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Trigonometric Identities

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AQA Level 2 Further Maths practice questions, each with a hint and a full worked solution.

Have a pen, paper and a calculator handy, to work out your answers.

Question 12 marks

Simplify fully \dfrac{\tan\theta\,(1-\sin^2\theta)}{\cos\theta}

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Replace 1-\sin^2\theta using \sin^2\theta+\cos^2\theta=1, and write \tan\theta as \dfrac{\sin\theta}{\cos\theta}

Worked solution
  1. 1-\sin^2\theta=\cos^2\theta
  2. So the expression is \dfrac{\tan\theta\cos^2\theta}{\cos\theta}=\tan\theta\cos\theta
  3. \tan\theta=\dfrac{\sin\theta}{\cos\theta}, so \tan\theta\cos\theta=\sin\theta
  4. Answer: \sin\theta

Question 22 marks

90^\circ<\theta<180^\circ

\sin\theta=\dfrac{\sqrt{7}}{4}

Work out the exact value of \cos\theta.

Hint

Use \sin^2\theta+\cos^2\theta=1, then think about whether cosine is positive or negative for an obtuse angle.

Worked solution
  1. \sin^2\theta=\dfrac{7}{16}
  2. \cos^2\theta=1-\dfrac{7}{16}=\dfrac{9}{16}
  3. \cos\theta=\pm\dfrac34
  4. \theta is between 90^\circ and 180^\circ, where cosine is negative
  5. Answer: \cos\theta=-\dfrac34

Question 3Challenge6 marks

\frac{8\sin^2\theta-5}{1-\sin^2\theta}\equiv a\tan^2\theta+b where a and b are integers.

(a)

Write \dfrac{8\sin^2\theta-5}{1-\sin^2\theta} in the form a\tan^2\theta+b

3 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

(b)

Hence solve \dfrac{8\sin^2\theta-5}{1-\sin^2\theta}=7 for 0^\circ\leqslant\theta\leqslant360^\circ

Give your answers to 1 decimal place.

3 marks

Give every value, separated by commas

Hint

Write the denominator as \cos^2\theta and split the 5 in the numerator as 5(\sin^2\theta+\cos^2\theta); when you square root, remember the negative value too.

Worked solution

Part (a)

  1. Denominator: 1-\sin^2\theta=\cos^2\theta
  2. Numerator: 8\sin^2\theta-5=8\sin^2\theta-5(\sin^2\theta+\cos^2\theta)=3\sin^2\theta-5\cos^2\theta
  3. Divide each term by \cos^2\theta: \dfrac{3\sin^2\theta}{\cos^2\theta}-\dfrac{5\cos^2\theta}{\cos^2\theta}
  4. \dfrac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta
  5. Answer: 3\tan^2\theta-5

Part (b)

  1. Using part (a): 3\tan^2\theta-5=7
  2. 3\tan^2\theta=12, so \tan^2\theta=4
  3. \tan\theta=2 or \tan\theta=-2
  4. \tan\theta=2: \theta=63.43\ldots^\circ and 180^\circ+63.43\ldots^\circ=243.43\ldots^\circ
  5. \tan\theta=-2: \theta=180^\circ-63.43\ldots^\circ=116.56\ldots^\circ and 360^\circ-63.43\ldots^\circ=296.56\ldots^\circ
  6. Answer: 63.4^\circ,\ 116.6^\circ,\ 243.4^\circ,\ 296.6^\circ

More on this topic: Trigonometric Identities worksheet with full solutions

All AQA Level 2 Further Maths practice questions

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