Quick Revision

Degrees and Radians

$\pi$ radians $= 180^\circ$ . Degrees to radians: $\times \frac{\pi}{180}$ ; radians to degrees: $\times \frac{180}{\pi}$ .
$1^\circ = 60'$ and $1' = 60''$ , so $22^\circ 30' = 22.5^\circ = \frac{\pi}{8}$ .
One radian is about $57.3^\circ$ .
Arc $l = r\theta$ and sector $\frac{1}{2}r^2\theta$ , with $\theta$ in radians.
Clock: the minute hand turns $6^\circ$ a minute, the hour hand $\frac{1}{2}^\circ$ . The hands are $|30h - 5.5m|$ degrees apart; above $180^\circ$ , take it from $360^\circ$ .

For example: in 20 minutes the minute hand turns $120^\circ$ , which is $\frac{2\pi}{3}$ .

$Right triangle with opposite adjacent and hypotenuse marked$

$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ , $\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin\theta}{\cos\theta}$ .
Cosec, sec and cot are the reciprocals of sin, cos and tan.
One ratio given: draw the triangle and find the third side with a triplet (5-12-13, 8-15-17, 7-24-25, 20-21-29, 9-40-41).
From $0^\circ$ to $90^\circ$ , sin and tan rise and cos falls; sin and cos never pass 1.

Sec goes with cos and cosec with sin, so the pairs cross over.

$\sin\theta$ at $0^\circ$ , $30^\circ$ , $45^\circ$ , $60^\circ$ , $90^\circ$ is $\sqrt{\frac{0}{4}}$ , $\sqrt{\frac{1}{4}}$ , $\sqrt{\frac{2}{4}}$ , $\sqrt{\frac{3}{4}}$ , $\sqrt{\frac{4}{4}}$ ; $\cos\theta$ is the same row backwards.
$\tan\theta$ : 0, $\frac{1}{\sqrt{3}}$ , 1, $\sqrt{3}$ , not defined.
With squares, use 0, $\frac{1}{4}$ , $\frac{1}{2}$ , $\frac{3}{4}$ , 1 for $\sin^2\theta$ and 0, $\frac{1}{3}$ , 1, 3 for $\tan^2\theta$ .
Not defined: $\tan 90^\circ$ , $\sec 90^\circ$ , $\cot 0^\circ$ , $\operatorname{cosec} 0^\circ$ .

[SSC Tip] $\frac{1}{\sqrt{3}}$ and $\sqrt{3}$ often sit side by side in the options. $\tan 30^\circ$ is the smaller one, less than 1.

$\sin(90^\circ - \theta) = \cos\theta$ , $\tan(90^\circ - \theta) = \cot\theta$ , $\sec(90^\circ - \theta) = \operatorname{cosec}\theta$ , and the other way round.
$\sin A = \cos B$ with both angles acute means $A + B = 90^\circ$ .
$\tan\theta\tan(90^\circ - \theta) = 1$ : long products of tan give 1.
$\sin^2\theta + \sin^2(90^\circ - \theta) = 1$ : a sum of squares is the number of pairs, plus $\frac{1}{2}$ for $45^\circ$ .
In a triangle, $\sin\frac{B + C}{2} = \cos\frac{A}{2}$ .

A product with $\cos 90^\circ$ in it is 0, however long it is.

Bring the equation to one ratio equal to a standard value, read off the angle, then solve for $\theta$ : $2\cos 3\theta = 1$ gives $3\theta = 60^\circ$ and $\theta = 20^\circ$ .
Two equations in $A + B$ and $A - B$ : find both sums, then add and subtract.
A sin equal to a cos: add the two angles and set the sum to $90^\circ$ .
A quadratic in $\sin\theta$ or $\cos\theta$ : factorise, and drop any root above 1.

Last step every time: check that the angle lies in the range the question allows.

$\sin^2\theta + \cos^2\theta = 1$ , $\sec^2\theta - \tan^2\theta = 1$ , $\operatorname{cosec}^2\theta - \cot^2\theta = 1$ .
$\sec\theta - \tan\theta = \frac{1}{\sec\theta + \tan\theta}$ , and the same for cosec and cot.
$\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta$ ; for sixth powers the 2 becomes 3.
$1 \pm \sin\theta$ leads to sec and tan; $1 \pm \cos\theta$ leads to cosec and cot.

To choose between options, test at $45^\circ$ . If two options agree there, test again at $30^\circ$ or $60^\circ$ .

$\sin\theta + \cos\theta = p$ gives $\sin\theta\cos\theta = \frac{p^2 - 1}{2}$ .
$\sec\theta + \tan\theta = p$ gives $\sec\theta - \tan\theta = \frac{1}{p}$ and $\sin\theta = \frac{p^2 - 1}{p^2 + 1}$ .
$\tan\theta$ given: divide the top and the bottom by $\cos\theta$ .
$\tan\theta + \cot\theta = 2$ gives $\theta = 45^\circ$ .
$a\cos\theta + b\sin\theta = c$ gives $a\sin\theta - b\cos\theta = \pm\sqrt{a^2 + b^2 - c^2}$ .
$\sin\theta + \sin^2\theta = 1$ gives $\sin\theta = \cos^2\theta$ .

$a\sin\theta + b\cos\theta$ runs from $-\sqrt{a^2 + b^2}$ to $\sqrt{a^2 + b^2}$ ; a constant outside shifts both ends.
$a\sin^2\theta + b\cos^2\theta$ lies between a and b.
$a\tan^2\theta + b\cot^2\theta \ge 2\sqrt{ab}$ and $a\sec^2\theta + b\operatorname{cosec}^2\theta \ge (\sqrt{a} + \sqrt{b})^2$ .
$\sin\theta\cos\theta \le \frac{1}{2}$ and $\sin^2\theta + \cos^4\theta \ge \frac{3}{4}$ .

The AM-GM value counts only if it can be reached: $\cos^2\theta + 36\sec^2\theta$ has least value 37, not 12.

Positive ratios: all in the first quadrant, sin and cosec in the second, tan and cot in the third, cos and sec in the fourth.
$180^\circ \pm \theta$ and $360^\circ - \theta$ keep the ratio; $90^\circ \pm \theta$ and $270^\circ \pm \theta$ switch to the co-ratio.
Take off whole turns of $360^\circ$ first. $\cos(-\theta) = \cos\theta$ but $\sin(-\theta) = -\sin\theta$ .

$Point on a circle of radius 1 at 135 degrees$

For example: at $135^\circ$ , $\sin 135^\circ = \frac{1}{\sqrt{2}}$ and $\cos 135^\circ = -\frac{1}{\sqrt{2}}$ , as the figure shows.

$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$ and $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$ .
$\tan 15^\circ = 2 - \sqrt{3}$ , $\tan 75^\circ = 2 + \sqrt{3}$ .
$A + B = 45^\circ$ : $(1 + \tan A)(1 + \tan B) = 2$ .
$\sin 2\theta = 2\sin\theta\cos\theta$ ; $\cos 2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$ .
[Tier 2] $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ , $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ , $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$ .

Write the standard table and the triplets on rough paper first.
Read the options before working; test at $45^\circ$ , then at $30^\circ$ or $60^\circ$ .
Watch for the usual slips: sin and cos swapped, $\frac{1}{\sqrt{3}}$ for $\sqrt{3}$ , the sign in the second quadrant, a missing $\pi$ , $\cos(A + B)$ split up, and $a + b$ as a maximum.
Angles of elevation and depression come in Chapter 16, using the values of $30^\circ$ , $45^\circ$ and $60^\circ$ from here.

Then work through the PYQ Practice Sets of this chapter at under a minute a question.