Hard A-Level The form R sin(θ + α) Questions

Challenging, exam-style A-Level The form R sin(θ + α) questions with worked solutions. Stretch yourself on the hardest R sin form, harmonic form problems.

R sin formharmonic form
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
The expression 8sin⁡θ+15cos⁡θ8\sin\theta + 15\cos\theta is written as Rsin⁡(θ+α)R\sin(\theta+\alpha). Which of the following is α\alpha (to 1 d.p.)?
Show worked solution

Worked solution

  1. Write down the expression to be rewritten

    8sin⁡θ+15cos⁡θ8\sin\theta + 15\cos\theta

    We want a single sine (or cosine) function instead of a sum.

  2. State the target form and expand it with the addition formula

    Rsin⁡(θ+α)=Rsin⁡θcos⁡α+Rcos⁡θsin⁡αR\sin(\theta+\alpha)=R\sin\theta\cos\alpha+R\cos\theta\sin\alpha

    Comparing this with the expression lets us find R and alpha.

  3. Equate the coefficients of sin(theta)

    Rcos⁡α=8R\cos\alpha=8

    The sin(theta) terms on each side must match.

  4. Equate the coefficients of cos(theta)

    Rsin⁡α=15R\sin\alpha=15

    The cos(theta) terms on each side must match.

  5. Square both equations and add them

    R2(cos⁡2α+sin⁡2α)=82+152R^2(\cos^2\alpha+\sin^2\alpha)=8^2+15^2

    Squaring removes alpha once we use the Pythagorean identity.

  6. Apply the identity cos^2+sin^2=1 and substitute the values

    R2=289R^2=289

    cos^2 alpha + sin^2 alpha equals 1 for every angle.

  7. Take the positive square root to find R

    R=289=17R=\sqrt{289}=17

    R is a length, so we take the positive root.

  8. Divide the coefficient equations to eliminate R

    Rsin⁡αRcos⁡α=158\frac{R\sin\alpha}{R\cos\alpha}=\frac{15}{8}

    Dividing cancels R and leaves a ratio for tan alpha.

  9. Simplify to obtain tan(alpha)

    tan⁡α=158\tan\alpha=\frac{15}{8}

    Since sin/cos = tan, this gives an equation for alpha.

  10. Solve for alpha

    α=61.9∘\alpha=61.9^\circ

    Take the inverse tangent; alpha is acute here.

  11. Write the expression in the required form

    17sin⁡(θ+61.9∘)17\sin(\theta + 61.9^\circ)

    Combine the values of R and alpha into a single sine term.

  12. State the range of the sine function

    −1≤sin⁡(θ+α)≤1-1\le\sin(\theta+\alpha)\le 1

    Sine always lies between -1 and 1.

  13. Deduce the range of the whole expression

    −17≤8sin⁡θ+15cos⁡θ≤17-17\le 8\sin\theta + 15\cos\theta\le 17

    Multiplying the sine bounds by R scales the range.

  14. Find the maximum value

    sin⁡(θ+α)=1 ⇒ max⁡=17\sin(\theta+\alpha)=1\ \Rightarrow\ \max=17

    The greatest value occurs when the sine equals 1.

  15. Identify the correct value of alpha

    α=61.9∘\alpha=61.9^\circ

    alpha = arctan(b/a).

Answer
61.9∘61.9^\circ
Question 2
8 markschallenging
What is the minimum value of 5sin⁡θ+12cos⁡θ5\sin\theta + 12\cos\theta?
Show worked solution

Worked solution

  1. Write down the expression to be rewritten

    5sin⁡θ+12cos⁡θ5\sin\theta + 12\cos\theta

    We want a single sine (or cosine) function instead of a sum.

  2. State the target form and expand it with the addition formula

    Rsin⁡(θ+α)=Rsin⁡θcos⁡α+Rcos⁡θsin⁡αR\sin(\theta+\alpha)=R\sin\theta\cos\alpha+R\cos\theta\sin\alpha

    Comparing this with the expression lets us find R and alpha.

  3. Equate the coefficients of sin(theta)

    Rcos⁡α=5R\cos\alpha=5

    The sin(theta) terms on each side must match.

  4. Equate the coefficients of cos(theta)

    Rsin⁡α=12R\sin\alpha=12

    The cos(theta) terms on each side must match.

  5. Square both equations and add them

    R2(cos⁡2α+sin⁡2α)=52+122R^2(\cos^2\alpha+\sin^2\alpha)=5^2+12^2

    Squaring removes alpha once we use the Pythagorean identity.

  6. Apply the identity cos^2+sin^2=1 and substitute the values

    R2=169R^2=169

    cos^2 alpha + sin^2 alpha equals 1 for every angle.

  7. Take the positive square root to find R

    R=169=13R=\sqrt{169}=13

    R is a length, so we take the positive root.

  8. Divide the coefficient equations to eliminate R

    Rsin⁡αRcos⁡α=125\frac{R\sin\alpha}{R\cos\alpha}=\frac{12}{5}

    Dividing cancels R and leaves a ratio for tan alpha.

  9. Simplify to obtain tan(alpha)

    tan⁡α=125\tan\alpha=\frac{12}{5}

    Since sin/cos = tan, this gives an equation for alpha.

  10. Solve for alpha

    α=67.4∘\alpha=67.4^\circ

    Take the inverse tangent; alpha is acute here.

  11. Write the expression in the required form

    13sin⁡(θ+67.4∘)13\sin(\theta + 67.4^\circ)

    Combine the values of R and alpha into a single sine term.

  12. State the range of the sine function

    −1≤sin⁡(θ+α)≤1-1\le\sin(\theta+\alpha)\le 1

    Sine always lies between -1 and 1.

  13. Deduce the range of the whole expression

    −13≤5sin⁡θ+12cos⁡θ≤13-13\le 5\sin\theta + 12\cos\theta\le 13

    Multiplying the sine bounds by R scales the range.

  14. Find the maximum value

    sin⁡(θ+α)=1 ⇒ max⁡=13\sin(\theta+\alpha)=1\ \Rightarrow\ \max=13

    The greatest value occurs when the sine equals 1.

  15. Identify the minimum value

    min⁡=−13\min=-13

    The minimum of R sin(...) is -R.

Answer
−13-13
Question 3
8 markschallenging
What is the maximum value of 7sin⁡θ+24cos⁡θ7\sin\theta + 24\cos\theta?
Show worked solution

Worked solution

  1. Write down the expression to be rewritten

    7sin⁡θ+24cos⁡θ7\sin\theta + 24\cos\theta

    We want a single sine (or cosine) function instead of a sum.

  2. State the target form and expand it with the addition formula

    Rsin⁡(θ+α)=Rsin⁡θcos⁡α+Rcos⁡θsin⁡αR\sin(\theta+\alpha)=R\sin\theta\cos\alpha+R\cos\theta\sin\alpha

    Comparing this with the expression lets us find R and alpha.

  3. Equate the coefficients of sin(theta)

    Rcos⁡α=7R\cos\alpha=7

    The sin(theta) terms on each side must match.

  4. Equate the coefficients of cos(theta)

    Rsin⁡α=24R\sin\alpha=24

    The cos(theta) terms on each side must match.

  5. Square both equations and add them

    R2(cos⁡2α+sin⁡2α)=72+242R^2(\cos^2\alpha+\sin^2\alpha)=7^2+24^2

    Squaring removes alpha once we use the Pythagorean identity.

  6. Apply the identity cos^2+sin^2=1 and substitute the values

    R2=625R^2=625

    cos^2 alpha + sin^2 alpha equals 1 for every angle.

  7. Take the positive square root to find R

    R=625=25R=\sqrt{625}=25

    R is a length, so we take the positive root.

  8. Divide the coefficient equations to eliminate R

    Rsin⁡αRcos⁡α=247\frac{R\sin\alpha}{R\cos\alpha}=\frac{24}{7}

    Dividing cancels R and leaves a ratio for tan alpha.

  9. Simplify to obtain tan(alpha)

    tan⁡α=247\tan\alpha=\frac{24}{7}

    Since sin/cos = tan, this gives an equation for alpha.

  10. Solve for alpha

    α=73.7∘\alpha=73.7^\circ

    Take the inverse tangent; alpha is acute here.

  11. Write the expression in the required form

    25sin⁡(θ+73.7∘)25\sin(\theta + 73.7^\circ)

    Combine the values of R and alpha into a single sine term.

  12. State the range of the sine function

    −1≤sin⁡(θ+α)≤1-1\le\sin(\theta+\alpha)\le 1

    Sine always lies between -1 and 1.

  13. Deduce the range of the whole expression

    −25≤7sin⁡θ+24cos⁡θ≤25-25\le 7\sin\theta + 24\cos\theta\le 25

    Multiplying the sine bounds by R scales the range.

  14. Find the maximum value

    sin⁡(θ+α)=1 ⇒ max⁡=25\sin(\theta+\alpha)=1\ \Rightarrow\ \max=25

    The greatest value occurs when the sine equals 1.

  15. Identify the maximum value

    max⁡=25\max=25

    The maximum of R sin(...) is R.

Answer
2525
Question 4
8 markschallenging
The expression 9sin⁡θ+12cos⁡θ9\sin\theta + 12\cos\theta is written in the form Rsin⁡(θ+α)R\sin(\theta+\alpha) with R>0R>0. Which of the following is the value of RR?
Show worked solution

Worked solution

  1. Write down the expression to be rewritten

    9sin⁡θ+12cos⁡θ9\sin\theta + 12\cos\theta

    We want a single sine (or cosine) function instead of a sum.

  2. State the target form and expand it with the addition formula

    Rsin⁡(θ+α)=Rsin⁡θcos⁡α+Rcos⁡θsin⁡αR\sin(\theta+\alpha)=R\sin\theta\cos\alpha+R\cos\theta\sin\alpha

    Comparing this with the expression lets us find R and alpha.

  3. Equate the coefficients of sin(theta)

    Rcos⁡α=9R\cos\alpha=9

    The sin(theta) terms on each side must match.

  4. Equate the coefficients of cos(theta)

    Rsin⁡α=12R\sin\alpha=12

    The cos(theta) terms on each side must match.

  5. Square both equations and add them

    R2(cos⁡2α+sin⁡2α)=92+122R^2(\cos^2\alpha+\sin^2\alpha)=9^2+12^2

    Squaring removes alpha once we use the Pythagorean identity.

  6. Apply the identity cos^2+sin^2=1 and substitute the values

    R2=225R^2=225

    cos^2 alpha + sin^2 alpha equals 1 for every angle.

  7. Take the positive square root to find R

    R=225=15R=\sqrt{225}=15

    R is a length, so we take the positive root.

  8. Divide the coefficient equations to eliminate R

    Rsin⁡αRcos⁡α=129\frac{R\sin\alpha}{R\cos\alpha}=\frac{12}{9}

    Dividing cancels R and leaves a ratio for tan alpha.

  9. Simplify to obtain tan(alpha)

    tan⁡α=129\tan\alpha=\frac{12}{9}

    Since sin/cos = tan, this gives an equation for alpha.

  10. Solve for alpha

    α=53.1∘\alpha=53.1^\circ

    Take the inverse tangent; alpha is acute here.

  11. Write the expression in the required form

    15sin⁡(θ+53.1∘)15\sin(\theta + 53.1^\circ)

    Combine the values of R and alpha into a single sine term.

  12. State the range of the sine function

    −1≤sin⁡(θ+α)≤1-1\le\sin(\theta+\alpha)\le 1

    Sine always lies between -1 and 1.

  13. Deduce the range of the whole expression

    −15≤9sin⁡θ+12cos⁡θ≤15-15\le 9\sin\theta + 12\cos\theta\le 15

    Multiplying the sine bounds by R scales the range.

  14. Find the maximum value

    sin⁡(θ+α)=1 ⇒ max⁡=15\sin(\theta+\alpha)=1\ \Rightarrow\ \max=15

    The greatest value occurs when the sine equals 1.

  15. Identify the correct value of R

    R=225=15R=\sqrt{225}=15

    R = sqrt(a^2+b^2).

Answer
1515
Question 5
8 markschallenging
Which of the following is equivalent to 5sin⁡θ+12cos⁡θ5\sin\theta + 12\cos\theta? (Angles to 1 d.p.)
Show worked solution

Worked solution

  1. Write down the expression to be rewritten

    5sin⁡θ+12cos⁡θ5\sin\theta + 12\cos\theta

    We want a single sine (or cosine) function instead of a sum.

  2. State the target form and expand it with the addition formula

    Rsin⁡(θ+α)=Rsin⁡θcos⁡α+Rcos⁡θsin⁡αR\sin(\theta+\alpha)=R\sin\theta\cos\alpha+R\cos\theta\sin\alpha

    Comparing this with the expression lets us find R and alpha.

  3. Equate the coefficients of sin(theta)

    Rcos⁡α=5R\cos\alpha=5

    The sin(theta) terms on each side must match.

  4. Equate the coefficients of cos(theta)

    Rsin⁡α=12R\sin\alpha=12

    The cos(theta) terms on each side must match.

  5. Square both equations and add them

    R2(cos⁡2α+sin⁡2α)=52+122R^2(\cos^2\alpha+\sin^2\alpha)=5^2+12^2

    Squaring removes alpha once we use the Pythagorean identity.

  6. Apply the identity cos^2+sin^2=1 and substitute the values

    R2=169R^2=169

    cos^2 alpha + sin^2 alpha equals 1 for every angle.

  7. Take the positive square root to find R

    R=169=13R=\sqrt{169}=13

    R is a length, so we take the positive root.

  8. Divide the coefficient equations to eliminate R

    Rsin⁡αRcos⁡α=125\frac{R\sin\alpha}{R\cos\alpha}=\frac{12}{5}

    Dividing cancels R and leaves a ratio for tan alpha.

  9. Simplify to obtain tan(alpha)

    tan⁡α=125\tan\alpha=\frac{12}{5}

    Since sin/cos = tan, this gives an equation for alpha.

  10. Solve for alpha

    α=67.4∘\alpha=67.4^\circ

    Take the inverse tangent; alpha is acute here.

  11. Write the expression in the required form

    13sin⁡(θ+67.4∘)13\sin(\theta + 67.4^\circ)

    Combine the values of R and alpha into a single sine term.

  12. State the range of the sine function

    −1≤sin⁡(θ+α)≤1-1\le\sin(\theta+\alpha)\le 1

    Sine always lies between -1 and 1.

  13. Deduce the range of the whole expression

    −13≤5sin⁡θ+12cos⁡θ≤13-13\le 5\sin\theta + 12\cos\theta\le 13

    Multiplying the sine bounds by R scales the range.

  14. Find the maximum value

    sin⁡(θ+α)=1 ⇒ max⁡=13\sin(\theta+\alpha)=1\ \Rightarrow\ \max=13

    The greatest value occurs when the sine equals 1.

  15. Identify the correct equivalent form

    5sin⁡θ+12cos⁡θ=13sin⁡(θ+67.4∘)5\sin\theta + 12\cos\theta=13\sin(\theta + 67.4^\circ)

    Only R sin(theta+alpha) matches both coefficients.

Answer
13sin⁡(θ+67.4∘)13\sin(\theta + 67.4^\circ)

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