\sin(\alpha)\cos(\beta)cos(α±β)=cos(α)cos(β)∓sin(α)cos(β)

27. tan⁡(α±β)\tan(\alpha \pm \beta)tan(α±β)

    tan⁡(α±β)=tan⁡(α)±tan⁡β1∓tan⁡(α)tan⁡(β)\tan(\alpha \pm \beta)=\frac{\tan(\alpha)\pm\tan{\beta}}{1\mp\tan(\alpha)\tan(\beta)}tan(α±β)=1∓tan(α)tan(β)tan(α)±tanβ​

28. sin⁡(2α)\sin(2\alpha)sin(2α)

    sin⁡(2α)=2sin⁡(α)cos⁡(α)\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)sin(2α)=2sin(α)cos(α)

29. cos⁡(2α)\cos(2\alpha)cos(2α)

    cos⁡(2α)=cos⁡2(α)−sin⁡2(α)=2cos⁡2(α)−1=1−2sin⁡2(α)\cos(2\alpha)=\cos^2(\alpha)-\sin^2(\alpha)=2\cos^2(\alpha)-1=1-2\sin^2(\alpha)cos(2α)=cos2(α)−sin2(α)=2cos2(α)−1=1−2sin2(α)

30. tan⁡(2α)\tan(2\alpha)tan(2α)

    tan⁡(2α)=2tan⁡(α)1−tan⁡2(α)\tan(2\alpha)=\frac{2\tan(\alpha)}{1-\tan^2(\alpha)}tan(2α)=1−tan2(α)2tan(α)​

31. sin⁡(α2)\sin(\frac{\alpha}{2})sin(2α​)

    sin⁡(α2)=±1−cos⁡(α)2\sin(\frac{\alpha}{2})=\pm\sqrt{\frac{1-\cos(\alpha)}{2}}sin(2α​)=±21−cos(α)​ ​

32. cos⁡(α2)\cos(\frac{\alpha}{2})cos(2α​)

    cos⁡(α2)=±1+cos⁡(α)2\cos(\frac{\alpha}{2})=\pm\sqrt{\frac{1+\cos(\alpha)}{2}}cos(2α​)=±21+cos(α)​ ​

33. tan⁡(α2)\tan(\frac{\alpha}{2})tan(2α​)

    tan⁡(α2)=±1−cos⁡(α)1+cos⁡(α)=sin⁡(α)1+cos⁡(α)=1−cos⁡(α)sin⁡α\tan(\frac{\alpha}{2})=\pm\sqrt{\frac{1-\cos(\alpha)}{1+\cos(\alpha)}}=\frac{\sin(\alpha)}{1+\cos(\alpha)}=\frac{1-\cos(\alpha)}{\sin{\alpha}}tan(2α​)=±1+cos(α)1−cos(α)​ ​=1+cos(α)sin(α)​=sinα1−cos(α)​

34. tan⁡(θ)\tan({\theta})tan(θ)

    tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan({\theta})=\frac{\sin({\theta})}{\cos({\theta})}tan(θ)=cos(θ)sin(θ)​

35. Misc

    Used in the following flash cards

36. sin⁡2(x)\sin^2(x)sin2(x)

    sin⁡2(x)=1−cos⁡(2x)2\sin^2(x)=\frac{1-\cos(2x)}{2}sin2(x)=21−cos(2x)​

37. cos⁡2(x)\cos^2(x)cos2(x)

    cos⁡2(x)=1+cos⁡(2x)2\cos^2(x)=\frac{1+\cos(2x)}{2}cos2(x)=21+cos(2x)​

38. 2sin⁡(A)sin⁡(B)2\sin(A)\sin(B)2sin(A)sin(B)

    2sin⁡(A)sin⁡(B)=cos⁡(A−B)−cos⁡(A+B)2\sin(A)\sin(B)=\cos(A-B)-\cos(A+B)2sin(A)sin(B)=cos(A−B)−cos(A+B)

39. 2sin⁡(A)cos⁡(B)2\sin(A)\cos(B)2sin(A)cos(B)

    2sin⁡(A)cos⁡(B)=sin⁡(A+B)+sin⁡(A−B)2\sin(A)\cos(B)=\sin(A+B)+\sin(A-B)2sin(A)cos(B)=sin(A+B)+sin(A−B)

40. 2cos⁡(A)cos⁡(B)2\cos(A)\cos(B)2cos(A)cos(B)

    2cos⁡(A)cos⁡(B)=cos⁡(A−B)+cos⁡(A+B)2\cos(A)\cos(B)=\cos(A-B)+\cos(A+B)2cos(A)cos(B)=cos(A−B)+cos(A+B)