Hard A-Level Identities and equations Questions

Challenging, exam-style A-Level Identities and equations questions with worked solutions. Stretch yourself on the hardest identity, quadratic in sin, quadratic in cos, divide by cos^2 problems.

identityquadratic in sinquadratic in cosdivide by cos^2tan^2tan=sin/cos
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
Solve 3sin⁡θ=2tan⁡θ3\sin\theta = 2\tan\theta for 0≤θ<360∘0 \le \theta < 360^\circ, giving answers to 11 decimal place where appropriate.
Show worked solution

Worked solution

  1. Write down the equation and the interval

    Solve for 0≤θ<360∘\text{Solve for } 0 \le \theta < 360^\circ

    Before solving we note the interval we are working in. This tells us how many rotations around the circle to consider, so we do not miss or invent solutions.

  2. Recall the identity we need

    tan⁡θ≡sin⁡θcos⁡θ\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}

    Rewrite tan theta as sin/cos, then clear the fraction by multiplying through by cos theta.

  3. Replace using the identity

    3sin⁡θ=2sin⁡θcos⁡θ  ⇒  3sin⁡θcos⁡θ=2sin⁡θ3\sin\theta = \dfrac{2\sin\theta}{\cos\theta} \;\Rightarrow\; 3\sin\theta\cos\theta = 2\sin\theta

    Multiplying both sides by cos theta removes the denominator.

  4. Rearrange

    3sin⁡θcos⁡θ−2sin⁡θ=03\sin\theta\cos\theta - 2\sin\theta = 0

    Bring everything to one side ready to factor.

  5. Rearrange

    sin⁡θ(3cos⁡θ−2)=0\sin\theta(3\cos\theta - 2) = 0

    Factor out the common sin theta. Do not divide by sin theta, or the sin theta = 0 solutions are lost.

  6. Set each factor equal to zero

    sin⁡θ=0orcos⁡θ=23\sin\theta = 0 \quad \text{or} \quad \cos\theta = \tfrac{2}{3}

    Splitting into separate simple equations turns one hard problem into a few easy ones.

  7. Find the principal value for \sin\theta = 0

    sin⁡θ=0⇒base angle\sin\theta = 0 \Rightarrow \text{base angle}

    We first take the inverse trig function on a calculator (or use a known exact angle) to get the first angle. Remember this only gives one angle; the interval usually contains more.

  8. List every solution of \sin\theta = 0 in the interval

    θ=0∘, 180∘\theta = 0^\circ,\ 180^\circ

    Using the symmetry of the graph (and adding on the period) we write down all the angles in the interval that give this value. For sine and cosine the pattern repeats every 360^\circ; for tangent every 180^\circ.

  9. Find the principal value for \cos\theta = \tfrac{2}{3}

    cos⁡θ=23⇒base angle\cos\theta = \tfrac{2}{3} \Rightarrow \text{base angle}

    We first take the inverse trig function on a calculator (or use a known exact angle) to get the first angle. Remember this only gives one angle; the interval usually contains more.

  10. List every solution of \cos\theta = \tfrac{2}{3} in the interval

    θ=48.2∘, 311.8∘\theta = 48.2^\circ,\ 311.8^\circ

    Using the symmetry of the graph (and adding on the period) we write down all the angles in the interval that give this value. For sine and cosine the pattern repeats every 360^\circ; for tangent every 180^\circ.

  11. Collect all the solutions together

    θ=0∘, 48.2∘, 180∘, 311.8∘\theta = 0^\circ,\ 48.2^\circ,\ 180^\circ,\ 311.8^\circ

    We gather every valid angle from each branch and put them in order. Always double-check each one lies inside the required interval.

  12. Plan the method

    Get everything in one trig function, then solve.\text{Get everything in one trig function, then solve.}

    A good first move is to write the whole equation using a single trig function so it becomes ordinary algebra. This keeps the working neat.

  13. Check a solution by substitution

    Substitute one answer back into the original equation.\text{Substitute one answer back into the original equation.}

    Putting an answer back into the starting equation is a quick way to catch mistakes. If both sides match, the solution is correct.

  14. Sketch the graph to count solutions

    The graph confirms how many times the curves cross.\text{The graph confirms how many times the curves cross.}

    Picturing the graph over the interval tells you how many solutions to expect, so you know when you have found them all.

  15. Check the solution 0^\circ

    θ=0∘  ⇒  both sides agree\theta = 0^\circ \;\Rightarrow\; \text{both sides agree}

    Substituting this angle back into the original equation confirms it is correct and lies inside the required interval. Checking guards against slips.

Answer
θ=0∘, 48.2∘, 180∘, 311.8∘\theta = 0^\circ,\ 48.2^\circ,\ 180^\circ,\ 311.8^\circ
Question 2
8 markschallenging
Prove that tan⁡θsin⁡θ+cos⁡θ≡1cos⁡θ\tan\theta\sin\theta + \cos\theta \equiv \dfrac{1}{\cos\theta}.
Show worked solution

Worked solution

  1. Understand what must be shown

    Show LHS≡RHS for all valid θ.\text{Show LHS} \equiv \text{RHS for all valid } \theta.

    An identity must hold for every allowed angle, not just one number. Our job is to turn one side into the other using identities we already know.

  2. Choose where to start

    Begin with the more complicated side.\text{Begin with the more complicated side.}

    It is usually easier to start with the messier side and simplify it down. In a proof we never move terms across the identity as if it were an equation to solve.

  3. List the tools we will use

    sin⁡2θ+cos⁡2θ≡1,tan⁡θ≡sin⁡θcos⁡θ\sin^2\theta+\cos^2\theta\equiv1,\quad \tan\theta\equiv\tfrac{\sin\theta}{\cos\theta}

    These two identities are the main tools for this whole topic. Keeping them in view tells us which swaps are allowed at each step.

  4. Begin with the left-hand side

    LHS=tan⁡θsin⁡θ+cos⁡θ\text{LHS} = \tan\theta\sin\theta + \cos\theta

    We change the left side into the right side. Start by writing tan theta as sin/cos.

  5. Replace tan with sin/cos

    =sin⁡θcos⁡θ⋅sin⁡θ+cos⁡θ= \dfrac{\sin\theta}{\cos\theta}\cdot\sin\theta + \cos\theta

    tan theta sin theta becomes sin theta times sin theta over cos theta.

  6. Simplify the product

    =sin⁡2θcos⁡θ+cos⁡θ= \dfrac{\sin^2\theta}{\cos\theta} + \cos\theta

    Multiplying gives sin^2 theta over cos theta, plus the cos theta term.

  7. Use a common denominator

    =sin⁡2θ+cos⁡2θcos⁡θ= \dfrac{\sin^2\theta + \cos^2\theta}{\cos\theta}

    Writing cos theta as cos^2 theta over cos theta lets us add the two terms over cos theta.

  8. Apply the Pythagorean identity

    =1cos⁡θ=RHS= \dfrac{1}{\cos\theta} = \text{RHS}

    The numerator sin^2 + cos^2 equals 1, so the expression becomes 1 over cos theta, the right-hand side. Proof complete.

  9. Note the domain

    Valid wherever the denominators are non-zero.\text{Valid wherever the denominators are non-zero.}

    The identity holds for every angle except those that make a denominator zero (for example where cos theta = 0). This is a normal restriction, not an error.

  10. Check with a test angle

    θ=60∘⇒both sides give the same value\theta = 60^\circ \Rightarrow \text{both sides give the same value}

    A quick confidence check is to put one angle into both sides. If they match, our working is very likely correct.

  11. Identify the key step

    The crucial move used sin⁡2θ+cos⁡2θ≡1.\text{The crucial move used } \sin^2\theta+\cos^2\theta\equiv1.

    Spotting where the Pythagorean identity fits is what makes these proofs work. Look out for '1 minus a square' or a 'sum of squares'.

  12. State the conclusion

    LHS≡RHS, as required.\text{LHS} \equiv \text{RHS, as required.}

    Because the left-hand side has been rewritten as the right-hand side, the identity is proven. This is where you would write 'as required' or 'QED'.

  13. Recap the strategy

    One side was simplified step by step.\text{One side was simplified step by step.}

    Looking back, we kept simplifying a single side until it matched the other. That is the standard plan for proving any identity.

  14. Common mistake to avoid

    Never cross-multiply across the ≡ sign.\text{Never cross-multiply across the } \equiv \text{ sign.}

    Treating an identity like an equation and moving terms from side to side is a common slip. Always transform just one side at a time.

  15. Link to earlier algebra

    Uses factorising, expanding and common denominators.\text{Uses factorising, expanding and common denominators.}

    This trig proof reuses the same algebra skills from earlier topics: factorising, expanding brackets, and adding fractions.

Answer
Write tan⁡θ=sin⁡θcos⁡θ\tan\theta=\tfrac{\sin\theta}{\cos\theta}; the LHS becomes sin⁡2θ+cos⁡2θcos⁡θ=1cos⁡θ\tfrac{\sin^2\theta+\cos^2\theta}{\cos\theta}=\tfrac{1}{\cos\theta}.
Question 3
8 markschallenging
Solve 8cos⁡2θ−2sin⁡θ−7=08\cos^2\theta - 2\sin\theta - 7 = 0 for 0≤θ<360∘0 \le \theta < 360^\circ, giving answers to 11 decimal place where appropriate.
Show worked solution

Worked solution

  1. Write down the equation and the interval

    Solve for 0≤θ<360∘\text{Solve for } 0 \le \theta < 360^\circ

    Before solving we note the interval we are working in. This tells us how many rotations around the circle to consider, so we do not miss or invent solutions.

  2. Recall the identity we need

    cos⁡2θ≡1−sin⁡2θ\cos^2\theta \equiv 1 - \sin^2\theta

    Convert cos^2 theta into sine to make a quadratic in sin theta.

  3. Replace using the identity

    8(1−sin⁡2θ)−2sin⁡θ−7=08(1 - \sin^2\theta) - 2\sin\theta - 7 = 0

    Replace cos^2 theta with 1 - sin^2 theta.

  4. Tidy up into one trig function

    8−8sin⁡2θ−2sin⁡θ−7=08 - 8\sin^2\theta - 2\sin\theta - 7 = 0

    Expand ready to rearrange.

  5. Rearrange

    8sin⁡2θ+2sin⁡θ−1=08\sin^2\theta + 2\sin\theta - 1 = 0

    Rearrange to the standard quadratic form (multiply by -1).

  6. See the hidden quadratic

    8u2+2u−1=0,u=sin⁡θ8u^2 + 2u - 1 = 0, \quad u = \sin\theta

    If we let a single letter stand for the trig function, this is just a quadratic equation like the ones from the earlier algebra topic.

  7. Factorise

    (4sin⁡θ−1)(2sin⁡θ+1)=0(4\sin\theta - 1)(2\sin\theta + 1) = 0

    We factorise exactly as we would for any quadratic. A product equals zero only when one of the brackets equals zero.

  8. Set each factor equal to zero

    sin⁡θ=14orsin⁡θ=−12\sin\theta = \tfrac{1}{4} \quad \text{or} \quad \sin\theta = -\tfrac{1}{2}

    Splitting into separate simple equations turns one hard problem into a few easy ones.

  9. Find the principal value for \sin\theta = \tfrac{1}{4}

    sin⁡θ=14⇒base angle\sin\theta = \tfrac{1}{4} \Rightarrow \text{base angle}

    We first take the inverse trig function on a calculator (or use a known exact angle) to get the first angle. Remember this only gives one angle; the interval usually contains more.

  10. List every solution of \sin\theta = \tfrac{1}{4} in the interval

    θ=14.5∘, 165.5∘\theta = 14.5^\circ,\ 165.5^\circ

    Using the symmetry of the graph (and adding on the period) we write down all the angles in the interval that give this value. For sine and cosine the pattern repeats every 360^\circ; for tangent every 180^\circ.

  11. Find the principal value for \sin\theta = -\tfrac{1}{2}

    sin⁡θ=−12⇒base angle\sin\theta = -\tfrac{1}{2} \Rightarrow \text{base angle}

    We first take the inverse trig function on a calculator (or use a known exact angle) to get the first angle. Remember this only gives one angle; the interval usually contains more.

  12. List every solution of \sin\theta = -\tfrac{1}{2} in the interval

    θ=210∘, 330∘\theta = 210^\circ,\ 330^\circ

    Using the symmetry of the graph (and adding on the period) we write down all the angles in the interval that give this value. For sine and cosine the pattern repeats every 360^\circ; for tangent every 180^\circ.

  13. Collect all the solutions together

    θ=14.5∘, 165.5∘, 210∘, 330∘\theta = 14.5^\circ,\ 165.5^\circ,\ 210^\circ,\ 330^\circ

    We gather every valid angle from each branch and put them in order. Always double-check each one lies inside the required interval.

  14. Plan the method

    Get everything in one trig function, then solve.\text{Get everything in one trig function, then solve.}

    A good first move is to write the whole equation using a single trig function so it becomes ordinary algebra. This keeps the working neat.

  15. Check a solution by substitution

    Substitute one answer back into the original equation.\text{Substitute one answer back into the original equation.}

    Putting an answer back into the starting equation is a quick way to catch mistakes. If both sides match, the solution is correct.

Answer
θ=14.5∘, 165.5∘, 210∘, 330∘\theta = 14.5^\circ,\ 165.5^\circ,\ 210^\circ,\ 330^\circ
Question 4
8 markschallenging
Solve tan⁡2x−3=0\tan^2 x - 3 = 0 for 0≤x≤2π0 \le x \le 2\pi, giving answers in radians.
Show worked solution

Worked solution

  1. Write down the equation and the interval

    Solve for 0≤x≤2π\text{Solve for } 0 \le x \le 2\pi

    Before solving we note the interval we are working in. This tells us how many rotations around the circle to consider, so we do not miss or invent solutions.

  2. Rearrange

    tan⁡2x=3\tan^2 x = 3

    Add 3 to both sides to isolate tan^2 x.

  3. Rearrange

    tan⁡x=±3\tan x = \pm\sqrt{3}

    Square-rooting gives a positive and a negative value; both are solved in radians.

  4. Set each factor equal to zero

    tan⁡x=3ortan⁡x=−3\tan x = \sqrt{3} \quad \text{or} \quad \tan x = -\sqrt{3}

    Splitting into separate simple equations turns one hard problem into a few easy ones.

  5. Find the principal value for \tan x = \sqrt{3}

    tan⁡x=3⇒base angle\tan x = \sqrt{3} \Rightarrow \text{base angle}

    We first take the inverse trig function on a calculator (or use a known exact angle) to get the first angle. Remember this only gives one angle; the interval usually contains more.

  6. List every solution of \tan x = \sqrt{3} in the interval

    x=π3, 4π3x = \frac{\pi}{3},\ \frac{4\pi}{3}

    Using the symmetry of the graph (and adding on the period) we write down all the angles in the interval that give this value. For sine and cosine the pattern repeats every 360^\circ; for tangent every 180^\circ.

  7. Find the principal value for \tan x = -\sqrt{3}

    tan⁡x=−3⇒base angle\tan x = -\sqrt{3} \Rightarrow \text{base angle}

    We first take the inverse trig function on a calculator (or use a known exact angle) to get the first angle. Remember this only gives one angle; the interval usually contains more.

  8. List every solution of \tan x = -\sqrt{3} in the interval

    x=2π3, 5π3x = \frac{2\pi}{3},\ \frac{5\pi}{3}

    Using the symmetry of the graph (and adding on the period) we write down all the angles in the interval that give this value. For sine and cosine the pattern repeats every 360^\circ; for tangent every 180^\circ.

  9. Collect all the solutions together

    x=π3, 2π3, 4π3, 5π3x = \frac{\pi}{3},\ \frac{2\pi}{3},\ \frac{4\pi}{3},\ \frac{5\pi}{3}

    We gather every valid angle from each branch and put them in order. Always double-check each one lies inside the required interval.

  10. Plan the method

    Get everything in one trig function, then solve.\text{Get everything in one trig function, then solve.}

    A good first move is to write the whole equation using a single trig function so it becomes ordinary algebra. This keeps the working neat.

  11. Check a solution by substitution

    Substitute one answer back into the original equation.\text{Substitute one answer back into the original equation.}

    Putting an answer back into the starting equation is a quick way to catch mistakes. If both sides match, the solution is correct.

  12. Sketch the graph to count solutions

    The graph confirms how many times the curves cross.\text{The graph confirms how many times the curves cross.}

    Picturing the graph over the interval tells you how many solutions to expect, so you know when you have found them all.

  13. Check the solution \frac{\pi}{3}

    x=π3  ⇒  both sides agreex = \frac{\pi}{3} \;\Rightarrow\; \text{both sides agree}

    Substituting this angle back into the original equation confirms it is correct and lies inside the required interval. Checking guards against slips.

  14. Check the solution \frac{2\pi}{3}

    x=2π3  ⇒  both sides agreex = \frac{2\pi}{3} \;\Rightarrow\; \text{both sides agree}

    Substituting this angle back into the original equation confirms it is correct and lies inside the required interval. Checking guards against slips.

  15. Check the solution \frac{4\pi}{3}

    x=4π3  ⇒  both sides agreex = \frac{4\pi}{3} \;\Rightarrow\; \text{both sides agree}

    Substituting this angle back into the original equation confirms it is correct and lies inside the required interval. Checking guards against slips.

Answer
x=π3, 2π3, 4π3, 5π3x = \frac{\pi}{3},\ \frac{2\pi}{3},\ \frac{4\pi}{3},\ \frac{5\pi}{3}
Question 5
8 markschallenging
Solve 5sin⁡2θ+3cos⁡θ−3=05\sin^2\theta + 3\cos\theta - 3 = 0 for 0≤θ<360∘0 \le \theta < 360^\circ, giving answers to 11 decimal place where appropriate.
Show worked solution

Worked solution

  1. Write down the equation and the interval

    Solve for 0≤θ<360∘\text{Solve for } 0 \le \theta < 360^\circ

    Before solving we note the interval we are working in. This tells us how many rotations around the circle to consider, so we do not miss or invent solutions.

  2. Recall the identity we need

    sin⁡2θ≡1−cos⁡2θ\sin^2\theta \equiv 1 - \cos^2\theta

    Change sin^2 theta into cosine to make a quadratic in cos theta.

  3. Replace using the identity

    5(1−cos⁡2θ)+3cos⁡θ−3=05(1 - \cos^2\theta) + 3\cos\theta - 3 = 0

    Replace sin^2 theta with 1 - cos^2 theta.

  4. Tidy up into one trig function

    5−5cos⁡2θ+3cos⁡θ−3=05 - 5\cos^2\theta + 3\cos\theta - 3 = 0

    Expand ready to rearrange.

  5. Rearrange

    5cos⁡2θ−3cos⁡θ−2=05\cos^2\theta - 3\cos\theta - 2 = 0

    Rearrange to standard quadratic form (multiply by -1).

  6. See the hidden quadratic

    5u2−3u−2=0,u=cos⁡θ5u^2 - 3u - 2 = 0, \quad u = \cos\theta

    If we let a single letter stand for the trig function, this is just a quadratic equation like the ones from the earlier algebra topic.

  7. Factorise

    (5cos⁡θ+2)(cos⁡θ−1)=0(5\cos\theta + 2)(\cos\theta - 1) = 0

    We factorise exactly as we would for any quadratic. A product equals zero only when one of the brackets equals zero.

  8. Set each factor equal to zero

    cos⁡θ=−25orcos⁡θ=1\cos\theta = -\tfrac{2}{5} \quad \text{or} \quad \cos\theta = 1

    Splitting into separate simple equations turns one hard problem into a few easy ones.

  9. Find the principal value for \cos\theta = -\tfrac{2}{5}

    cos⁡θ=−25⇒base angle\cos\theta = -\tfrac{2}{5} \Rightarrow \text{base angle}

    We first take the inverse trig function on a calculator (or use a known exact angle) to get the first angle. Remember this only gives one angle; the interval usually contains more.

  10. List every solution of \cos\theta = -\tfrac{2}{5} in the interval

    θ=113.6∘, 246.4∘\theta = 113.6^\circ,\ 246.4^\circ

    Using the symmetry of the graph (and adding on the period) we write down all the angles in the interval that give this value. For sine and cosine the pattern repeats every 360^\circ; for tangent every 180^\circ.

  11. Find the principal value for \cos\theta = 1

    cos⁡θ=1⇒base angle\cos\theta = 1 \Rightarrow \text{base angle}

    We first take the inverse trig function on a calculator (or use a known exact angle) to get the first angle. Remember this only gives one angle; the interval usually contains more.

  12. List every solution of \cos\theta = 1 in the interval

    θ=0∘\theta = 0^\circ

    Using the symmetry of the graph (and adding on the period) we write down all the angles in the interval that give this value. For sine and cosine the pattern repeats every 360^\circ; for tangent every 180^\circ.

  13. Collect all the solutions together

    θ=0∘, 113.6∘, 246.4∘\theta = 0^\circ,\ 113.6^\circ,\ 246.4^\circ

    We gather every valid angle from each branch and put them in order. Always double-check each one lies inside the required interval.

  14. Plan the method

    Get everything in one trig function, then solve.\text{Get everything in one trig function, then solve.}

    A good first move is to write the whole equation using a single trig function so it becomes ordinary algebra. This keeps the working neat.

  15. Check a solution by substitution

    Substitute one answer back into the original equation.\text{Substitute one answer back into the original equation.}

    Putting an answer back into the starting equation is a quick way to catch mistakes. If both sides match, the solution is correct.

Answer
θ=0∘, 113.6∘, 246.4∘\theta = 0^\circ,\ 113.6^\circ,\ 246.4^\circ

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