Further Maths The t-formulae Practice Questions

Free Further Maths The t-formulae practice questions with full step-by-step worked solutions. Covers t-formulae, weierstrass-substitution, half-angle, expressing-in-terms-of-t. Practise exam-style problems and check your method.

t-formulaeweierstrass-substitutionhalf-angleexpressing-in-terms-of-talgebraic-manipulationexact-values
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Given that t=tan⁡θ2t=\tan\frac{\theta}{2}, express sin⁡θ\sin\theta in terms of tt.
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Worked solution

  1. Write down the substitution and the t-formulae

    sin⁡θ=2t1+t2,cos⁡θ=1−t21+t2\sin\theta=\frac{2t}{1+t^{2}},\quad\cos\theta=\frac{1-t^{2}}{1+t^{2}}

    Every trigonometric function of θ\theta is replaced by a rational function of tt.

  2. Write down the expression to be converted

    sin⁡θ\sin\theta

    The whole expression must be written in terms of tt alone.

  3. State the final expression

    sin⁡θ=2tt2+1\sin\theta=\frac{2t}{t^{2}+1}

    The expression is now written entirely in terms of tt.

Answer
2tt2+1\frac{2t}{t^{2}+1}
Question 2
2 markseasy
Given that t=tan⁡θt=\tan\theta, express cos⁡2θ\cos2\theta in terms of tt. Which of the options gives the correct expression?
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Worked solution

  1. Write down the substitution and the t-formulae

    sin⁡2θ=2t1+t2,cos⁡2θ=1−t21+t2\sin2\theta=\frac{2t}{1+t^{2}},\quad\cos2\theta=\frac{1-t^{2}}{1+t^{2}}

    Every trigonometric function of θ\theta is replaced by a rational function of tt.

  2. Write down the expression to be converted

    cos⁡2θ\cos2\theta

    The whole expression must be written in terms of tt alone.

  3. Expand the denominator

    t2+1=t2+1t^{2}+1=t^{2}+1

    The denominator is a power of 1+t21+t^{2}, which is never zero.

  4. Select the matching expression

    cos⁡2θ=1−t2t2+1\cos2\theta=\frac{1-t^{2}}{t^{2}+1}

    The expression is now written entirely in terms of tt.

Answer
1−t2t2+1\frac{1-t^{2}}{t^{2}+1}
Question 3
4 marksintermediate
Given that t=tan⁡θ2t=\tan\frac{\theta}{2}, express 2cos⁡θ+1\frac{2}{\cos\theta+1} in terms of tt. Which of the options gives the correct expression?
Show worked solution

Worked solution

  1. Write down the substitution and the t-formulae

    sin⁡θ=2t1+t2,cos⁡θ=1−t21+t2\sin\theta=\frac{2t}{1+t^{2}},\quad\cos\theta=\frac{1-t^{2}}{1+t^{2}}

    Every trigonometric function of θ\theta is replaced by a rational function of tt.

  2. Write down the expression to be converted

    2cos⁡θ+1\frac{2}{\cos\theta+1}

    The whole expression must be written in terms of tt alone.

  3. Substitute the t-formulae into the expression

    2cos⁡θ+1=t2+1\frac{2}{\cos\theta+1}=t^{2}+1

    Each trigonometric term becomes a quotient with denominator 1+t21+t^{2}.

  4. Check the result when θ=60∘\theta=60^{\circ}

    t=33,LHS=43,RHS=43t=\frac{\sqrt{3}}{3},\quad\text{LHS}=\frac{4}{3},\quad\text{RHS}=\frac{4}{3}

    The original expression and the expression in tt agree.

  5. Check the result when θ=120∘\theta=120^{\circ}

    t=3,LHS=4,RHS=4t=\sqrt{3},\quad\text{LHS}=4,\quad\text{RHS}=4

    The original expression and the expression in tt agree.

  6. Check the result when θ=240∘\theta=240^{\circ}

    t=−3,LHS=4,RHS=4t=-\sqrt{3},\quad\text{LHS}=4,\quad\text{RHS}=4

    The original expression and the expression in tt agree.

  7. Select the matching expression

    2cos⁡θ+1=t2+1\frac{2}{\cos\theta+1}=t^{2}+1

    The expression is now written entirely in terms of tt.

Answer
t2+1t^{2}+1
Question 4
6 markshard
Use the substitution t=tan⁡θ2t=\tan\frac{\theta}{2} to solve the equation 3sin⁡θ−cos⁡θ=1\sqrt{3}\sin\theta-\cos\theta=1 in the interval 0≤θ<2π0\le\theta<2\pi. Which of the options gives the complete solution set?
Show worked solution

Worked solution

  1. Write down the equation and the substitution

    3sin⁡θ−cos⁡θ=1,t=tan⁡θ2\sqrt{3}\sin\theta-\cos\theta=1,\quad t=\tan\frac{\theta}{2}

    The substitution converts the equation into a polynomial equation in tt.

  2. Write down the t-formulae

    sin⁡θ=2t1+t2,cos⁡θ=1−t21+t2\sin\theta=\frac{2t}{1+t^{2}},\quad\cos\theta=\frac{1-t^{2}}{1+t^{2}}

    These are the only two formulae needed here.

  3. Substitute the t-formulae into the equation

    3⋅2t1+t2−1−t21+t2=1\sqrt{3}\cdot\frac{2t}{1+t^{2}}-\frac{1-t^{2}}{1+t^{2}}=1

    Both trigonometric terms now share the denominator 1+t21+t^{2}.

  4. Multiply through by 1+t21+t^{2}

    3⋅(2t)−(1−t2)=(1+t2)\sqrt{3}\cdot\left(2t\right)-\left(1-t^{2}\right)=\left(1+t^{2}\right)

    Since 1+t2>01+t^{2}>0 for all real tt, this step neither gains nor loses roots.

  5. Collect all the terms on one side

    −23t+2=0-2\sqrt{3}t+2=0

    The trigonometric equation has become a polynomial equation in tt.

  6. Notice that the coefficient of t2t^{2} is zero

    b+c=−1+(1)=0b+c=-1+\left(1\right)=0

    The equation in tt is only linear, so it can supply at most one value of θ\theta.

  7. Solve the linear equation for tt

    t=33t=\frac{\sqrt{3}}{3}

    A linear equation has exactly one root.

  8. Recover θ\theta from t=33t=\frac{\sqrt{3}}{3}

    tan⁡θ2=33  ⟹  θ=2arctan⁡(33)=π3\tan\frac{\theta}{2}=\frac{\sqrt{3}}{3}\implies \theta=2\arctan\left(\frac{\sqrt{3}}{3}\right)=\frac{\pi}{3}

    The principal value is taken first; every other angle differs from it by a whole number of turns.

  9. Test the excluded angle θ=π\theta=\pi separately

    θ=π:1 = 1\theta=\pi:\quad 1\ =\ 1

    This angle IS a solution, but t=tan⁡θ2t=\tan\frac{\theta}{2} is undefined there, so the substitution cannot find it: it must be added by hand.

  10. List every angle in the interval

    0≤θ<2π:θ=π3, θ=π0\le\theta<2\pi:\quad \theta=\frac{\pi}{3},\ \theta=\pi

    Add multiples of a full turn to the principal values, then include any excluded angle that works.

  11. Select the complete solution set

    θ=π3, θ=π\theta=\frac{\pi}{3},\ \theta=\pi

    Every solution in the given interval is listed, including any angle the substitution cannot reach.

Answer
θ=π3, θ=π\theta=\frac{\pi}{3},\ \theta=\pi
Question 5
9 markschallenging
Use the substitution t=tan⁡θt=\tan\theta to solve the equation 3sin⁡2θ−cos⁡2θ=1\sqrt{3}\sin2\theta-\cos2\theta=1 in the interval 0≤θ<2π0\le\theta<2\pi. Which of the options gives the complete solution set?
Show worked solution

Worked solution

  1. Write down the equation and the substitution

    3sin⁡2θ−cos⁡2θ=1,t=tan⁡θ\sqrt{3}\sin2\theta-\cos2\theta=1,\quad t=\tan\theta

    The substitution converts the equation into a polynomial equation in tt.

  2. Write down the t-formulae

    sin⁡2θ=2t1+t2,cos⁡2θ=1−t21+t2\sin2\theta=\frac{2t}{1+t^{2}},\quad\cos2\theta=\frac{1-t^{2}}{1+t^{2}}

    These are the only two formulae needed here.

  3. Substitute the t-formulae into the equation

    3⋅2t1+t2−1−t21+t2=1\sqrt{3}\cdot\frac{2t}{1+t^{2}}-\frac{1-t^{2}}{1+t^{2}}=1

    Both trigonometric terms now share the denominator 1+t21+t^{2}.

  4. Multiply through by 1+t21+t^{2}

    3⋅(2t)−(1−t2)=(1+t2)\sqrt{3}\cdot\left(2t\right)-\left(1-t^{2}\right)=\left(1+t^{2}\right)

    Since 1+t2>01+t^{2}>0 for all real tt, this step neither gains nor loses roots.

  5. Collect all the terms on one side

    −23t+2=0-2\sqrt{3}t+2=0

    The trigonometric equation has become a polynomial equation in tt.

  6. Notice that the coefficient of t2t^{2} is zero

    b+c=−1+(1)=0b+c=-1+\left(1\right)=0

    The equation in tt is only linear, so it can supply at most one value of θ\theta.

  7. Solve the linear equation for tt

    t=33t=\frac{\sqrt{3}}{3}

    A linear equation has exactly one root.

  8. Recover 2θ2\theta from t=33t=\frac{\sqrt{3}}{3}

    tan⁡θ=33  ⟹  2θ=2arctan⁡(33)=π3,θ=π6\tan\theta=\frac{\sqrt{3}}{3}\implies 2\theta=2\arctan\left(\frac{\sqrt{3}}{3}\right)=\frac{\pi}{3},\quad\theta=\frac{\pi}{6}

    The principal value is taken first; every other angle differs from it by a whole number of turns.

  9. Test the excluded angle θ=π2\theta=\frac{\pi}{2} separately

    θ=π2:1 = 1\theta=\frac{\pi}{2}:\quad 1\ =\ 1

    This angle IS a solution, but t=tan⁡θ2t=\tan\frac{\theta}{2} is undefined there, so the substitution cannot find it: it must be added by hand.

  10. Test the excluded angle θ=3π2\theta=\frac{3\pi}{2} separately

    θ=3π2:1 = 1\theta=\frac{3\pi}{2}:\quad 1\ =\ 1

    This angle IS a solution, but t=tan⁡θ2t=\tan\frac{\theta}{2} is undefined there, so the substitution cannot find it: it must be added by hand.

  11. List every angle in the interval

    0≤θ<2π:θ=π6, θ=π2, θ=7π6, θ=3π20\le\theta<2\pi:\quad \theta=\frac{\pi}{6},\ \theta=\frac{\pi}{2},\ \theta=\frac{7\pi}{6},\ \theta=\frac{3\pi}{2}

    Add multiples of a full turn to the principal values, then include any excluded angle that works.

  12. Verify θ=π6\theta=\frac{\pi}{6} in the original equation

    3⋅sin⁡(π3)−cos⁡(π3)=1\sqrt{3}\cdot\sin\left(\frac{\pi}{3}\right)-\cos\left(\frac{\pi}{3}\right)=1

    The left-hand side equals the right-hand side, so this angle really is a solution.

  13. Verify θ=π2\theta=\frac{\pi}{2} in the original equation

    3⋅sin⁡(π)−cos⁡(π)=1\sqrt{3}\cdot\sin\left(\pi\right)-\cos\left(\pi\right)=1

    The left-hand side equals the right-hand side, so this angle really is a solution.

  14. Verify θ=7π6\theta=\frac{7\pi}{6} in the original equation

    3⋅sin⁡(7π3)−cos⁡(7π3)=1\sqrt{3}\cdot\sin\left(\frac{7\pi}{3}\right)-\cos\left(\frac{7\pi}{3}\right)=1

    The left-hand side equals the right-hand side, so this angle really is a solution.

  15. Verify θ=3π2\theta=\frac{3\pi}{2} in the original equation

    3⋅sin⁡(3π)−cos⁡(3π)=1\sqrt{3}\cdot\sin\left(3\pi\right)-\cos\left(3\pi\right)=1

    The left-hand side equals the right-hand side, so this angle really is a solution.

  16. Select the complete solution set

    θ=π6, θ=π2, θ=7π6, θ=3π2\theta=\frac{\pi}{6},\ \theta=\frac{\pi}{2},\ \theta=\frac{7\pi}{6},\ \theta=\frac{3\pi}{2}

    Every solution in the given interval is listed, including any angle the substitution cannot reach.

Answer
θ=π6, θ=π2, θ=7π6, θ=3π2\theta=\frac{\pi}{6},\ \theta=\frac{\pi}{2},\ \theta=\frac{7\pi}{6},\ \theta=\frac{3\pi}{2}

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