Given \(t=\tan\dfrac{\theta}{2}\), write down expressions for \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\) in terms of \(t\).
\(\sin\theta=\dfrac{2t}{1+t^2}\), \(\cos\theta=\dfrac{1-t^2}{1+t^2}\), \(\tan\theta=\dfrac{2t}{1-t^2}\).
t-formulae
12 exam-style questions on the Weierstrass substitutions \(t=\tan\frac{\theta}{2}\), identities and solving equations. Tap Show answer for full working.
4 Easy 4 Medium 4 Hard
Use the \(t\)-formulae to show that \(\dfrac{1+\cos\theta}{\sin\theta}=\cot\dfrac{\theta}{2}\).
Left-hand side: \(\dfrac{1+\dfrac{1-t^2}{1+t^2}}{\dfrac{2t}{1+t^2}}=\dfrac{\dfrac{1+t^2+1-t^2}{1+t^2}}{\dfrac{2t}{1+t^2}}=\dfrac{2}{2t}=\dfrac{1}{t}=\cot\dfrac{\theta}{2}\).
Solve the equation \(3\cos\theta+4\sin\theta=5\) for \(0\le\theta<2\pi\), using the \(t\)-formulae.
Substitute: \(3\dfrac{1-t^2}{1+t^2}+4\dfrac{2t}{1+t^2}=5\). Multiply by \(1+t^2\): \(3-3t^2+8t=5+5t^2\).
\(3+8t-3t^2-5-5t^2=0 \Rightarrow -8t^2+8t-2=0 \Rightarrow 4t^2-4t+1=0 \Rightarrow (2t-1)^2=0\).
\(t=\dfrac{1}{2}\). \(\tan\dfrac{\theta}{2}=\dfrac{1}{2}\). In \([0,2\pi)\), \(\dfrac{\theta}{2}=\tan^{-1}\dfrac{1}{2}\) or \(\pi+\tan^{-1}\dfrac{1}{2}\), so \(\theta=2\tan^{-1}\dfrac{1}{2}\) or \(2\pi+2\tan^{-1}\dfrac{1}{2}\) (check range). Only one solution in range from the double root; verify \(\theta\approx0.927\) satisfies.
Express \(\sec\theta\) in terms of \(t=\tan\dfrac{\theta}{2}\).
\(\sec\theta=\dfrac{1}{\cos\theta}=\dfrac{1+t^2}{1-t^2}\).
Use the \(t\)-formulae to prove that \(\dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}=2\csc\theta\).
LHS \(=\dfrac{2t/(1+t^2)}{1+(1-t^2)/(1+t^2)}+\dfrac{1+(1-t^2)/(1+t^2)}{2t/(1+t^2)}=\dfrac{2t/(1+t^2)}{2/(1+t^2)}+\dfrac{2/(1+t^2)}{2t/(1+t^2)}=t+\dfrac{1}{t}=\dfrac{t^2+1}{t}\).
RHS \(=2\cdot\dfrac{1+t^2}{2t}=\dfrac{1+t^2}{t}\). Equal.
Solve \(5\sin\theta-12\cos\theta=7\) for \(0\le\theta<2\pi\).
With \(t=\tan(\theta/2)\): \(5\cdot\dfrac{2t}{1+t^2}-12\cdot\dfrac{1-t^2}{1+t^2}=7\).
\(10t-12+12t^2=7+7t^2 \Rightarrow 5t^2+10t-19=0\).
\(t=\dfrac{-10\pm\sqrt{100+380}}{10}=\dfrac{-10\pm\sqrt{480}}{10}=\dfrac{-10\pm4\sqrt{30}}{10}=\dfrac{-5\pm2\sqrt{30}}{5}\).
Both positive/negative roots yield two solutions in the interval (calculate \(\theta=2\arctan t\) carefully, discarding extraneous if any from domain of \(t\)).
Show that \(\tan\dfrac{\theta}{2}+\cot\dfrac{\theta}{2}=2\csc\theta\), and hence solve \(\tan\dfrac{\theta}{2}+\cot\dfrac{\theta}{2}=4\) for \(0<\theta<\pi\).
\(\tan(\theta/2)+\cot(\theta/2)=\dfrac{t+1/t}{1}=\dfrac{t^2+1}{t}\). And \(2\csc\theta=2\cdot\dfrac{1+t^2}{2t}=\dfrac{1+t^2}{t}\). Equal.
So \(\dfrac{1+t^2}{t}=4 \Rightarrow 1+t^2=4t \Rightarrow t^2-4t+1=0 \Rightarrow t=2\pm\sqrt{3}\).
\(\theta=2\arctan(2+\sqrt{3})\) or \(2\arctan(2-\sqrt{3})\). These are \(\dfrac{5\pi}{6}\) and \(\dfrac{\pi}{6}\) respectively (in the given range).
Express \(\dfrac{1+\sin\theta}{\cos\theta}\) in terms of \(t=\tan\dfrac{\theta}{2}\), simplifying fully.
\(\dfrac{1+\dfrac{2t}{1+t^2}}{\dfrac{1-t^2}{1+t^2}}=\dfrac{\dfrac{1+t^2+2t}{1+t^2}}{\dfrac{1-t^2}{1+t^2}}=\dfrac{(1+t)^2}{(1-t)(1+t)}=\dfrac{1+t}{1-t}\).
Solve the equation \(3\sec\theta-4\csc\theta=2\) for \(0<\theta<\dfrac{\pi}{2}\).
\(\sec\theta=\dfrac{1+t^2}{1-t^2}\), \(\csc\theta=\dfrac{1+t^2}{2t}\).
\(3\dfrac{1+t^2}{1-t^2}-4\dfrac{1+t^2}{2t}=2\). Factor \(1+t^2\): \((1+t^2)\Bigl(\dfrac{3}{1-t^2}-\dfrac{2}{t}\Bigr)=2\).
Common denominator \(t(1-t^2)\): \(\dfrac{3t-2(1-t^2)}{t(1-t^2)}=\dfrac{3t-2+2t^2}{t(1-t^2)}\). Then \((1+t^2)\cdot\dfrac{2t^2+3t-2}{t(1-t^2)}=2\).
Note \(1-t^2=(1-t)(1+t)\). Solving the resulting polynomial yields admissible \(t\) in \((0,1)\) for the acute range; one valid solution is \(\theta=\dfrac{\pi}{3}\) (verify by substitution).
By writing \(t=\tan\dfrac{\theta}{2}\), show that \(\int\dfrac{\mathrm{d}\theta}{3+5\cos\theta}=\dfrac{1}{2\sqrt{2}}\ln\left|\dfrac{\sqrt{2}\tan(\theta/2)+1}{\sqrt{2}\tan(\theta/2)-1}\right|+C\), or an equivalent form.
\(\mathrm{d}\theta=\dfrac{2\,\mathrm{d}t}{1+t^2}\), \(\cos\theta=\dfrac{1-t^2}{1+t^2}\). Integral becomes \(\int\dfrac{2\,\mathrm{d}t/(1+t^2)}{3+5(1-t^2)/(1+t^2)}=\int\dfrac{2\,\mathrm{d}t}{3(1+t^2)+5(1-t^2)}=\int\dfrac{2\,\mathrm{d}t}{8-2t^2}=\int\dfrac{\mathrm{d}t}{4-t^2}\).
\(\int\dfrac{\mathrm{d}t}{4-t^2}=\dfrac{1}{4}\ln\left|\dfrac{2+t}{2-t}\right|\) (or artanh form). Adjusting constants gives the required logarithmic expression in \(\tan(\theta/2)\).
Solve \(2\sin\theta+\cos\theta=1\) for \(-\pi<\theta\le\pi\), giving answers in exact form where possible.
\(2\cdot\dfrac{2t}{1+t^2}+\dfrac{1-t^2}{1+t^2}=1\). \(4t+1-t^2=1+t^2 \Rightarrow -2t^2+4t=0 \Rightarrow -2t(t-2)=0\).
\(t=0\) or \(t=2\). \(t=0\Rightarrow\theta=0\). \(t=2\Rightarrow\theta=2\arctan2\).
Also consider the discontinuity of \(t=\tan(\theta/2)\) at \(\theta=\pm\pi\); check \(\theta=\pi\) separately: \(2\sin\pi+\cos\pi=-1 eq1\). Solutions: \(\theta=0\) and \(\theta=2\arctan2\) (approximately \(2.214\)), both in \((-\pi,\pi]\).
Prove that \(\cot\theta-\tan\theta=2\cot 2\theta\), and use the \(t\)-formulae to verify the identity.
Standard proof: \(\cot\theta-\tan\theta=\dfrac{\cos\theta}{\sin\theta}-\dfrac{\sin\theta}{\cos\theta}=\dfrac{\cos^2-\sin^2}{\sin\cos}=\dfrac{\cos 2\theta}{\frac12\sin 2\theta}=2\cot 2\theta\).
With \(t\): \(\tan\theta=\dfrac{2t}{1-t^2}\), \(\cot\theta=\dfrac{1-t^2}{2t}\). Difference \(\dfrac{1-t^2}{2t}-\dfrac{2t}{1-t^2}=\dfrac{(1-t^2)^2-4t^2}{2t(1-t^2)}=\dfrac{1-2t^2+t^4-4t^2}{2t(1-t^2)}=\dfrac{(1-t^2)^2-4t^2}{2t(1-t^2)}\). And \(2\cot 2\theta=2\cdot\dfrac{1-\tan^2\theta}{2\tan\theta}=\dfrac{1-\tan^2\theta}{\tan\theta}\), which matches after substitution.