The t-formulae Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths The t-formulae questions. See exactly how to solve problems on t-formulae, weierstrass-substitution, half-angle, expressing-in-terms-of-t.

t-formulaeweierstrass-substitutionhalf-angleexpressing-in-terms-of-talgebraic-manipulationexact-values
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
Given that t=tanθ2t=\tan\frac{\theta}{2}, express sinθ\sin\theta in terms of tt.

Worked solution

  1. Write down the substitution and the t-formulae

    sinθ=2t1+t2,cosθ=1t21+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

    sinθ\sin\theta

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

  3. State the final expression

    sinθ=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=tanθ2t=\tan\frac{\theta}{2}, express cosθ\cos\theta in terms of tt.

Worked solution

  1. Write down the substitution and the t-formulae

    sinθ=2t1+t2,cosθ=1t21+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

    cosθ\cos\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. State the final expression

    cosθ=1t2t2+1\cos\theta=\frac{1-t^{2}}{t^{2}+1}

    The expression is now written entirely in terms of tt.

Answer
1t2t2+1\frac{1-t^{2}}{t^{2}+1}
Question 3
2 markseasy
Given that t=tanθ2t=\tan\frac{\theta}{2}, express tanθ\tan\theta in terms of tt.

Worked solution

  1. Write down the substitution and the t-formulae

    sinθ=2t1+t2,cosθ=1t21+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

    tanθ\tan\theta

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

  3. State the final expression

    tanθ=2t1t2\tan\theta=\frac{2t}{1-t^{2}}

    The expression is now written entirely in terms of tt.

Answer
2t1t2\frac{2t}{1-t^{2}}
Question 4
2 markseasy
Given that t=tanθ2t=\tan\frac{\theta}{2}, express cosθ+1\cos\theta+1 in terms of tt.

Worked solution

  1. Write down the substitution and the t-formulae

    sinθ=2t1+t2,cosθ=1t21+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

    cosθ+1\cos\theta+1

    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. State the final expression

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

    The expression is now written entirely in terms of tt.

Answer
2t2+1\frac{2}{t^{2}+1}
Question 5
2 markseasy
Given that t=tanθ2t=\tan\frac{\theta}{2}, express 1cosθ1-\cos\theta in terms of tt.

Worked solution

  1. Write down the substitution and the t-formulae

    sinθ=2t1+t2,cosθ=1t21+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

    1cosθ1-\cos\theta

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

  3. State the final expression

    1cosθ=2t2t2+11-\cos\theta=\frac{2t^{2}}{t^{2}+1}

    The expression is now written entirely in terms of tt.

Answer
2t2t2+1\frac{2t^{2}}{t^{2}+1}

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