Test Problems about Trig like triangles (Algebraic expressions) and flashcards all about it

Basic Inverse Trig Evaluation

How do you evaluate arcsin⁡(x),arccos⁡(x),arctan⁡(x)

Flashcard 5: General Problem-Solving Steps

Front:
What is the overall strategy for solving any inverse trig problem?

Back:

1. Identify the function: arcsin, arccos, arctan

2. Check domain: is the input valid?

3. Set up a triangle if needed for visual understanding

4. Use trig definitions or identities: sin = opp/hyp, cos = adj/hyp, tan = opp/adj

5. Apply inverse function → get θ

6. Check range → adjust using reflection if needed

7. Simplify using Pythagorean identity if necessary

8. Final answer: ensure in correct range

• arcsin(x): Angle θ such that sin⁡θ=x\sin\theta = xsinθ=x

  • Domain: [−1,1][-1,1][−1,1]

  • Range: [−π/2,π/2][-π/2, π/2][−π/2,π/2]

• arccos(x): Angle θ such that cos⁡θ=x\cos\theta = xcosθ=x

  • Domain: [−1,1][-1,1][−1,1]

  • Range: [0,π][0, π][0,π]

• arctan(x): Angle θ such that tan⁡θ=x\tan\theta = xtanθ=x

  • Domain: (−∞,∞)(-\infty, ∞)(−∞,∞)

  • Range: (−π/2,π/2)(-π/2, π/2)(−π/2,π/2)

• Tip: Check if the number is in the domain; otherwise, undefined.

Flashcard 2: Using Right Triangles

Front:
How do you evaluate arcsin⁡(x)\arcsin(x)arcsin(x), using a triangle?

Steps 

1. Draw a right triangle. Label sides: opposite, adjacent, hypotenuse.

2. For arcsin: sin⁡θ=opposite/hypotenuse

3. For arccos: cos⁡θ=adjacent/hypotenuse

4. For arctan: tan⁡θ=opposite/adjacent

5. Check the range: arcsin [-π/2, π/2], arccos [0, π], arctan (-π/2, π/2)

Flashcard 3: Inverse Trig of Trig (Reducing Angles)

How do you evaluate arccos⁡(cos⁡θ),arcsin⁡(sin⁡θ),arctan⁡(tan⁡θ)

Evaluation steps 

Check if is in the range: First, determine if the given 𝜃

is already within the principal value range for the respective inverse function. If it is, the expression simplifies to θ

Find an equivalent angle: If θ is outside the range, find a new angle, let's call it θ′theta prime that is coterminal or otherwise equivalent to and falls within the correct range. 

Flashcard 5: General Problem-Solving Steps 

• Front: What is the overall strategy for solving any inverse trig problem?

Analyze the Expression and Identify the Function: What is the function you are evaluating? Is it a single inverse function like

arcsin(x) , or a composition like tan(arccos(x))

Check the Domain: Ensure the input value is within the defined domain for that inverse function. If it is not, there is no real-number solution.

Draw and Label a Reference Triangle: For expressions involving nested functions, assign the inverse function to a variable. For example, let θ=arccos(x)theta equals arc cosine x 𝜃=arccos(𝑥)

. Then, use the definition of the inverse function to create a right triangle with sides labeled appropriately.

Use Trigonometric Definitions to Find Missing Sides: Use the Pythagorean theorem 𝑎2+𝑏2=𝑐2 to solve for any unknown sides of the reference triangle.

Evaluate the Final Expression: Use the values from your triangle and the appropriate trigonometric definition (SOH CAH TOA) to find the answer.

Verify the Range: Ensure your final answer is in the correct range for the specific inverse trig function in the original problem. 

cos(11π/4)

(-square root 2)/2

tan(-2π/3)

square root 3

csc(19π/6)

-2

cot(π)

undefined 

sec (600degrees)

-2

arccos(3)

undefined

sin(arcsin (-2/7))

-2/7

cos(arccot(-3/10))

-(3√109)/109

arctan(-1)

-π/4

tan(arcsin(x/4)

x(√16-x²)/ 16-x²

HOMEWORK ORACTICE PROBLEMS

7. Evaluate arccos⁡(−root3/2)

(Answer / Steps):

1. Recall: arccos⁡(x) = “angle θ\thetaθ such that cos⁡θ=x

2. Range: 0≤θ≤π

3. Solve: cos⁡θ=-(root3)/2

4. It equals -π/6 Negative → Quadrant II

5. Range is 0 to π so cannot work. add π to it

6. Check range: 5π/6 is between 0 and π correct range
   ✅ Answer: θ=5π/6 or 150°

1. Why add π:

   • Cosine is even: cos⁡(−θ)=cos⁡(θ)

   • Arccos cannot output negative angles

   • To get the angle in Quadrant II (range of arccos), reflect into [0,π][θ=π−∣angle∣=π−π/6=5π/6

8. arcsin⁡(−root2/2)

Steps 

1. Recall: arcsin⁡(x)= “angle θ such that sin⁡θ=x

2. Range: −π/2≤θ≤π/2

3. Solve: sin⁡θ=(−root2/2)

4. sine → -π/4 . Negative → Quadrant IV (arcsin allows negative angles)

5. Quadrant IV angle: θ=−π/4

6. Check range: −π/4 is between −π/2 and π/2 ✅ Correct range

✅ Answer: θ=−π/4

9. arctan⁡(−root 3)

Steps

1. Recall: arctan⁡(x)= “angle θ tan⁡θ=x

2. Range: −π/2<θ<π/2

3. Solve: tan⁡θ=−root3

4. tan = -π/3 Negative → Quadrant IV

5. Check range: −π/3 is between −π/2 and π/2

✅ Answer: θ=−π/3

10. Evaluate arccos⁡(−1/2)

Steps

1. Solve: cos⁡θ=−1/2

2. Range: 0≤θ≤π

3. cosine= -π/3

4. Check range: -π/3 is not between 0 and π

5. So add π and get 2π/3

6. That is in the range

✅ Answer: θ=2π/3

11. Evaluate arcsin⁡(root2/2)

Steps

• Recall: arcsin⁡(x)=\arcsin(x) “angle θ such that sin⁡θ=x

• Range: −π/2≤θ≤π/2

• Solve: sin⁡θ=root2/2

• θ=π/4

• Range check: π/4\pi/4π/4 is between −π/2-\pi/2−π/2 and π/2\pi/2π/2 ✅ valid
  ✅ Final Answer: θ=π/4

12. arccos⁡(1)

Steps 

• Recall: arccos⁡(x)= “angle θ  such that cos⁡θ=x

• Range: 0≤θ≤π

• Solve: cos⁡θ=1

• theta= 0

• Range check: 0 is between 0 and π ✅ valid

• Why no adjustment: Already in range
  ✅ Final Answer: θ=0

13. Evaluate arctan⁡(−root3/3)

Steps 

• Recall: arctan⁡(x)= “angle θsuch that tan⁡θ=x

• Range: −π/2<θ<π/2

• Solve: tan⁡θ=−root3/3

•  θ=−π/6

• Range check: −π/2<−π/6<π/2 ✅ valid

• Why no adjustment: Already in range, no need to add π
  ✅ Final Answer: θ=−π/6\theta = -\pi/6θ=−π/6 or -30°

Homework:

Using inverse trig function to write theta as a function of X

Write θ as a function of x, given opposite = x adjacent = 4.

Steps

• Recall: tan⁡θ=opposite/adjacent=x/4

• Range: −π/2<θ<π/2

• Solve: θ=arctan⁡(x/4)

• Range check: arctan always outputs in −π/2,π/2-\pi/2 ✅ already in range

• Why no adjustment: arctan function automatically gives angles in range
  ✅ Final Answer: θ=arctan⁡(x/4)

Write θ as a function of x, given opposite = x+2 and hypotenuse=5

Steps 

• Recall: sinθ=opposite/ hypotenuse=(x+2)/5

• Range: π/2≤0 ≤π/2 

• Solve: θ=arccos⁡(x+2)/5

• Range check: Must ensure (x+2)/5​ is between -1 and 1

  • ✅ valid

• Why no adjustment needed: arcsin outputs in correct range by definition
  ✅ Final Answer:arcsin(theta)= (x+2)/5

Write θ as a function of x, given opposite = x−1 adjacent = x^2−1

STEPS

• Recall: tan⁡θ=opposite/adjacent=(x-1)/(x²-1)

• Simplify: factor out the x-1 and get 1/(x+1)

• Range: −π/2<θ<π/2

• Range check: arctan always outputs in range → ✅ no adjustment needed
  ✅ Final Answer: θ=arctan⁡1/(x+1)

HW, use the properties of inverse trigono-

metric functions to evaluate the expression.

sin(arcsin.3)

Steps

arcsin⁡(x)=the angle θ such that sin⁡θ=x\

• Range: −π/2≤θ≤π/2

• Domain: −1≤x≤1

Step 2: Check the input

• Input: 0.30

• 0.3∈[−1,1] ✅ valid

Step 3: Result sin⁡(arcsin⁡(0.3))=0.3

✅ Answer:.3

arccos(-.5)

Steps

• Range: 0≤θ≤π only Quadrant I & II angles allowed

• Solve: cos⁡θ=−0.5

• θ=−π/3

• Range check: −π/3<0-\pi/3 < 0−π/3<0 → outside [0, π]

• Why add π:

  • Arccos cannot output negative angles

  • Reflect into Quadrant II: get 2π/3

• Check range: 2π/3 0 and π ✅ valid

✅ Final Answer: θ=2π/3\

arccos(cos 7pie/2)

Steps

first find the value of cos(7π/2) and then find the angle whose arccosine is that value. 

Simplify

cos(7π/2) : The cosine function is periodic with a period of

2π so We can subtract multiples of 2π from the angle until it lies within the interval [0,2π). Subtract 2pie and get 3pie/2

Cos( 3pie/2) : 0 

arccos(0) The arccos function is defined to have a range of

[0,π]open bracket 0 comma pi close bracket We need to find the angle θtheta in this range such that cos(θ)=0

The angle is  𝜋/2

arccos(cos(7𝜋/2))=arccos(0)=𝜋/2

. 

sin(arctan 3/4)

draw a right triangle do pythagorean theorem because it says arctan (3/4) and so that would be a triangle with an angle and opposite 3 and adjacent 4 and then hypotenuse 5 and the sin there would be 3/5

Sec(arcsin 4/5)

draw a right triangle do pythagorean theorem because it says arc sin (4/5) and so that would be a triangle with an angle and opposite 4 and hypotenuse 5 and then hypotenuse would be 5 and the sec would be 5/3

cos(arctan 2)

draw a right triangle do pythagorean theorem because it says arc tan (2/1) and so that would be a triangle with an angle and opposite 2 and hypotenuse adjancet 1 and then hypotenuse would be root 5 and cos would be root5/5

sec(arctan(-3/5)

draw a right triangle do pythagorean theorem because it says arc sin (-3/5) and so that would be a triangle with an angle and opposite -3 and adjance of 5 and then hypotenuse would be root 34 and the sec would be (root 39)/5

Other problems on page 369

write an algebraic expression that is equivalent to the expression. (Hint: Sketch a right triangle, as demonstrated in Example 7.)

cot(arctanx)

Draw right triangle and opposite would be x and adjacent would be 1 and so hypotenuse would be squarre root of (1+x²) and then cot would be 1/x

Sin(arctanx)

Draw right triangle and opposite would be x and adjacent would be 1 and so hypotenuse would be squarre root of (1+x²) and then sin woul be x/ squarre root (1+x²)

csc(arctan x/root2)

Draw right triangle and opposite would be x and adjacent would be root2 and opposite would be x and so hypotenuse would be squarre root of (2+x²) and then csc would be squarre root of (x²+2)/x

cos(arcsin ((x-h)/r))

Draw right triangle and opposite would be (x-h) and hypotenuse would be r and so adjacent side would be r²-(x-h)² so then cos would be squarre root of is this also right squre root of (1-((x-h)/(r))^2)