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

Free A-Level The form R sin(θ + α) practice questions with full step-by-step worked solutions. Covers R sin form, harmonic form. Practise exam-style problems and check your method.

R sin formharmonic form
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Express 3sin⁡θ+4cos⁡θ3\sin\theta + 4\cos\theta in the form Rsin⁡(θ+α)R\sin(\theta+\alpha) where R>0R>0 and 0∘<α<90∘0^\circ<\alpha<90^\circ. State the exact value of RR.
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Worked solution

  1. Set up the identity for the harmonic form

    3sin⁡θ+4cos⁡θ=Rsin⁡(θ+α)3\sin\theta + 4\cos\theta=R\sin(\theta+\alpha)

    Introduce R and alpha to be found.

  2. Compare coefficients of sin and cos

    Rcos⁡α=3, Rsin⁡α=4R\cos\alpha=3,\ R\sin\alpha=4

    Matching each term gives two equations for R and alpha.

  3. Find R by squaring and adding the coefficients

    R=25=5R=\sqrt{25}=5

    R = sqrt(a^2+b^2) as an exact surd.

Answer
55
Question 2
2 markseasy
Which of the following is equivalent to 5sin⁡θ+12cos⁡θ5\sin\theta + 12\cos\theta? (Angles to 1 d.p.)
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Worked solution

  1. Set up the identity for the harmonic form

    5sin⁡θ+12cos⁡θ=Rsin⁡(θ+α)5\sin\theta + 12\cos\theta=R\sin(\theta+\alpha)

    Introduce R and alpha to be found.

  2. Compare coefficients of sin and cos

    Rcos⁡α=5, Rsin⁡α=12R\cos\alpha=5,\ R\sin\alpha=12

    Matching each term gives two equations for R and alpha.

  3. 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)
Question 3
3 marksintermediate
Which of the following is equivalent to sin⁡θ+3cos⁡θ\sin\theta + 3\cos\theta? (Angles to 1 d.p.)
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Worked solution

  1. Write down the expression to be rewritten

    sin⁡θ+3cos⁡θ\sin\theta + 3\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=1R\cos\alpha=1

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

  4. Equate the coefficients of cos(theta)

    Rsin⁡α=3R\sin\alpha=3

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

  5. Square both equations and add them

    R2(cos⁡2α+sin⁡2α)=12+32R^2(\cos^2\alpha+\sin^2\alpha)=1^2+3^2

    Squaring removes alpha once we use the Pythagorean identity.

  6. Identify the correct equivalent form

    sin⁡θ+3cos⁡θ=10sin⁡(θ+71.6∘)\sin\theta + 3\cos\theta=\sqrt{10}\sin(\theta + 71.6^\circ)

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

Answer
10sin⁡(θ+71.6∘)\sqrt{10}\sin(\theta + 71.6^\circ)
Question 4
5 markshard
Which of the following is equivalent to 3sin⁡θ+2cos⁡θ3\sin\theta + 2\cos\theta? (Angles to 1 d.p.)
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Worked solution

  1. Write down the expression to be rewritten

    3sin⁡θ+2cos⁡θ3\sin\theta + 2\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=3R\cos\alpha=3

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

  4. Equate the coefficients of cos(theta)

    Rsin⁡α=2R\sin\alpha=2

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

  5. Square both equations and add them

    R2(cos⁡2α+sin⁡2α)=32+22R^2(\cos^2\alpha+\sin^2\alpha)=3^2+2^2

    Squaring removes alpha once we use the Pythagorean identity.

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

    R2=13R^2=13

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

  7. Take the positive square root to find R

    R=13R=\sqrt{13}

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

  8. Divide the coefficient equations to eliminate R

    Rsin⁡αRcos⁡α=23\frac{R\sin\alpha}{R\cos\alpha}=\frac{2}{3}

    Dividing cancels R and leaves a ratio for tan alpha.

  9. Simplify to obtain tan(alpha)

    tan⁡α=23\tan\alpha=\frac{2}{3}

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

  10. Identify the correct equivalent form

    3sin⁡θ+2cos⁡θ=13sin⁡(θ+33.7∘)3\sin\theta + 2\cos\theta=\sqrt{13}\sin(\theta + 33.7^\circ)

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

Answer
13sin⁡(θ+33.7∘)\sqrt{13}\sin(\theta + 33.7^\circ)
Question 5
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

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