Trigonometry: Simplification, Eliminating Theta and Radians

Choosing the option by trying one easy angle, how to choose it, and the three situations where the trick can mislead.

If an identity is true for every angle, it is also true for the one angle we choose. So instead of simplifying with algebra we pick an easy angle, find the value of the given expression, find the value of every option at that same angle, and the option that matches is the answer. This is called putting a value, and in option-type questions it turns minutes into seconds.

The rule for choosing the angle is simple: take one where the functions in the question have easy values. Choose from the table below, and remember never to use an angle where a function is undefined (like $\tan 90^{\circ }$) or where some denominator becomes zero.

Three situations where one angle is not enough: (1) two or more options match at that value, like $\sin \theta =\cos \theta$ and $\tan \theta =\cot \theta$ at $45^{\circ }$; then put a second angle ($30^{\circ }$ is the best partner because $\sin \ne \cos$ there). (2) A denominator becomes zero or the form $\frac{0}{0}$ appears; change the angle. (3) The question says “prove that” or has no options; then a number is not a proof, you must show it by algebra.

1. Look at the options: which functions do they use? Choose the angle from that (table above).
2. Check that at the chosen angle no function in the question or an option is undefined and no denominator is zero.
3. Find the exact value of the given expression ($\sqrt{2}$, $\frac{1}{2}$ and so on); do not go to decimals.
4. Find the value of every option at the same angle and compare.
5. If more than one option matches, put a second angle and filter again. When only one is left, that is the answer.

Using double-angle formulas to turn every layer into a perfect square, and fixing the sign from the range of the angle.

A square root disappears only when what is inside is a perfect square. Trigonometry gives four tools to make perfect squares, all from the double-angle formulas: $1+\cos 2\theta =2{\cos }^2\theta$, $1-\cos 2\theta =2{\sin }^2\theta$, and $1\pm \sin 2\theta =(\sin \theta \pm \cos \theta {)}^2$ (because ${\sin }^2+{\cos }^2=1$ and $2\sin \theta \cos \theta =\sin 2\theta$).

1. Find the innermost root and look at what is under it.
2. Write it as $2(1\pm \cos 2x)$ or $(\sin x\pm \cos x{)}^2$ (using $1={\sin }^2+{\cos }^2$ or pulling a $2$ out).
3. Take the root. Fix the sign from the range of the angle (modulus).
4. Put the result into the next layer and repeat, until all roots are gone.
5. Test with one angle ($30^{\circ }$ or $15^{\circ }$).

Calling $\sin \theta +\cos \theta$ by the name $x$ and writing the rest in $x$, products of the form $(a+1)(a-1)$, and the chain of ${\sin }^n+{\cos }^n$.

When an expression keeps showing $\sin \theta +\cos \theta$ (or $\tan \theta +\cot \theta$), give it a name $x$ and write everything else in $x$. The key is one square: $(\sin \theta +\cos \theta {)}^2={\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta =1+2\sin \theta \cos \theta$. So the product $\sin \theta \cos \theta$ is tied to the sum $x$.

Letters in the table: $x=\sin \theta +\cos \theta$ (the given sum). The last column shows how each form comes from $\sin \theta \cos \theta =\frac{x^2-1}{2}$.

Second trick: $(a+1)(a-1)=a^2-1$. A product whose two brackets are the same $a$ with $+1$ and $-1$ becomes $a^2-1$. Example: $(1+\tan \theta +\sec \theta )(1+\tan \theta -\sec \theta )=(1+\tan \theta {)}^2-{\sec }^2\theta =1+2\tan \theta +{\tan }^2\theta -{\sec }^2\theta =2\tan \theta$. Third trick (hidden square): if $\sin \theta +\mathrm{cosec}\theta =2$, since $a+\frac{1}{a}\ge 2$ with equality only at $a=1$, we get $\sin \theta =1$.

Why the chain is true. $({\sin }^{n-2}\theta +{\cos }^{n-2}\theta )({\sin }^2\theta +{\cos }^2\theta )={\sin }^n\theta +{\cos }^n\theta +{\sin }^{n-2}\theta {\cos }^2\theta +{\cos }^{n-2}\theta {\sin }^2\theta$; the first factor is $u_{n-2}$, the second is $1$, and the last two terms are ${\sin }^2\theta {\cos }^2\theta ({\sin }^{n-4}\theta +{\cos }^{n-4}\theta )=p{u}_{n-4}$. Hence $u_{n-2}=u_n+p{u}_{n-4}$.

1. Spot the combination that keeps coming ($\sin +\cos$, $\tan +\cot$, $\sin -\cos$) and name it $x$.
2. Square it and write $\sin \theta \cos \theta$ in $x$ ($\frac{x^2-1}{2}$).
3. Break the asked expression into fractions/powers (table) and put in $x$.
4. If a product has the $(a+1)(a-1)$ shape, turn it into $a^2-1$ first.
5. Check the answer by putting one angle ($45^{\circ }$ or $30^{\circ }$).

When the question says the angles are in AP, the middle angle is $60^{\circ }$; solve by writing $A=60^{\circ }-d$, $C=60^{\circ }+d$ or by putting angles that obey the condition.

If three numbers are in AP (arithmetic progression), the middle one is their average and the difference stays the same: $B-d,B,B+d$. If these are the angles $A,B,C$ of a triangle then $A+B+C=180^{\circ }$ gives $3B=180^{\circ }$, i.e. $B=60^{\circ }$, and $A=60^{\circ }-d$, $C=60^{\circ }+d$. In the picture $d$ changes, angle $A$ shrinks and $C$ grows, but $B$ stays $60^{\circ }$.

The useful fact: the sum of the $\sin$ (or $\cos$) of $A$ and $C$ becomes simple. $\sin (B-d)+\sin (B+d)=(\sin B\cos d-\cos B\sin d)+(\sin B\cos d+\cos B\sin d)=2\sin B\cos d$ (the middle terms cancel); likewise $\cos (B-d)+\cos (B+d)=2\cos B\cos d$. Both have $\cos d$, so dividing cancels it and $\tan B$ is left.

Putting values: choose angles that obey the condition, like $(30^{\circ },60^{\circ },90^{\circ })$ ($d=30^{\circ }$) or $(40^{\circ },60^{\circ },80^{\circ })$ ($d=20^{\circ }$). $(60^{\circ },60^{\circ },60^{\circ })$ is also an AP ($d=0$), but with $A=C$ many expressions give $\frac{0}{0}$ or the same values; do not choose it. Always check with two different $d$.

1. Look at the sum of the angles. For a triangle the middle angle is $B=60^{\circ }$.
2. Write $A=60^{\circ }-d$, $C=60^{\circ }+d$ (or find $d$ from the condition, like “the largest angle is three times the smallest”).
3. Replace sums of $\sin ,\cos$ by $2\sin B\cos d$, $2\cos B\cos d$; cancel $\cos d$.
4. Check by putting angles with two different $d$.

Removing $\theta$ from $x=f(\theta )$, $y=g(\theta )$: when to square and add, when to square and subtract, when to add/subtract first and then multiply.

The forms that come again and again, their eliminations and the curves the point traces, with the moving $\sec ,\tan$ picture.

Solving for $\sin \theta$ and $\cos \theta$ when they are mixed in both equations; and removing one of two angles $\alpha ,\beta$ to find the other.

When $\tan \theta$ is given, the value of an expression in $\sin$ and $\cos$: by dividing top and bottom by a power of $\cos$.

Bringing a whole equation to one function with ${\sin }^2=1-{\cos }^2$ and so on, solving the quadratic and rejecting impossible roots; tricks for proving identities.

Understanding the radian (cut the arc into radius-long pieces), converting both ways, and the exact value of $1$ radian.

In radians $l=r\theta$, sector area $\frac{1}{2}{r}^2\theta =\frac{1}{2}rl$ and perimeter, with pictures.

The distance of a rolling wheel, the rate of turning in radians per second, and the relation for belts and gears.

The minute hand moves $6^{\circ }$ per minute and the hour hand $0.5^{\circ }$ per minute; from these comes $|30H-5.5m|$ and its traps.

When the hands are at a given angle (together, right angle, opposite) and how many times in a day, from the relative speed.

Every method: how to recognise the question and which formula or shortcut to use.

Mixed questions like SSC CGL/CHSL/CPO — first spot the method, then solve.

The key formulas and rules in small question-and-answer cards.