A-Level Identities and equations Practice Questions

Free A-Level Identities and equations practice questions with full step-by-step worked solutions. Covers Pythagorean identity, tan = sin/cos, solving sin, solving cos. Practise exam-style problems and check your method.

Pythagorean identitytan = sin/cossolving sinsolving cossolving tannegative value
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Given that sin⁡θ=35\sin\theta = \tfrac{3}{5} and θ\theta is acute, find the exact value of cos⁡θ\cos\theta.
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Worked solution

  1. Write down the identity linking sine and cosine

    sin⁡2θ+cos⁡2θ≡1\sin^2\theta + \cos^2\theta \equiv 1

    This is the most important trig identity. It works for every angle, so we can use it to swap between sine and cosine.

  2. Substitute the known value of sine

    (35)2+cos⁡2θ=1\left(\tfrac{3}{5}\right)^2 + \cos^2\theta = 1

    We put in sin theta = 3/5. Squaring 3/5 gives 9/25, which we deal with next.

  3. Make cos-squared the subject

    cos⁡2θ=1−925=1625\cos^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}

    Taking 9/25 from a whole (25/25) leaves 16/25. We now just need the square root.

  4. Take the square root

    cos⁡θ=45\cos\theta = \tfrac{4}{5}

    Since theta is acute (between 0 and 90 degrees) cosine is positive, so we keep only the positive root.

Answer
45\dfrac{4}{5}
Question 2
2 markseasy
Simplify 1−sin⁡2θcos⁡θ\dfrac{1 - \sin^2\theta}{\cos\theta}.
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Worked solution

  1. Replace the numerator using the identity

    1−sin⁡2θ≡cos⁡2θ1 - \sin^2\theta \equiv \cos^2\theta

    From sin^2 + cos^2 = 1 we know 1 - sin^2 theta is cos^2 theta. Swapping it makes the fraction much simpler.

  2. Rewrite the fraction

    cos⁡2θcos⁡θ\dfrac{\cos^2\theta}{\cos\theta}

    The top is now cos^2 theta, which is cos theta times cos theta.

  3. Cancel one cosine

    =cos⁡θ= \cos\theta

    One factor of cos theta on top cancels with the cos theta on the bottom, leaving cos theta.

Answer
cos⁡θ\cos\theta
Question 3
4 marksintermediate
Explain why the equation sin⁡θ=2\sin\theta = 2 has no solutions.
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Worked solution

  1. Read the question carefully

    Decide exactly what is being asked.\text{Decide exactly what is being asked.}

    Before answering, be clear whether the question wants a count, a reason, or a value. Rushing this is a common cause of lost marks.

  2. Recall the relevant fact

    −1≤sin⁡θ≤1,−1≤cos⁡θ≤1-1\le\sin\theta\le1,\quad -1\le\cos\theta\le1

    The key facts here are the ranges of sine and cosine and the shapes of their graphs. These decide how many solutions are possible.

  3. Recall the range of the sine function

    −1≤sin⁡θ≤1-1 \le \sin\theta \le 1

    The value of sine can never be bigger than 1 or smaller than -1. This is because it comes from the y-coordinate on the unit circle.

  4. Compare with the required value

    2>12 > 1

    The number 2 lies outside the possible range of sine, so no angle can ever make sine equal to 2.

  5. Conclude

    No solutions.\text{No solutions.}

    There is simply no angle theta with sin theta = 2, so the equation has no solutions at all.

  6. Picture the graph

    A quick sketch shows the crossings.\text{A quick sketch shows the crossings.}

    Drawing the trig graph over the interval lets you literally count where it meets the horizontal line, which confirms the answer.

Answer
sin⁡θ\sin\theta can only take values between −1-1 and 11, and 22 is outside this range.
Question 4
6 markshard
Solve tan⁡2θ−2tan⁡θ−3=0\tan^2\theta - 2\tan\theta - 3 = 0 for 0≤θ<360∘0 \le \theta < 360^\circ, giving answers to 11 decimal place where appropriate.
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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. See the hidden quadratic

    u2−2u−3=0,u=tan⁡θu^2 - 2u - 3 = 0, \quad u = \tan\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.

  3. Factorise

    (tan⁡θ−3)(tan⁡θ+1)=0(\tan\theta - 3)(\tan\theta + 1) = 0

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

  4. Set each factor equal to zero

    tan⁡θ=3ortan⁡θ=−1\tan\theta = 3 \quad \text{or} \quad \tan\theta = -1

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

  5. Find the principal value for \tan\theta = 3

    tan⁡θ=3⇒base angle\tan\theta = 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\theta = 3 in the interval

    θ=71.6∘, 251.6∘\theta = 71.6^\circ,\ 251.6^\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.

  7. Find the principal value for \tan\theta = -1

    tan⁡θ=−1⇒base angle\tan\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.

  8. List every solution of \tan\theta = -1 in the interval

    θ=135∘, 315∘\theta = 135^\circ,\ 315^\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. Collect all the solutions together

    θ=71.6∘, 135∘, 251.6∘, 315∘\theta = 71.6^\circ,\ 135^\circ,\ 251.6^\circ,\ 315^\circ

    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.

Answer
θ=71.6∘, 135∘, 251.6∘, 315∘\theta = 71.6^\circ,\ 135^\circ,\ 251.6^\circ,\ 315^\circ
Question 5
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.
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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

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