Trigonometry – Test 1 Review, Inverse‐Trig Equations (3.1) & Compositions (3.2)

Administrative & Course Logistics

• Homework as Bonus Credit
  • Each homework set is worth 5 % added to the quiz category (and, if quizzes are already full, to test scores).
  • Submit as many as possible – about 20 sets exist, so finishing all can raise the quiz average by an entire letter grade.
  • Instructor does not grade home-work for accuracy (there is no feedback) because it is treated strictly as extra credit.
• Graphing on Online Tests (D2L)
  • Do not attempt to draw sin/cos graphs inside the answer box – it is slow and often unreadable.
  • Instead, sketch graphs or tables on scratch paper, photograph/scan, and upload; those images are what get graded.
• Replacement Policy
  • After the final exam, the final-exam grade replaces the single lowest unit-test grade (Test 1, 2, or 3).
  • Replacement happens only at the end of the semester once all grades are in.
  • A high final (≈ 85 – 100) produces a dramatic bump; a marginal final (≈ 70) produces only minor change.
• Time-Management Advice
  • Trigonometry requires 3–5 hrs of dedicated study per week in addition to class time.
  • Re-watch class videos multiple times, pausing to re-work every solved example.
  • Begin Sections 3.3 and 3.4 before the lecture – Section 3.4 is regarded as one of the hardest topics.

Review of Test #1 Problems

Problem 15 – Graphing a Secant Function

Given the red cosine curve (periodic guide), find the blue secant function.

1. Identify that secant is the reciprocal of cosine – it shares the same period, phase shift, and vertical shift as the parent cosine.
2. Amplitude / Vertical Stretch
   • Distance from peak to trough: $6\text{ units}\Rightarrow \text{amplitude}=\pm3$.
3. Vertical Shift
   • Expected extrema at $\pm3$ moved down 2 units to $+1$ and $-5$. ⇒ vertical translation $-2$.
4. Period
   • First maximum at $x=1$, last maximum at $x=4$ so one full cycle length $=3$.
   • Standard cosine period $2\pi$ rescaled by a factor $B$: $\displaystyle B\cdot2\pi=3\;\rightarrow\;B=\frac\pi2$.
5. Reflection? None – cosine opens downward, secant inherits the same orientation.
6. Equation
   $\boxed{\displaystyle f(x)=3\sec\left(\frac\pi2x-2\right)}$

Problem 6 – Exact Value

Evaluate $\csc^{2}\left(\frac{11\pi}{6}\right)-\sec^{2}\left(-\frac{4\pi}{3}\right)+\tan^{2}\left(-\frac{3\pi}{4}\right)$

1. Even/Odd clean-up
   • $\csc^{2}\theta$ and $\sec^{2}\theta$ are even ⇒ drop negatives.
   • $\tan^{2}\theta$ is even automatically because of the square.
   • Expression becomes $\csc^{2}!\bigl(\tfrac{11\pi}{6}\bigr)-\sec^{2}!\bigl(\tfrac{4\pi}{3}\bigr)+\tan^{2}!\bigl(\tfrac{3\pi}{4}\bigr)$.
2. Convert to reciprocals
   $\frac{1}{\sin^{2}!\theta}-\frac{1}{\cos^{2}!\phi}+\tan^{2}!\psi$
3. Substitute exact values
   • $\sin\frac{11\pi}{6}=-\tfrac12\;\Rightarrow\;\csc^{2}=\tfrac{1}{(-1/2)^{2}}=4$
   • $\cos\frac{4\pi}{3}=-\tfrac12\;\Rightarrow\;\sec^{2}=\tfrac{1}{(-1/2)^{2}}=4$
   • $\tan\frac{3\pi}{4}=-1\;\Rightarrow\;\tan^{2}=1$
4. Arithmetic $4-4+1=\boxed1$

Problem 7 – Identity with Periodic & Even/Odd Properties

Expression: $\sin^{2}!\theta-\cos^{2}!\theta-\tan^{2}!\phi+\sin^{2}!\alpha$ (specific angles were $-330^{\circ},\;510^{\circ},\; -330^{\circ},\;510^{\circ}$).
Key steps:

1. Apply even/odd to clear negatives for (\sin,\cos) and change sign for (\tan).
2. Apply periodic identities $\sin(\theta\pm360^{\circ})=\sin\theta$ etc.
3. Use $\sec^{2}\theta-\tan^{2}\theta=1$ to collapse the reciprocal pair.
4. Final result $=\boxed{-1}$.

Problem 10 – Mixed Even/Odd plus Reduction

Evaluate $\cos!\left(-\frac{11\pi}{4}\right)+\csc!\left(\frac{3\pi}{2}\right)$.

1. Even property of cosine ⇒ $\cos(-\alpha)=\cos\alpha$.
2. Reduce $\frac{11\pi}{4}-2\pi=\frac{3\pi}{4}$.
3. $\cos\frac{3\pi}{4}=-\tfrac{\sqrt2}{2}\,,\;\csc\frac{3\pi}{2}=-1$.
4. Common denominator 2 ⇒ $\displaystyle -\frac{\sqrt2+2}{2}$.

Problem 1 – Degree ⟹ Decimal Conversion & Rounding

Given $43^{\circ}\,34'\,55''$ convert to decimal degrees correct to 4 decimal places.
$\displaystyle 43+\frac{34}{60}+\frac{55}{3600}=43+0.5666667+0.0152778=43.5819445\approx\boxed{43.5819^{\circ}}$.

Section 3.1 – Solving Equations Containing Inverse Trig Functions

General Strategy

For an equation of the form $A\,f^{-1}(x)+B=C\,f^{-1}(x)+D$ ( (f^{-1}) can be (\sin^{-1},\cos^{-1},\tan^{-1})).

1. Collect like terms (all inverse-trig terms on one side, constants on the other).
2. Isolate the inverse function: $f^{-1}(x)=\dfrac{D-B}{A-C}$.
3. Apply the direct function to both sides:
   $x=f!\left(\dfrac{D-B}{A-C}\right).$
4. Evaluate using the exact value on the principal branch.

Worked Example 1

Solve $8\,\sin^{-1}x+5\pi=4\,\sin^{-1}x+2\pi$.

• Move terms ⇒ $4\,\sin^{-1}x=3\pi$.
• $\sin^{-1}x=\dfrac{3\pi}{4}$.
• Apply sine ⇒ $x=\sin\frac{3\pi}{4}=\frac{\sqrt2}{2}$.

Worked Example 2

Solve $7\,\tan^{-1}x+9\pi = 13\pi+\tan^{-1}x$.

• Rearranged ⇒ $6\,\tan^{-1}x=-\pi$.
• $\tan^{-1}x=-\dfrac{\pi}{6}$.
• $x=\tan\bigl(-\dfrac{\pi}{6}\bigr)=-\dfrac{\sqrt3}{3}$.

(Several class practice problems followed the same template.)

Section 3.2 – Compositions with Inverse Trig (Reference Triangle Method)

Reference-Triangle Construction Rules

1. Determine Quadrant:
   • Use sign and the principal-value range of the inverse function.
   • (\sin^{-1}!u\in[-\frac\pi2,\frac\pi2]), (\tan^{-1}!u\in(-\frac\pi2,\frac\pi2)), (\cos^{-1}!u\in[0,\pi]), etc.
2. Label the Triangle:
   • For (\tan^{-1}(\tfrac{1}{2})) set opposite = 1, adjacent = 2.
   • For (\csc^{-1}(-\tfrac{15}{7})) set hypotenuse = 15, opposite = −7, etc.
3. Find the Missing Side via Pythagorean Theorem.
4. Compute the Requested Function on that triangle.
5. Rationalize if a radical remains in the denominator.

Example 1

Evaluate $\displaystyle \sin\Bigl(\tan^{-1}\tfrac12\Bigr)$.

1. (\tan^{-1}(\tfrac12)) is in Quadrant I (positive).
2. Triangle: opposite 1, adjacent 2 ⇒ hyp $=\sqrt{1^{2}+2^{2}}=\sqrt5$.
3. $\sin=\dfrac{1}{\sqrt5}=\frac{\sqrt5}{5}$ (after rationalization).

Example 2

Evaluate $\displaystyle \sec\Bigl(\tan^{-1}\tfrac12\Bigr)$ using the same triangle.
$\sec=\dfrac{\sqrt5}{2}$.

Example 3

Evaluate $\displaystyle \cos\Bigl(\sin^{-1}(-\tfrac13)\Bigr)$.

1. (\sin^{-1}) negative ⇒ Quadrant IV.
2. Triangle: opposite = −1, hypotenuse = 3 ⇒ adjacent $=\sqrt{3^{2}-1^{2}}=\sqrt8=2\sqrt2$.
3. $\cos=\dfrac{2\sqrt2}{3}$.

Example 4

Evaluate $\displaystyle \cot\Bigl(\csc^{-1}(-\tfrac{15}{7})\Bigr)$.

1. (\csc^{-1}) negative ⇒ Quadrant IV (sine negative).
2. Hyp = 15, opposite = −7 ⇒ adjacent $=\sqrt{15^{2}-7^{2}}=\sqrt{225-49}=\sqrt{176}=4\sqrt{11}$.
3. $\cot=\dfrac{\text{adjacent}}{\text{opposite}}=\dfrac{4\sqrt{11}}{-7}= -\dfrac{4\sqrt{11}}{7}$.

(A few numeric values in the live session were adjusted on-the-fly; the method above is the clean, correct template.)

Key Identities & Properties Used Throughout

• Reciprocal identities
  $\csc\theta=\frac1{\sin\theta},\;\sec\theta=\frac1{\cos\theta},\;\cot\theta=\frac{\cos\theta}{\sin\theta}$
• Pythagorean identities
  $\sin^{2}\theta+\cos^{2}\theta=1$
  $1+\tan^{2}\theta=\sec^{2}\theta$
  $1+\cot^{2}\theta=\csc^{2}\theta$
• Even/Odd
  $\cos(-\theta)=\cos\theta,\quad\sec(-\theta)=\sec\theta$
  $\sin(-\theta)=-\sin\theta,\;\tan(-\theta)=-\tan\theta,$ etc.
• Periodicity
  $\sin(\theta\pm360^{\circ})=\sin\theta,\quad\tan(\theta\pm180^{\circ})=\tan\theta,$ etc.
• Quadrant Signs (CAST rule)
  Quadrant I: all +, QII: only sin/csc +, QIII: only tan/cot +, QIV: only cos/sec +.

Study Recommendations Before the Next Class

1. Finish Homework 3.1 (due Tuesday).
2. Pre-read Sections 3.3 & 3.4, attempting at least odd-numbered exercises.
3. Memorize the core identities above – they are used constantly in Test 2 and Quiz 2.
4. Practice constructing reference triangles until the procedure is automatic.

"Never give up." – Prof. R