Hard Further Maths The t-formulae Questions

Challenging, exam-style Further Maths The t-formulae questions with worked solutions. Stretch yourself on the hardest t-formulae, solving-trig-equations, excluded-values, half-angle problems.

t-formulaesolving-trig-equationsexcluded-valueshalf-angledouble-angleexpressing-in-terms-of-t
Further Maths34 questionsStep-by-step solutions
Question 1
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}
Question 2
9 markschallenging
Given that t=tan⁡θ2t=\tan\frac{\theta}{2}, express 1−cos⁡θsin⁡θ+sin⁡θ1−cos⁡θ\frac{1-\cos\theta}{\sin\theta}+\frac{\sin\theta}{1-\cos\theta} 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

    1−cos⁡θsin⁡θ+sin⁡θ1−cos⁡θ\frac{1-\cos\theta}{\sin\theta}+\frac{\sin\theta}{1-\cos\theta}

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

  3. Substitute the t-formulae into the expression

    1−cos⁡θsin⁡θ+sin⁡θ1−cos⁡θ=t2+1t\frac{1-\cos\theta}{\sin\theta}+\frac{\sin\theta}{1-\cos\theta}=\frac{t^{2}+1}{t}

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

  4. Expand the denominator

    t=tt=t

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

  5. Cancel any common factors

    1−cos⁡θsin⁡θ+sin⁡θ1−cos⁡θ=t2+1t\frac{1-\cos\theta}{\sin\theta}+\frac{\sin\theta}{1-\cos\theta}=\frac{t^{2}+1}{t}

    This is the required expression in terms of tt.

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

    t=33,LHS=433,RHS=433t=\frac{\sqrt{3}}{3},\quad\text{LHS}=\frac{4\sqrt{3}}{3},\quad\text{RHS}=\frac{4\sqrt{3}}{3}

    The original expression and the expression in tt agree.

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

    t=3,LHS=433,RHS=433t=\sqrt{3},\quad\text{LHS}=\frac{4\sqrt{3}}{3},\quad\text{RHS}=\frac{4\sqrt{3}}{3}

    The original expression and the expression in tt agree.

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

    t=−3,LHS=−433,RHS=−433t=-\sqrt{3},\quad\text{LHS}=-\frac{4\sqrt{3}}{3},\quad\text{RHS}=-\frac{4\sqrt{3}}{3}

    The original expression and the expression in tt agree.

  9. Check the result when θ=300∘\theta=300^{\circ}

    t=−33,LHS=−433,RHS=−433t=-\frac{\sqrt{3}}{3},\quad\text{LHS}=-\frac{4\sqrt{3}}{3},\quad\text{RHS}=-\frac{4\sqrt{3}}{3}

    The original expression and the expression in tt agree.

  10. Check the result when θ=30∘\theta=30^{\circ}

    t=2−3,LHS=4,RHS=4t=2-\sqrt{3},\quad\text{LHS}=4,\quad\text{RHS}=4

    The original expression and the expression in tt agree.

  11. Recall the half-angle substitution

    t=tan⁡θ2t=\tan\frac{\theta}{2}

    The Weierstrass substitution turns any rational trigonometric equation into a rational equation in tt.

  12. Quote the t-formula for sin⁡θ\sin\theta

    sin⁡θ=2t1+t2\sin\theta=\frac{2t}{1+t^{2}}

    This follows from sin⁡θ=2sin⁡θ2cos⁡θ2\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}.

  13. Quote the t-formula for cos⁡θ\cos\theta

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

    This follows from cos⁡θ=cos⁡2θ2−sin⁡2θ2\cos\theta=\cos^{2}\frac{\theta}{2}-\sin^{2}\frac{\theta}{2}.

  14. Quote the t-formula for tan⁡θ\tan\theta

    tan⁡θ=2t1−t2\tan\theta=\frac{2t}{1-t^{2}}

    Dividing the formula for sin⁡θ\sin\theta by the formula for cos⁡θ\cos\theta gives this result.

  15. Check the t-formulae satisfy the Pythagorean identity

    (2t1+t2)2+(1−t21+t2)2=4t2+1−2t2+t4(1+t2)2=1\left(\frac{2t}{1+t^{2}}\right)^{2}+\left(\frac{1-t^{2}}{1+t^{2}}\right)^{2}=\frac{4t^{2}+1-2t^{2}+t^{4}}{\left(1+t^{2}\right)^{2}}=1

    The t-formulae are consistent with sin⁡2θ+cos⁡2θ=1\sin^{2}\theta+\cos^{2}\theta=1.

  16. Note the double-angle version of the substitution

    t=tan⁡θ  ⟹  sin⁡2θ=2t1+t2,cos⁡2θ=1−t21+t2t=\tan\theta\implies\sin2\theta=\frac{2t}{1+t^{2}},\quad\cos2\theta=\frac{1-t^{2}}{1+t^{2}}

    The same formulae work with θ\theta replaced by 2θ2\theta throughout.

  17. Note that the substitution is undefined at the half-turn

    tan⁡θ2→∞asθ→180∘\tan\frac{\theta}{2}\to\infty\quad\text{as}\quad\theta\to180^{\circ}

    No finite value of tt corresponds to θ=180∘\theta=180^{\circ}, so that angle must always be tested separately.

  18. Select the matching expression

    1−cos⁡θsin⁡θ+sin⁡θ1−cos⁡θ=t2+1t\frac{1-\cos\theta}{\sin\theta}+\frac{\sin\theta}{1-\cos\theta}=\frac{t^{2}+1}{t}

    The expression is now written entirely in terms of tt.

Answer
t2+1t\frac{t^{2}+1}{t}
Question 3
9 markschallenging
Use the substitution t=tan⁡θ2t=\tan\frac{\theta}{2} to solve the equation sin⁡θ−cos⁡θ=−1\sin\theta-\cos\theta=-1 in the interval 0≤θ<360∘0\le\theta<360^{\circ}. Which of the options gives the complete solution set?
Show worked solution

Worked solution

  1. Write down the equation and the substitution

    sin⁡θ−cos⁡θ=−1,t=tan⁡θ2\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

    2t1+t2−1−t21+t2=−1\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}

    (2t)−(1−t2)=−(1+t2)\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

    −2t2−2t=0-2t^{2}-2t=0

    The trigonometric equation has become a polynomial equation in tt.

  6. Compute the discriminant

    Δ=(−2)2−4(−2)(0)=4\Delta=\left(-2\right)^{2}-4\left(-2\right)\left(0\right)=4

    The discriminant tells us how many values of tt to expect.

  7. Solve the quadratic for tt

    t=−1,t=0t=-1,\quad t=0

    These are the only finite values of tt that can satisfy the equation.

  8. Recover θ\theta from t=−1t=-1

    tan⁡θ2=−1  ⟹  θ=2arctan⁡(−1)=−π2\tan\frac{\theta}{2}=-1\implies \theta=2\arctan\left(-1\right)=-\frac{\pi}{2}

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

  9. Recover θ\theta from t=0t=0

    tan⁡θ2=0  ⟹  θ=2arctan⁡(0)=0\tan\frac{\theta}{2}=0\implies \theta=2\arctan\left(0\right)=0

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

  10. Test the excluded angle θ=180∘\theta=180^{\circ} separately

    θ=180∘:1 ≠ −1\theta=180^{\circ}:\quad 1\ \neq\ -1

    This angle is not a solution, so nothing is lost by the substitution being undefined there.

  11. List every angle in the interval

    0≤θ<360∘:θ=0∘, θ=270∘0\le\theta<360^{\circ}:\quad \theta=0^{\circ},\ \theta=270^{\circ}

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

  12. Verify θ=0∘\theta=0^{\circ} in the original equation

    sin⁡(0∘)−cos⁡(0∘)=−1\sin\left(0^{\circ}\right)-\cos\left(0^{\circ}\right)=-1

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

  13. Verify θ=270∘\theta=270^{\circ} in the original equation

    sin⁡(270∘)−cos⁡(270∘)=−1\sin\left(270^{\circ}\right)-\cos\left(270^{\circ}\right)=-1

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

  14. Cross-check with the RR-formula: find RR

    R=(1)2+(−1)2=2R=\sqrt{\left(1\right)^{2}+\left(-1\right)^{2}}=\sqrt{2}

    A completely different method should give the same solutions.

  15. Select the complete solution set

    θ=0∘, θ=270∘\theta=0^{\circ},\ \theta=270^{\circ}

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

Answer
θ=0∘, θ=270∘\theta=0^{\circ},\ \theta=270^{\circ}
Question 4
9 markschallenging
Use the substitution t=tan⁡θt=\tan\theta to solve the equation sin⁡2θ−cos⁡2θ=1\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

    sin⁡2θ−cos⁡2θ=1,t=tan⁡θ\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

    2t1+t2−1−t21+t2=1\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}

    (2t)−(1−t2)=(1+t2)\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

    2−2t=02-2t=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=1t=1

    A linear equation has exactly one root.

  8. Recover 2θ2\theta from t=1t=1

    tan⁡θ=1  ⟹  2θ=2arctan⁡(1)=π2,θ=π4\tan\theta=1\implies 2\theta=2\arctan\left(1\right)=\frac{\pi}{2},\quad\theta=\frac{\pi}{4}

    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=π4, θ=π2, θ=5π4, θ=3π20\le\theta<2\pi:\quad \theta=\frac{\pi}{4},\ \theta=\frac{\pi}{2},\ \theta=\frac{5\pi}{4},\ \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=π4\theta=\frac{\pi}{4} in the original equation

    sin⁡(π2)−cos⁡(π2)=1\sin\left(\frac{\pi}{2}\right)-\cos\left(\frac{\pi}{2}\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

    sin⁡(π)−cos⁡(π)=1\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=5π4\theta=\frac{5\pi}{4} in the original equation

    sin⁡(5π2)−cos⁡(5π2)=1\sin\left(\frac{5\pi}{2}\right)-\cos\left(\frac{5\pi}{2}\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

    sin⁡(3π)−cos⁡(3π)=1\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. Cross-check with the RR-formula: find RR

    R=(1)2+(−1)2=2R=\sqrt{\left(1\right)^{2}+\left(-1\right)^{2}}=\sqrt{2}

    A completely different method should give the same solutions.

  17. Select the complete solution set

    θ=π4, θ=π2, θ=5π4, θ=3π2\theta=\frac{\pi}{4},\ \theta=\frac{\pi}{2},\ \theta=\frac{5\pi}{4},\ \theta=\frac{3\pi}{2}

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

Answer
θ=π4, θ=π2, θ=5π4, θ=3π2\theta=\frac{\pi}{4},\ \theta=\frac{\pi}{2},\ \theta=\frac{5\pi}{4},\ \theta=\frac{3\pi}{2}
Question 5
9 markschallenging
Given that t=tan⁡θ2t=\tan\frac{\theta}{2}, express cos⁡2θ\cos2\theta in terms of tt.
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

    cos⁡2θ\cos2\theta

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

  3. Substitute the t-formulae into the expression

    cos⁡2θ=(t2−2t−1)(t2+2t−1)(t2+1)2\cos2\theta=\frac{\left(t^{2}-2t-1\right)\left(t^{2}+2t-1\right)}{\left(t^{2}+1\right)^{2}}

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

  4. Expand the numerator

    t4−6t2+1=(t2−2t−1)(t2+2t−1)t^{4}-6t^{2}+1=\left(t^{2}-2t-1\right)\left(t^{2}+2t-1\right)

    Multiplying out lets any common factors be spotted.

  5. Expand the denominator

    t4+2t2+1=(t2+1)2t^{4}+2t^{2}+1=\left(t^{2}+1\right)^{2}

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

  6. Cancel any common factors

    cos⁡2θ=(t2−2t−1)(t2+2t−1)(t2+1)2\cos2\theta=\frac{\left(t^{2}-2t-1\right)\left(t^{2}+2t-1\right)}{\left(t^{2}+1\right)^{2}}

    This is the required expression in terms of tt.

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

    t=33,LHS=−12,RHS=−12t=\frac{\sqrt{3}}{3},\quad\text{LHS}=-\frac{1}{2},\quad\text{RHS}=-\frac{1}{2}

    The original expression and the expression in tt agree.

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

    t=3,LHS=−12,RHS=−12t=\sqrt{3},\quad\text{LHS}=-\frac{1}{2},\quad\text{RHS}=-\frac{1}{2}

    The original expression and the expression in tt agree.

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

    t=−3,LHS=−12,RHS=−12t=-\sqrt{3},\quad\text{LHS}=-\frac{1}{2},\quad\text{RHS}=-\frac{1}{2}

    The original expression and the expression in tt agree.

  10. Check the result when θ=300∘\theta=300^{\circ}

    t=−33,LHS=−12,RHS=−12t=-\frac{\sqrt{3}}{3},\quad\text{LHS}=-\frac{1}{2},\quad\text{RHS}=-\frac{1}{2}

    The original expression and the expression in tt agree.

  11. Check the result when θ=30∘\theta=30^{\circ}

    t=2−3,LHS=12,RHS=12t=2-\sqrt{3},\quad\text{LHS}=\frac{1}{2},\quad\text{RHS}=\frac{1}{2}

    The original expression and the expression in tt agree.

  12. Recall the half-angle substitution

    t=tan⁡θ2t=\tan\frac{\theta}{2}

    The Weierstrass substitution turns any rational trigonometric equation into a rational equation in tt.

  13. Quote the t-formula for sin⁡θ\sin\theta

    sin⁡θ=2t1+t2\sin\theta=\frac{2t}{1+t^{2}}

    This follows from sin⁡θ=2sin⁡θ2cos⁡θ2\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}.

  14. Quote the t-formula for cos⁡θ\cos\theta

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

    This follows from cos⁡θ=cos⁡2θ2−sin⁡2θ2\cos\theta=\cos^{2}\frac{\theta}{2}-\sin^{2}\frac{\theta}{2}.

  15. Quote the t-formula for tan⁡θ\tan\theta

    tan⁡θ=2t1−t2\tan\theta=\frac{2t}{1-t^{2}}

    Dividing the formula for sin⁡θ\sin\theta by the formula for cos⁡θ\cos\theta gives this result.

  16. State the final expression

    cos⁡2θ=(t2−2t−1)(t2+2t−1)(t2+1)2\cos2\theta=\frac{\left(t^{2}-2t-1\right)\left(t^{2}+2t-1\right)}{\left(t^{2}+1\right)^{2}}

    The expression is now written entirely in terms of tt.

Answer
(t2−2t−1)(t2+2t−1)(t2+1)2\frac{\left(t^{2}-2t-1\right)\left(t^{2}+2t-1\right)}{\left(t^{2}+1\right)^{2}}

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