Resources

In a physics laboratory experiment measuring wave interference, a detector records an optical intensity proportional to the squared amplitude I=(A1cos⁡(θ)−A2sin⁡(θ))2+(A1sin⁡(θ)+A2cos⁡(θ))2I = (A_1 \cos(\theta) - A_2 \sin(\theta))^2 + (A_1 \sin(\theta) + A_2 \cos(\theta))^2. Expand both terms using perfect square binomial patterns and simplify the expression to show that the intensity is completely independent of the phase angle θ\theta. State the simplified expression in terms of A1A_1 and A2A_2.

Answer and steps

Final answer

A_1^2 + A_2^2

Steps

01

Expand each squared term

To expand and simplify the expression for optical intensity, apply the perfect square binomial formulas: (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

  1. For the first term, let a=A1cos⁡(θ)a = A_1 \cos(\theta) and b=A2sin⁡(θ)b = A_2 \sin(\theta): (A1cos⁡(θ)−A2sin⁡(θ))2=A12cos⁡2(θ)−2A1A2cos⁡(θ)sin⁡(θ)+A22sin⁡2(θ)(A_1 \cos(\theta) - A_2 \sin(\theta))^2 = A_1^2 \cos^2(\theta) - 2A_1 A_2 \cos(\theta)\sin(\theta) + A_2^2 \sin^2(\theta)
  2. For the second term, let a=A1sin⁡(θ)a = A_1 \sin(\theta) and b=A2cos⁡(θ)b = A_2 \cos(\theta): (A1sin⁡(θ)+A2cos⁡(θ))2=A12sin⁡2(θ)+2A1A2sin⁡(θ)cos⁡(θ)+A22cos⁡2(θ)(A_1 \sin(\theta) + A_2 \cos(\theta))^2 = A_1^2 \sin^2(\theta) + 2A_1 A_2 \sin(\theta)\cos(\theta) + A_2^2 \cos^2(\theta)

Common Pitfall: Be careful not to simply square the individual parts inside the parentheses, such as writing (a−b)2=a2−b2(a - b)^2 = a^2 - b^2. The cross-term ±2ab\pm 2ab must always be included.

02

Sum the two expansions and cancel opposite terms

Add both expanded expressions together: I=[A12cos⁡2(θ)−2A1A2cos⁡(θ)sin⁡(θ)+A22sin⁡2(θ)]+[A12sin⁡2(θ)+2A1A2sin⁡(θ)cos⁡(θ)+A22cos⁡2(θ)]I = \left[A_1^2 \cos^2(\theta) - 2A_1 A_2 \cos(\theta)\sin(\theta) + A_2^2 \sin^2(\theta)\right] + \left[A_1^2 \sin^2(\theta) + 2A_1 A_2 \sin(\theta)\cos(\theta) + A_2^2 \cos^2(\theta)\right]

The middle cross-terms are opposites and sum to zero: −2A1A2cos⁡(θ)sin⁡(θ)+2A1A2sin⁡(θ)cos⁡(θ)=0-2A_1 A_2 \cos(\theta)\sin(\theta) + 2A_1 A_2 \sin(\theta)\cos(\theta) = 0

This leaves: I=A12cos⁡2(θ)+A12sin⁡2(θ)+A22sin⁡2(θ)+A22cos⁡2(θ)I = A_1^2 \cos^2(\theta) + A_1^2 \sin^2(\theta) + A_2^2 \sin^2(\theta) + A_2^2 \cos^2(\theta)

03

Factor and apply the fundamental trigonometric identity

Group like coefficients together: I=A12(cos⁡2(θ)+sin⁡2(θ))+A22(sin⁡2(θ)+cos⁡2(θ))I = A_1^2\left(\cos^2(\theta) + \sin^2(\theta)\right) + A_2^2\left(\sin^2(\theta) + \cos^2(\theta)\right)

Using the Pythagorean trigonometric identity cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1: I=A12(1)+A22(1)=A12+A22I = A_1^2(1) + A_2^2(1) = A_1^2 + A_2^2

Since the variable θ\theta is completely eliminated, the intensity II is independent of the phase angle.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student misapplies the binomial square pattern as (a ± b)² = a² ± b², omitting both cross terms completely.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    (A_1 \cos(θ\theta) - A_2 \sin(θ\theta))^2 = A_1^2 \cos^2(θ\theta) - A_2^2 \sin^2(θ\theta)   ⟹  \implies (A_1 \sin(θ\theta) + A_2 \cos(θ\theta))^2 = A_1^2 \sin^2(θ\theta) + A_2^2 \cos^2(θ\theta)

    Correct

    (A_1 \cos(θ\theta) - A_2 \sin(θ\theta))^2 = A_1^2 \cos^2(θ\theta) - 2A_1 A_2 \cos(θ\theta)\sin(θ\theta) + A_2^2 \sin^2(θ\theta)   ⟹  \implies (A_1 \sin(θ\theta) + A_2 \cos(θ\theta))^2 = A_1^2 \sin^2(θ\theta) + 2A_1 A_2 \sin(θ\theta)\cos(θ\theta) + A_2^2 \cos^2(θ\theta)

    When squaring a binomial, remember to include the middle product term ±2ab\pm 2ab and note that (−b)2=+b2(-b)^2 = +b^2. Expanding (a−b)2(a - b)^2 yields a2−2ab+b2a^2 - 2ab + b^2, not a2−b2a^2 - b^2.

  2. Step 2

    I = A_1^2 \cos^2(θ\theta) - A_2^2 \sin^2(θ\theta) + A_1^2 \sin^2(θ\theta) + A_2^2 \cos^2(θ\theta)   ⟹  \implies I = A_1^2(\cos^2(θ\theta) + \sin^2(θ\theta)) + A_2^2(\cos^2(θ\theta) - \sin^2(θ\theta))

    Correct

    I = A_1^2(\cos^2(θ\theta) + \sin^2(θ\theta)) + A_2^2(\sin^2(θ\theta) + \cos^2(θ\theta))

    Because the cross terms were missing in the expansion, the opposite terms −2A1A2cos⁡(θ)sin⁡(θ)-2A_1 A_2 \cos(\theta)\sin(\theta) and +2A1A2sin⁡(θ)cos⁡(θ)+2A_1 A_2 \sin(\theta)\cos(\theta) were never canceled, leaving an incorrect difference of squared terms for A22A_2^2.

  3. Step 3

    I = A_1^2 + A_2^2 \cos(2θ\theta)

    Correct

    I = A_1^2 + A_2^2

    This result carried over the earlier expansion error. With the correct expansions, both groups simplify via cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1, eliminating θ\theta entirely.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly factored common coefficients A12A_1^2 and A22A_2^2 from the combined terms.

You correctly recalled and applied the Pythagorean identity cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1.

You correctly identified and applied the cosine double-angle identity cos⁡2(θ)−sin⁡2(θ)=cos⁡(2θ)\cos^2(\theta) - \sin^2(\theta) = \cos(2\theta) to your grouped expression.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Omission of Binomial Cross Terms

Treating exponentiation as distributive over addition and subtraction by writing (a±b)2=a2±b2(a \pm b)^2 = a^2 \pm b^2.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the square of a binomial (a±b)2=a2±2ab+b2(a \pm b)^2 = a^2 \pm 2ab + b^2 with distributing powers over individual terms, leading to the complete omission of the cross terms.

REASONING PATH
Your ACEpath teacher’s suggested approach
Check the Middle Term

  • (a±b)2=a2±2ab+b2(a \pm b)^2 = a^2 \pm 2ab + b^2

  • Always write out the three terms of a squared binomial: first term squared, twice the product of both terms, and second term squared.

Check the Last Sign

  • The square of any real term is non-negative: (±b)2=+b2(\pm b)^2 = +b^2

  • Ensure the third term in (a−b)2(a - b)^2 always has a positive sign before applying subsequent identities.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Expand the first binomial: (A1cos⁡(θ)−A2sin⁡(θ))2=A12cos⁡2(θ)−2A1A2cos⁡(θ)sin⁡(θ)+A22sin⁡2(θ)(A_1 \cos(\theta) - A_2 \sin(\theta))^2 = A_1^2 \cos^2(\theta) - 2A_1 A_2 \cos(\theta)\sin(\theta) + A_2^2 \sin^2(\theta).

  2. Expand the second binomial: (A1sin⁡(θ)+A2cos⁡(θ))2=A12sin⁡2(θ)+2A1A2sin⁡(θ)cos⁡(θ)+A22cos⁡2(θ)(A_1 \sin(\theta) + A_2 \cos(\theta))^2 = A_1^2 \sin^2(\theta) + 2A_1 A_2 \sin(\theta)\cos(\theta) + A_2^2 \cos^2(\theta).

  3. Add both expansions together and cancel the opposite cross terms: −2A1A2cos⁡(θ)sin⁡(θ)+2A1A2sin⁡(θ)cos⁡(θ)=0-2A_1 A_2 \cos(\theta)\sin(\theta) + 2A_1 A_2 \sin(\theta)\cos(\theta) = 0.

  4. Factor the remaining terms: I=A12(cos⁡2(θ)+sin⁡2(θ))+A22(sin⁡2(θ)+cos⁡2(θ))I = A_1^2(\cos^2(\theta) + \sin^2(\theta)) + A_2^2(\sin^2(\theta) + \cos^2(\theta)).

  5. Substitute cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1 into both parts.

Final answer

A_1^2 + A_2^2

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the geometric area model for (a+b)2(a + b)^2 and (a−b)2(a - b)^2 to reinforce why the 2ab2ab cross term is required.

  • Practice expanding binomials with compound expressions such as (3x−2y)2(3x - 2y)^2 before applying them inside trigonometric identities.

OverviewACEpath Teacher's Diagnosis

02 The student forgets to double the cross term when expanding the binomials, writing ab instead of 2ab.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    (A1cos⁡(θ)−A2sin⁡(θ))2=A12cos⁡2(θ)−A1A2cos⁡(θ)sin⁡(θ)+A22sin⁡2(θ)(A_1 \cos(\theta) - A_2 \sin(\theta))^2 = A_1^2 \cos^2(\theta) - A_1 A_2 \cos(\theta)\sin(\theta) + A_2^2 \sin^2(\theta)

    Correct

    (A1cos⁡(θ)−A2sin⁡(θ))2=A12cos⁡2(θ)−2A1A2cos⁡(θ)sin⁡(θ)+A22sin⁡2(θ)(A_1 \cos(\theta) - A_2 \sin(\theta))^2 = A_1^2 \cos^2(\theta) - 2A_1 A_2 \cos(\theta)\sin(\theta) + A_2^2 \sin^2(\theta)

    When expanding (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2, the middle cross-term must be doubled. You wrote −A1A2cos⁡(θ)sin⁡(θ)-A_1 A_2 \cos(\theta)\sin(\theta) instead of −2A1A2cos⁡(θ)sin⁡(θ)-2A_1 A_2 \cos(\theta)\sin(\theta) after crossing out the coefficient 2.

  2. Step 2

    (A1sin⁡(θ)+A2cos⁡(θ))2=A12sin⁡2(θ)+A1A2sin⁡(θ)cos⁡(θ)+A22cos⁡2(θ)(A_1 \sin(\theta) + A_2 \cos(\theta))^2 = A_1^2 \sin^2(\theta) + A_1 A_2 \sin(\theta)\cos(\theta) + A_2^2 \cos^2(\theta)

    Correct

    (A1sin⁡(θ)+A2cos⁡(θ))2=A12sin⁡2(θ)+2A1A2sin⁡(θ)cos⁡(θ)+A22cos⁡2(θ)(A_1 \sin(\theta) + A_2 \cos(\theta))^2 = A_1^2 \sin^2(\theta) + 2A_1 A_2 \sin(\theta)\cos(\theta) + A_2^2 \cos^2(\theta)

    When expanding (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2, the middle cross-term must be doubled to +2ab+2ab. You wrote +A1A2sin⁡(θ)cos⁡(θ)+A_1 A_2 \sin(\theta)\cos(\theta) instead of +2A1A2sin⁡(θ)cos⁡(θ)+2A_1 A_2 \sin(\theta)\cos(\theta).

  3. Step 3

    I=A12(cos⁡2(θ)+sin⁡2(θ))+A22(sin⁡2(θ)+cos⁡2(θ))  ⟹  I=A12(1)+A22(1)I = A_1^2 (\cos^2(\theta) + \sin^2(\theta)) + A_2^2 (\sin^2(\theta) + \cos^2(\theta)) \implies I = A_1^2 (1) + A_2^2 (1)

    Correct

    I=A12(cos⁡2(θ)+sin⁡2(θ))+A22(sin⁡2(θ)+cos⁡2(θ))  ⟹  I=A12(1)+A22(1)I = A_1^2 (\cos^2(\theta) + \sin^2(\theta)) + A_2^2 (\sin^2(\theta) + \cos^2(\theta)) \implies I = A_1^2 (1) + A_2^2 (1)

    Your regrouping and use of cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1 is mathematically sound. Because both cross-terms were symmetrical opposites, they canceled out to 0 even without the factor of 2, so this step carried forward the previous expansion error without affecting the cancellation.

  4. Step 4

    I=A12+A22I = A_1^2 + A_2^2

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly squared the individual first and last terms inside both binomial expansions.

You factored out A12A_1^2 and A22A_2^2 accurately when collecting like terms.

You successfully applied the fundamental Pythagorean identity cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1 to eliminate the angle θ\theta.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Missing Cross-Term Doubling in Binomial Expansion

You initially wrote 2A1A22A_1 A_2 in the first expansion but crossed it out, omitting the factor of 2 from both middle terms.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You treated the middle term of a binomial square (a±b)2(a \pm b)^2 as simply the product ±ab\pm ab rather than the doubled product ±2ab\pm 2ab.

REASONING PATH
Your ACEpath teacher’s suggested approach
Recall the Full Expansion

  • Multiplying (a±b)(a±b)(a \pm b)(a \pm b) yields a2±ab±ab+b2=a2±2ab+b2a^2 \pm ab \pm ab + b^2 = a^2 \pm 2ab + b^2.

  • Always verify that the middle term has a factor of 2 before proceeding.

Cross-Term Check

  • Every binomial square produces three distinct terms: first squared, twice the product, and last squared.

  • Mentally compute 2×(first term)×(second term)2 \times (\text{first term}) \times (\text{second term}) whenever expanding (a±b)2(a \pm b)^2.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Expand the first term correctly: (A1cos⁡(θ)−A2sin⁡(θ))2=A12cos⁡2(θ)−2A1A2cos⁡(θ)sin⁡(θ)+A22sin⁡2(θ)(A_1 \cos(\theta) - A_2 \sin(\theta))^2 = A_1^2 \cos^2(\theta) - 2A_1 A_2 \cos(\theta)\sin(\theta) + A_2^2 \sin^2(\theta).

  2. Expand the second term correctly: (A1sin⁡(θ)+A2cos⁡(θ))2=A12sin⁡2(θ)+2A1A2sin⁡(θ)cos⁡(θ)+A22cos⁡2(θ)(A_1 \sin(\theta) + A_2 \cos(\theta))^2 = A_1^2 \sin^2(\theta) + 2A_1 A_2 \sin(\theta)\cos(\theta) + A_2^2 \cos^2(\theta).

  3. Add both expanded expressions: −2A1A2cos⁡(θ)sin⁡(θ)+2A1A2sin⁡(θ)cos⁡(θ)=0-2A_1 A_2 \cos(\theta)\sin(\theta) + 2A_1 A_2 \sin(\theta)\cos(\theta) = 0.

  4. Group coefficients: I=A12(cos⁡2(θ)+sin⁡2(θ))+A22(sin⁡2(θ)+cos⁡2(θ))=A12(1)+A22(1)=A12+A22I = A_1^2(\cos^2(\theta) + \sin^2(\theta)) + A_2^2(\sin^2(\theta) + \cos^2(\theta)) = A_1^2(1) + A_2^2(1) = A_1^2 + A_2^2.

Final answer

A12+A22A_1^2 + A_2^2

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the special product formulas (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2 and (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.

  • Practice multiplying out binomials with trigonometric coefficients using FOIL or an area model to visually track both inner and outer products.

  • Solve the practice problems below to master expanding binomial squares with compound terms.

OverviewACEpath Teacher's Diagnosis

03 The student incorrectly squares the negative term in (a - b)² as -b² instead of +b², leading to an uncancelled sine and cosine term.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    (A_1 \cos(θ\theta) - A_2 \sin(θ\theta))^2 = A_1^2 \cos^2(θ\theta) - 2A_1 A_2 \cos(θ\theta)\sin(θ\theta) - A_2^2 \sin^2(θ\theta)

    Correct

    (A_1 \cos(θ\theta) - A_2 \sin(θ\theta))^2 = A_1^2 \cos^2(θ\theta) - 2A_1 A_2 \cos(θ\theta)\sin(θ\theta) + A_2^2 \sin^2(θ\theta)

    When expanding a difference of two terms using (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2, squaring the last term yields +(−A2sin⁡(θ))2=+A22sin⁡2(θ)+(-A_2 \sin(\theta))^2 = +A_2^2 \sin^2(\theta), not a negative term.

  2. Step 2

    (A_1 \sin(θ\theta) + A_2 \cos(θ\theta))^2 = A_1^2 \sin^2(θ\theta) + 2A_1 A_2 \sin(θ\theta)\cos(θ\theta) + A_2^2 \cos^2(θ\theta)

  3. Step 3

    I = A_1^2(\cos^2(θ\theta) + \sin^2(θ\theta)) + A_2^2(\cos^2(θ\theta) - \sin^2(θ\theta))

    Correct

    I = A_1^2(\cos^2(θ\theta) + \sin^2(θ\theta)) + A_2^2(\sin^2(θ\theta) + \cos^2(θ\theta))

    The cross-terms −2A1A2cos⁡(θ)sin⁡(θ)-2A_1 A_2 \cos(\theta)\sin(\theta) and +2A1A2sin⁡(θ)cos⁡(θ)+2A_1 A_2 \sin(\theta)\cos(\theta) cancel out correctly, but the −A22sin⁡2(θ)-A_2^2 \sin^2(\theta) from Step 1 caused the grouped term for A22A_2^2 to become cos⁡2(θ)−sin⁡2(θ)\cos^2(\theta) - \sin^2(\theta) instead of cos⁡2(θ)+sin⁡2(θ)\cos^2(\theta) + \sin^2(\theta).

  4. Step 4

    I = A_1^2 + A_2^2 \cos(2θ\theta)

    Correct

    I = A_1^2 + A_2^2

    Although cos⁡2(θ)−sin⁡2(θ)=cos⁡(2θ)\cos^2(\theta) - \sin^2(\theta) = \cos(2\theta) is a valid double-angle identity, it resulted directly from the earlier sign error and left an uncancelled θ\theta term.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly recognized the perfect square binomial pattern and included both cross-terms.

You accurately cancelled the equal and opposite cross-terms during summation.

You recognized the Pythagorean identity cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1 and correctly replaced it with 11.

You demonstrated advanced recall of the double-angle formula cos⁡(2θ)=cos⁡2(θ)−sin⁡2(θ)\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta).

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Sign Error in Binomial Expansion

You treated the last term in (a−b)2(a - b)^2 as −b2-b^2 rather than +b2+b^2.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Conflating the difference of squares identity (a−b)(a+b)=a2−b2(a - b)(a + b) = a^2 - b^2 with the square of a binomial difference (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2, or forgetting that squaring any real quantity always yields a non-negative result ((−b)2=+b2(-b)^2 = +b^2).

REASONING PATH
Your ACEpath teacher’s suggested approach
Square of a Difference Rule

  • (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2

  • Always verify that the leading and trailing squared terms are both positive (+a2+a^2 and +b2+b^2). Only the middle product carries the negative sign.

Sign Sanity Check

  • (−x)2=+x2(-x)^2 = +x^2

  • Whenever a negative term is squared, write parentheses around the entire quantity, such as (−A2sin⁡(θ))2(-A_2 \sin(\theta))^2, to ensure the squared result evaluates to positive.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Correct the trailing term of the first expansion to +A22sin⁡2(θ)+ A_2^2 \sin^2(\theta).

  2. Combine like terms from both expansions: I=A12(cos⁡2(θ)+sin⁡2(θ))+A22(sin⁡2(θ)+cos⁡2(θ))I = A_1^2(\cos^2(\theta) + \sin^2(\theta)) + A_2^2(\sin^2(\theta) + \cos^2(\theta)).

  3. Apply the Pythagorean identity cos⁡2(θ)+sin⁡2(θ)=1\cos^2(\theta) + \sin^2(\theta) = 1 to both groups: I=A12(1)+A22(1)I = A_1^2(1) + A_2^2(1).

Final answer

A_1^2 + A_2^2

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the distinct algebraic patterns for (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 versus (a−b)(a+b)=a2−b2(a - b)(a + b) = a^2 - b^2.

  • When evaluating physics expressions where a quantity should be phase-independent or conserved, check the signs of your squared terms immediately if an angle remains.

Lock it in

Question

What are the three terms in the expansion of (a−b)2(a - b)^2?

Answer

a2−2ab+b2a^2 - 2ab + b^2. Remember: squaring a binomial always produces three terms, and the squared second term is always positive.

Similar question

01

Expand and simplify the algebraic expression (3x−4y)2+(4x+3y)2(3x - 4y)^2 + (4x + 3y)^2.

Answer

25x2+25y225x^2 + 25y^2

  1. Expand (3x−4y)2=9x2−24xy+16y2(3x - 4y)^2 = 9x^2 - 24xy + 16y^2.
  2. Expand (4x+3y)2=16x2+24xy+9y2(4x + 3y)^2 = 16x^2 + 24xy + 9y^2.
  3. Add the two expressions: cross terms −24xy+24xy=0-24xy + 24xy = 0.
  4. Combine like terms: (9x2+16x2)+(16y2+9y2)=25x2+25y2(9x^2 + 16x^2) + (16y^2 + 9y^2) = 25x^2 + 25y^2.
02

Simplify the trigonometric expression (cos⁡(θ)−2sin⁡(θ))2+(2cos⁡(θ)+sin⁡(θ))2(\cos(\theta) - 2\sin(\theta))^2 + (2\cos(\theta) + \sin(\theta))^2 to a single constant.

Answer

55

  1. Expand the first term: cos⁡2(θ)−4cos⁡(θ)sin⁡(θ)+4sin⁡2(θ)\cos^2(\theta) - 4\cos(\theta)\sin(\theta) + 4\sin^2(\theta).
  2. Expand the second term: 4cos⁡2(θ)+4cos⁡(θ)sin⁡(θ)+sin⁡2(θ)4\cos^2(\theta) + 4\cos(\theta)\sin(\theta) + \sin^2(\theta).
  3. Cancel cross terms: −4cos⁡(θ)sin⁡(θ)+4cos⁡(θ)sin⁡(θ)=0-4\cos(\theta)\sin(\theta) + 4\cos(\theta)\sin(\theta) = 0.
  4. Group terms: 5cos⁡2(θ)+5sin⁡2(θ)=5(cos⁡2(θ)+sin⁡2(θ))=5(1)=55\cos^2(\theta) + 5\sin^2(\theta) = 5(\cos^2(\theta) + \sin^2(\theta)) = 5(1) = 5.
03

In an AC electrical circuit, two orthogonal voltage components combine as V2=(V1cos⁡(ωt+ϕ)−V2sin⁡(ωt+ϕ))2+(V1sin⁡(ωt+ϕ)+V2cos⁡(ωt+ϕ))2V^2 = (V_1 \cos(\omega t + \phi) - V_2 \sin(\omega t + \phi))^2 + (V_1 \sin(\omega t + \phi) + V_2 \cos(\omega t + \phi))^2. Determine whether the total squared voltage V2V^2 varies with time tt, and state its simplified form.

Answer

It is constant (does not vary with time), and V2=V12+V22V^2 = V_1^2 + V_2^2.

  1. Let α=ωt+ϕ\alpha = \omega t + \phi.
  2. Expand the first square: V12cos⁡2(α)−2V1V2cos⁡(α)sin⁡(α)+V22sin⁡2(α)V_1^2 \cos^2(\alpha) - 2V_1 V_2 \cos(\alpha)\sin(\alpha) + V_2^2 \sin^2(\alpha).
  3. Expand the second square: V12sin⁡2(α)+2V1V2sin⁡(α)cos⁡(α)+V22cos⁡2(α)V_1^2 \sin^2(\alpha) + 2V_1 V_2 \sin(\alpha)\cos(\alpha) + V_2^2 \cos^2(\alpha).
  4. The cross terms cancel to 00, giving V12(cos⁡2(α)+sin⁡2(α))+V22(sin⁡2(α)+cos⁡2(α))=V12+V22V_1^2(\cos^2(\alpha) + \sin^2(\alpha)) + V_2^2(\sin^2(\alpha) + \cos^2(\alpha)) = V_1^2 + V_2^2.
Loading workspace…