Precalculus Analytic Trigonometry Syllabus

Analytic Trigonometry syllabus — full chapter and topic list

Expand any chapter to see its topics and sub-topics. This is the whole examinable outline for Analytic Trigonometry in Precalculus, not a summary of it.

1. 1

   Fundamental Identities

   3 topics

   • Reciprocal and Quotient Identities
   • Pythagorean Identities
   • Simplifying Trigonometric Expressions
2. 2

   Verifying Trigonometric Identities

   2 topics

   • Strategies for Proofs
   • Working with Both Sides
3. 3

   Solving Trigonometric Equations

   3 topics

   • Equations of Linear Type
   • Equations of Quadratic Type
   • Solutions on an Interval and General Solutions
4. 4

   Sum and Difference Formulas

   2 topics

   • Sine, Cosine, and Tangent Formulas
   • Evaluating Exact Values
5. 5

   Multiple-Angle and Product-Sum Formulas

   4 topics

   • Double-Angle Formulas
   • Half-Angle Formulas
   • Power-Reducing Formulas
   • Product-to-Sum and Sum-to-Product Formulas

Analytic Trigonometry flashcards for Precalculus

24 of 50 cards from the Analytic Trigonometry deck — real questions with worked answers.

1. State the three reciprocal identities relating $\sin\theta$, $\cos\theta$, and $\tan\theta$ to their reciprocals.

   $$\csc\theta=\frac{1}{\sin\theta},\quad \sec\theta=\frac{1}{\cos\theta},\quad \cot\theta=\frac{1}{\tan\theta}$$

2. Write the two quotient identities expressing $\tan\theta$ and $\cot\theta$ in terms of $\sin\theta$ and $\cos\theta$.

   $$\tan\theta=\frac{\sin\theta}{\cos\theta},\qquad \cot\theta=\frac{\cos\theta}{\sin\theta}$$

3. What is the fundamental Pythagorean identity relating sine and cosine?

   $$\sin^{2}\theta+\cos^{2}\theta=1$$

4. State the Pythagorean identity involving $\tan\theta$ and $\sec\theta$.

   $$1+\tan^{2}\theta=\sec^{2}\theta$$

5. State the Pythagorean identity involving $\cot\theta$ and $\csc\theta$.

   $$1+\cot^{2}\theta=\csc^{2}\theta$$

6. Using the Pythagorean identity, solve for $\sin^{2}\theta$ and for $\cos^{2}\theta$ individually.

   $$\sin^{2}\theta=1-\cos^{2}\theta,\qquad \cos^{2}\theta=1-\sin^{2}\theta$$

7. State the even/odd identities for $\sin(-\theta)$, $\cos(-\theta)$, and $\tan(-\theta)$.

   $$\sin(-\theta)=-\sin\theta,\quad \cos(-\theta)=\cos\theta,\quad \tan(-\theta)=-\tan\theta$$ Sine and tangent are odd; cosine is even.

8. When simplifying trigonometric expressions, why is it often useful to rewrite everything in terms of $\sin\theta$ and $\cos\theta$?

   Because all six functions can be expressed through $\sin\theta$ and $\cos\theta$, doing so exposes common factors and lets the Pythagorean and quotient identities combine and cancel terms.

9. Simplify $\dfrac{\sec\theta}{\csc\theta}$ to a single trigonometric function.

   $$\frac{\sec\theta}{\csc\theta}=\frac{1/\cos\theta}{1/\sin\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta$$

10. Simplify the expression $\sin\theta\,\cot\theta$.

    $$\sin\theta\cdot\frac{\cos\theta}{\sin\theta}=\cos\theta$$

11. Simplify $\dfrac{1-\cos^{2}\theta}{\sin\theta}$.

    $$\frac{1-\cos^{2}\theta}{\sin\theta}=\frac{\sin^{2}\theta}{\sin\theta}=\sin\theta$$

12. When proving a trigonometric identity, what is the general strategy for choosing which side to transform?

    Start with the more complicated side and transform it into the simpler side; work with only one side at a time, never treating the identity as an equation to solve across the equals sign.

13. List three common techniques used to prove trigonometric identities.

    Convert all functions to sines and cosines; combine fractions over a common denominator; factor or expand; multiply numerator and denominator by a conjugate; and substitute a Pythagorean identity.

14. In the 'working with both sides' proof strategy, what must both sides be reduced to for the proof to be valid?

    Both sides must be independently transformed until they reach the exact same intermediate expression; you may never move terms across the equals sign as if solving an equation.

15. Why is multiplying by a conjugate (e.g. multiplying $\dfrac{1}{1+\sin\theta}$ by $\dfrac{1-\sin\theta}{1-\sin\theta}$) helpful in proofs?

    It creates a difference of squares in the denominator, such as $1-\sin^{2}\theta=\cos^{2}\theta$, converting a sum into a single Pythagorean term that simplifies.

16. What defines a trigonometric equation of 'linear type', and how do you solve it?

    It contains a single trig function raised to the first power, e.g. $2\sin x-1=0$. Isolate the trig function ($\sin x=\tfrac12$) then find all angles satisfying it.

17. Solve $2\cos x-\sqrt{3}=0$ on the interval $[0,2\pi)$.

    $$\cos x=\frac{\sqrt{3}}{2}\;\Rightarrow\; x=\frac{\pi}{6},\;\frac{11\pi}{6}$$

18. What defines a trigonometric equation of 'quadratic type', and what is the key solving step?

    It has the form $a\,u^{2}+b\,u+c=0$ where $u$ is a trig function (e.g. $2\sin^{2}x-\sin x-1=0$). Factor or use the quadratic formula in $u$, then solve each resulting basic equation.

19. Solve $2\sin^{2}x-\sin x-1=0$ on $[0,2\pi)$.

    Factor: $(2\sin x+1)(\sin x-1)=0$. So $\sin x=-\tfrac12$ or $\sin x=1$, giving $$x=\frac{7\pi}{6},\;\frac{11\pi}{6},\;\frac{\pi}{2}$$

20. How do you handle a trig equation containing two different functions, such as $2\cos^{2}x+\sin x=1$?

    Use a Pythagorean identity to write everything in one function, e.g. replace $\cos^{2}x=1-\sin^{2}x$, giving a quadratic in $\sin x$, then solve.

21. What is the difference between finding solutions 'on an interval' versus finding the 'general solution' of a trig equation?

    Interval solutions list only angles within a bounded range like $[0,2\pi)$; the general solution adds the period multiple to cover every coterminal angle, e.g. $+2\pi n$ for sine/cosine or $+\pi n$ for tangent.

22. Write the general solution of $\sin x=\tfrac12$.

    $$x=\frac{\pi}{6}+2\pi n\quad\text{or}\quad x=\frac{5\pi}{6}+2\pi n,\quad n\in\mathbb{Z}$$

23. Write the general solution of $\tan x=1$.

    $$x=\frac{\pi}{4}+\pi n,\quad n\in\mathbb{Z}$$ (period $\pi$)

24. State the sum and difference formulas for cosine.

    $$\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$$

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