Identities and equations Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Identities and equations questions. See exactly how to solve problems on Pythagorean identity, tan = sin/cos, solving sin, solving cos.

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.

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
Given that cos⁡θ=513\cos\theta = \tfrac{5}{13} and θ\theta is acute, find the exact value of tan⁡θ\tan\theta.

Worked solution

  1. Find sine using the Pythagorean identity

    sin⁡2θ=1−(513)2=144169\sin^2\theta = 1 - \left(\tfrac{5}{13}\right)^2 = \tfrac{144}{169}

    We use sin^2 + cos^2 = 1 to get sine from cosine. 1 - 25/169 = 144/169.

  2. Square root to find sine

    sin⁡θ=1213\sin\theta = \tfrac{12}{13}

    As theta is acute, sine is positive, so we take the positive root 12/13.

  3. Use the definition of tangent

    tan⁡θ≡sin⁡θcos⁡θ=12/135/13\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta} = \dfrac{12/13}{5/13}

    Remember tan theta means sin theta divided by cos theta. The /13 parts cancel.

  4. Simplify the fraction

    tan⁡θ=125\tan\theta = \tfrac{12}{5}

    Dividing 12/13 by 5/13 leaves 12/5. This is the final exact value.

Answer
125\dfrac{12}{5}
Question 3
2 markseasy
Simplify sin⁡θcos⁡θ\dfrac{\sin\theta}{\cos\theta}.

Worked solution

  1. Recognise the quotient

    sin⁡θcos⁡θ\dfrac{\sin\theta}{\cos\theta}

    This is sine divided by cosine. Whenever you see this exact shape, a trig identity is waiting to be used.

  2. Recall the definition of tangent

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

    One of the key identities in this topic says tan theta is defined as sin theta over cos theta. Our expression matches the right-hand side exactly.

  3. Write the simplified answer

    sin⁡θcos⁡θ=tan⁡θ\dfrac{\sin\theta}{\cos\theta} = \tan\theta

    So the whole expression simplifies to tan theta, a single tidy term.

Answer
tan⁡θ\tan\theta
Question 4
2 markseasy
Simplify 1−cos⁡2θ1 - \cos^2\theta.

Worked solution

  1. Start from the Pythagorean identity

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

    The identity sin^2 + cos^2 = 1 can be rearranged to describe 1 minus a square.

  2. Rearrange to isolate sin-squared

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

    Subtracting cos^2 theta from both sides shows that 1 - cos^2 theta is exactly sin^2 theta.

  3. Write the simplified result

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

    So the expression simplifies neatly to sin^2 theta.

Answer
sin⁡2θ\sin^2\theta
Question 5
2 markseasy
Write down the exact value of sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta.

Worked solution

  1. Recognise the expression

    sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta

    We are asked for the value of sine squared plus cosine squared. This is one of the most famous expressions in trigonometry.

  2. Recall the Pythagorean identity

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

    This identity is true for every possible angle theta. It comes from Pythagoras' theorem applied to a point on the unit circle.

  3. State the value

    sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

    No matter what theta is, the answer is always exactly 1.

Answer
11

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