A LevelFurther MathsEdexcelFurther Pure 1FlashcardsFurther TrigonometryThe t-formulae

The t-formulae (Edexcel A Level Further Maths: Further Pure 1): Flashcards

Exam code: 9FM0

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FrontTrig identities using t-formulae
If $t = \tan \frac{θ}{2}$ , complete all three of the t-formulae:
$\sin θ = \frac{______}{1 + t^{2}}, \cos θ = \frac{______}{1 + t^{2}}, \tan θ = \frac{2 t}{______}$

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Cards in this collection (19)

If $t = \tan \frac{θ}{2}$ , complete all three of the t-formulae:
$\sin θ = \frac{______}{1 + t^{2}}, \cos θ = \frac{______}{1 + t^{2}}, \tan θ = \frac{2 t}{______}$
The completed t-formulae are:
$\sin θ = \frac{2 t}{1 + t^{2}}, \cos θ = \frac{1 - t^{2}}{1 + t^{2}}, \tan θ = \frac{2 t}{1 - t^{2}}$
The two easily confused are $\cos θ$ and $\tan θ$ : $1 - t^{2}$ is on the top for $\cos θ$ and on the bottom for $\tan θ$ .
What do the t-formulae achieve that ordinary trig identities do not?
They express $\sin θ$ , $\cos θ$ and $\tan θ$ in terms of a single variable $t$ , and one that is algebraic rather than trigonometric.
A trigonometric equation or identity therefore becomes an ordinary algebraic one in $t$ , which can be solved or rearranged with nothing more than fraction work and factorising.
True or False?
The t-substitution is always $t = \tan \frac{θ}{2}$ , whatever angles the identity happens to contain.
False.
The substitution is adapted so that $t$ is the tangent of half the angle you are working with.
An identity in $\sin 4 θ$ and $\cos 4 θ$ therefore wants $t = \tan 2 θ$ , and one in $\tan \frac{x}{3}$ and $\sin \frac{x}{3}$ wants $t = \tan \frac{x}{6}$ .
How do you get the t-formulae for $cosec θ$ , $\sec θ$ and $\cot θ$ ?
Turn each of the three fractions upside down, since these are just the reciprocals, giving $cosec θ = \frac{1 + t^{2}}{2 t}$ and $\sec θ = \frac{1 + t^{2}}{1 - t^{2}}$ .
The third is $\cot θ = \frac{1 - t^{2}}{2 t}$ , and none of them has to be derived separately.
Which double-angle identity does each t-formula derivation start from?
Each one takes $A = \frac{θ}{2}$ in a different double-angle identity:
- $\sin θ$ starts from $\sin 2 A \equiv 2 \sin A \cos A$
- $\cos θ$ starts from $\cos 2 A \equiv \cos^{2} A - \sin^{2} A$
- $\tan θ$ starts from $\tan 2 A \equiv \frac{2 \tan A}{1 - \tan^{2} A}$
The $\tan$ one is quickest, because $\tan 2 A$ already has $\tan A$ on the right-hand side and so needs no further work at all.
Deriving $\sin θ = \frac{2 t}{1 + t^{2}}$ starts from $\sin θ \equiv 2 \sin \frac{θ}{2} \cos \frac{θ}{2}$ . What is the next move?
Multiply by $\frac{\sec^{2} \frac{θ}{2}}{\sec^{2} \frac{θ}{2}}$ , which changes nothing but sets up both halves of the fraction at once.
In the denominator $1 + \tan^{2} A \equiv \sec^{2} A$ gives $1 + t^{2}$ , while the numerator simplifies as $2 s c \times \frac{1}{c^{2}} = \frac{2 s}{c} = 2 \tan \frac{θ}{2}$ , which is $2 t$ .
How do you prove a trigonometric identity using the t-formulae?
Convert the trig functions on one side, usually the more complicated one, into algebraic fractions in $t$ , then rearrange with ordinary algebra until the two sides match.
Convert back to $θ$ through $t = \tan \frac{θ}{2}$ as the final step, so the result is stated in the same terms as the identity you were given.
In $t$ form, $\sec θ + \tan θ$ becomes $\frac{1 + 2 t + t^{2}}{1 - t^{2}}$ . How does this simplify?
The numerator is the perfect square ${(1 + t)}^{2}$ and the denominator is the difference of two squares $(1 + t) (1 - t)$ , so cancelling one factor of $1 + t$ leaves $\frac{1 + t}{1 - t}$ .
Expressions built from t-formulae very often factorise like this, so look for a perfect square or a difference of two squares before reaching for anything heavier.
After substituting the t-formulae, what kind of equation are you left with?
A polynomial equation in $t$ , most often a quadratic, though it can be cubic or quartic.
Multiplying through by the common denominator is what produces it, since every t-formula is a fraction and clearing them leaves nothing but powers of $t$ .
A t-substitution has given $t = 1$ and $t = - \frac{1}{3}$ . How do you turn these into values of $x$ ?
Undo $t = \tan \frac{x}{2}$ with $\frac{x}{2} = \arctan t + n π$ , then double to reach $x$ .
Add the $n π$ before doubling rather than after, because $\tan$ has period $π$ in $\frac{x}{2}$ .
Here $\arctan (- \frac{1}{3})$ is negative and outside $0 < x < 2 π$ , but adding $π$ first and then doubling gives $x = 5.64$ to 3 significant figures.
True or False?
Solving a trigonometric equation with the t-substitution finds every solution in the range.
False.
The substitution $t = \tan \frac{θ}{2}$ is undefined at $θ = π$ , so a solution sitting there can never be produced by it.
Try it on $\cos θ = - 1$ : substituting gives $\frac{1 - t^{2}}{1 + t^{2}} = - 1$ , which reduces to $1 = - 1$ and yields nothing, yet $θ = π$ plainly solves the original equation.
Why must you find every root of the equation in $t$ , not just one?
Each root produces its own family of $θ$ values, so stopping early loses a whole family rather than a single answer.
A quadratic in $t$ typically gives two roots, and for $2 \cos x + \sin x = 1$ in $0 < x < 2 π$ each of them contributes exactly one solution.
An equation contains $\sec θ$ rather than $\cos θ$ . Does that change how you use the substitution?
The method is unchanged: substitute the $t$ form of $\sec θ$ straight in and carry on exactly as before.
What changes is only the common denominator needed to combine the fractions, so the algebra is a little heavier while the plan of forming and solving a polynomial in $t$ is identical.
A model is to be rewritten in terms of $t$ and also differentiated. In which order should you do those two things?
Differentiate first, then substitute for $t$ .
Differentiating the $t$ form instead means a chain rule wrapped around a quotient rule, because $t$ is itself a function of $x$ , whereas differentiating $h = \sin 8 x + \cos 8 x$ directly gives $\frac{d h}{d x} = 8 \cos 8 x - 8 \sin 8 x$ in a single line.
Having found $\frac{d h}{d x} = 8 \cos 8 x - 8 \sin 8 x$ , how do you express it in terms of $t = \tan 4 x$ ?
Substitute the $t$ form of each trig term, giving $8 (\frac{1 - t^{2}}{1 + t^{2}}) - 8 (\frac{2 t}{1 + t^{2}}) = \frac{8 (1 - 2 t - t^{2})}{1 + t^{2}}$ .
Both terms already share the denominator $1 + t^{2}$ here, so combining them is only a matter of collecting the numerators.
True or False?
A model containing both $\sin \frac{x}{2}$ and $\sin x$ still needs only one substitution.
True.
Once $\sin \frac{x}{2}$ and $\cos \frac{x}{2}$ are in $t$ form, the double-angle identity $\sin x \equiv 2 \sin \frac{x}{2} \cos \frac{x}{2}$ supplies $\sin x = \frac{4 t (1 - t^{2})}{{(1 + t^{2})}^{2}}$ from the same $t$ .
Note the squared denominator, which is what makes this different from the t-formulae themselves.
A model's derivative comes out as $\frac{2 (1 - t^{2}) (1 + t^{2}) + 4 t (1 - t^{2})}{{(1 + t^{2})}^{2}}$ . What is the first move?
Take out the common factor $2 (1 - t^{2})$ from the numerator, giving $\frac{2 (1 - t^{2}) [(1 + t^{2}) + 2 t]}{{(1 + t^{2})}^{2}}$ .
Tidying the bracket and then the difference of two squares collects everything into $\frac{2 (1 - t) {(1 + t)}^{3}}{{(1 + t^{2})}^{2}}$ .
For a model with $\frac{d A}{d x} = \frac{2 (1 - t) {(1 + t)}^{3}}{{(1 + t^{2})}^{2}}$ , how do you find its stationary points?
Set the numerator equal to zero, which is safe here because ${(1 + t^{2})}^{2}$ can never be zero, giving $t = 1$ and $t = - 1$ .
Each value then has to be converted back into $x$ through $t = \tan \frac{x}{4}$ and tested against the range, which for $0 < x < 2 π$ leaves only $x = π$ .
You have $\frac{d^{2} A}{d x^{2}}$ written in terms of $t$ , and a stationary point at $t = 1$ . How do you classify it?
Substitute $t = 1$ straight into the second derivative and look only at the sign, without converting back to $x$ at all.
Here that gives $\frac{2^{2} \times (- 2)}{2^{2}} = - 2$ , which is negative, so the point is a maximum and the model is at a crest rather than a trough.

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