Trigonometric Functions — MHT-CET Maths Formulas and Traps

Trigonometric Equations — General Solutions, Counting Roots and a cos x + b sin x

General and Principal Solutions of sin θ = k, cos θ = k, tan θ = k

The three general-solution patterns

\sin\theta=\sin\alpha \Rightarrow \theta=n\pi+(-1)^n\alpha \qquad \cos\theta=\cos\alpha \Rightarrow \theta=2n\pi\pm\alpha \qquad \tan\theta=\tan\alpha \Rightarrow \theta=n\pi+\alpha

$\alpha$ any one solution — usually the principal value
$n$ any integer

Equations That Reduce to a Quadratic in One Ratio — and the Roots to Reject

Reduce, then check

\cos^2 x = 1-\sin^2 x \qquad -1 \le \sin x,\ \cos x \le 1 \qquad \tan x,\ \sec x \text{ need } \cos x \ne 0

a cos x + b sin x = c — the R Form, When a Solution Exists, and Roots as a Pair

The R form and the existence condition

a\cos x+b\sin x=\sqrt{a^2+b^2}\,\cos(x-\phi) \qquad \text{solvable} \iff |c|\le\sqrt{a^2+b^2}

$\phi$ the angle with $\cos\phi = a/R$ , $\sin\phi = b/R$

Solving by Factorisation — Sum-to-Product and Multiple Angles

Sum-to-product

\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \qquad \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}

Range Arguments — No Solution, Exponential Forms and Trigonometric Inequalities

Bounds that decide an equation

\sin^4 x+\cos^4 x = 1-\tfrac12\sin^2 2x \in \left[\tfrac12,\,1\right] \qquad a^{\sin^2 x}\cdot a^{\cos^2 x}=a

Common traps

Using the sine pattern for cosine

The

(-1)^n

belongs to SINE only.

\cos\theta = \cos\alpha

gives

2n\pi \pm \alpha

;

\tan

gives

n\pi + \alpha

. The options always include the wrong pattern for the right

\alpha

.

Counting the root the equation cannot hold

When the equation contains

\tan

,

\sec

,

\cot

or

\csc

, clearing denominators can create a root where one of them is undefined. The option that counts it (3 instead of 2 for

\tan x + \sec x = 2\cos x

) is always there.

Counting the integers at the boundary

\sqrt{74} \approx 8.6

is not an integer, so

-8.6 \le 2k + 1 \le 8.6

gives

-4.8 \le k \le 3.8

: the integers are

-4

to

3

, eight of them. Rounding the bound before solving for

k

is the usual way to lose one.

Dividing by a factor that can be zero

Cancelling

\sin 3x

from both sides of

2\sin 3x\cos 2x = \sin 3x

loses every root of

\sin 3x = 0

. Move everything to one side and factor instead.

Counting two roots per value of sin²x instead of four

\sin^2 x = \frac14

means

\sin x = \frac12

OR

-\frac12

, and each has two roots in

[0, 2\pi]

. Over

[0, \pi]

only the positive value counts, and it has two.

A second root of an equation the key counts once

(1 - \tan^2\theta)\sec^2\theta + 2^{\tan^2\theta} = 0

is keyed 2 values, from

\tan^2\theta = 3

. The equation

2^t = t^2 - 1

also crosses near

t \approx 3.41

. The paper's answer is 2; know that it counts only the exact root.

Solution of a Triangle — the Sine, Cosine and Projection Rules

Learn this subtopic in the notes

The Sine Rule and the Circumradius

Sine rule

\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R

$R$ radius of the circumcircle

The Cosine Rule — an Angle from Three Sides, a Side from Two and the Included Angle

Cosine rule

c^2=a^2+b^2-2ab\cos C \qquad \cos C=\frac{a^2+b^2-c^2}{2ab}

Reading an Angle off a Relation Among the Sides

Target form

a^2+b^2-c^2=k\,ab \iff \cos C=\frac{k}{2}

The Projection Rule

Projection rule

a=b\cos C+c\cos B \qquad b=c\cos A+a\cos C \qquad c=a\cos B+b\cos A

Common traps

Putting the sides in the ratio of the angles

Angles 2 : 3 : 7 do NOT give sides 2 : 3 : 7. The sides follow the sines —

\sqrt2 : 2 : (\sqrt3 + 1)

here — and the option that copies the angle ratio is the distractor.

Computing the angle opposite the wrong side

The largest angle is opposite the largest side. Plugging the sides into

\cos C

in the order printed, rather than putting the largest side as

c

, gives a positive cosine and an acute answer that looks plausible.

Losing the sign of k

a^2 + b^2 - c^2 = -ab

is an obtuse

C = 120^\circ

, not

60^\circ

. The stems that give the sum and product of two sides (

x^2 - c^2 = y

) lead exactly there.

Solution of a Triangle — Half-Angle Formulas, Napier's Analogy and Area

Learn this subtopic in the notes

Half-Angle Formulas in Terms of s

Half-angle formulas

\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \qquad \tan\frac{A}{2}\tan\frac{C}{2}=\frac{s-b}{s} \qquad \cot\frac{B}{2}\cot\frac{C}{2}=\frac{s}{s-a}

$s$ semi-perimeter, (a + b + c)/2

Napier's Analogy — the Difference of Two Angles

Napier's analogy

\tan\frac{B-C}{2}=\frac{b-c}{b+c}\cot\frac{A}{2}

Right Triangle: tan of the Two Half-Angles as Roots of a Quadratic

The relation

\tan\left(\tfrac{A}{2}+\tfrac{B}{2}\right)=1 \;\Rightarrow\; p+q=r \quad\text{for } px^2+qx+r=0

Area — Heron's Formula and Its Consequences

Heron's formula

\Delta=\sqrt{s(s-a)(s-b)(s-c)}=\tfrac12\,bc\sin A

Common traps

Using the side you were given instead of the one left over

\tan\frac{A}{2}\tan\frac{C}{2}

leaves

s - b

, not

s - a

or

s - c

. Name the angle that is missing from the product — its side is the one in the numerator.

Inverse Trigonometric Functions — Principal Values and Evaluating Expressions

Learn this subtopic in the notes

A Trigonometric Ratio of an Inverse Value — the Right-Triangle Conversion

Two conversions worth memorising

\sin(\cot^{-1}x)=\cos(\tan^{-1}x)=\frac{1}{\sqrt{1+x^2}} \qquad \sin(2\sin^{-1}x)=2x\sqrt{1-x^2}

Principal Ranges, Negative Arguments and f⁻¹(f(x))

Function	Domain	Principal range
$\sin^{-1}x$	$[-1, 1]$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ Q
$\cos^{-1}x$	$[-1, 1]$	$[0, \pi]$ Q
$\tan^{-1}x$	$\mathbb{R}$	$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ Q
$\cot^{-1}x$	$\mathbb{R}$	$(0, \pi)$ Q
$\sec^{-1}x$	$\|x\| \ge 1$	$[0, \pi]$ , not $\frac{\pi}{2}$ Q
$\csc^{-1}x$	$\|x\| \ge 1$	$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , not 0Q

Common traps

Taking the negative out of cos⁻¹

\cos^{-1}\left(-\frac12\right)

is

\frac{2\pi}{3}

, not

-\frac{\pi}{3}

. Negative angles are outside the range of

\cos^{-1}

,

\cot^{-1}

and

\sec^{-1}

; for these three the rule is

\pi

minus the positive value.

Several options can be true

With

\alpha = 3\sin^{-1}\frac{6}{11}\approx 1.74

and

\beta = 3\cos^{-1}\frac49 \approx 3.33

, the statements

\sin\beta < 0

,

\cos(\alpha + \beta) > 0

and

\cos\alpha < 0

are all true. The paper keys

\sin\beta < 0

; place each angle by estimate before choosing.

Doubling the ratio instead of the angle

\sin(2\sin^{-1}0.8)

is

2(0.8)(0.6) = 0.96

, not

2 \times 0.8

. The double-angle formula needs the cosine too, and the option

0.16

or

1.6

catches the shortcut.

Inverse Trigonometric Identities — Complementary Pairs, the Addition Formula, Substitution and Telescoping

Learn this subtopic in the notes

Complementary Pairs — sin⁻¹x + cos⁻¹x = π/2

The three complementary pairs

\sin^{-1}x+\cos^{-1}x=\tan^{-1}x+\cot^{-1}x=\sec^{-1}x+\csc^{-1}x=\frac{\pi}{2}

The Addition Formula — tan⁻¹x ± tan⁻¹y, 2 tan⁻¹x and Three-Term Sums

Addition formula

\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\ (xy<1) \qquad 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}

Simplifying by Substitution — x = tan θ, cos θ or cos 2θ

The tan θ substitution

x=\tan\theta:\quad \frac{2x}{1+x^2}=\sin2\theta,\quad \frac{1-x^2}{1+x^2}=\cos2\theta,\quad \frac{2x}{1-x^2}=\tan2\theta

Telescoping Sums of Inverse Tangents

The split

\tan^{-1}\frac{A-B}{1+AB}=\tan^{-1}A-\tan^{-1}B

Common traps

Forgetting the xy < 1 condition

When

xy > 1

and

x, y > 0

, the sum is

\pi + \tan^{-1}\frac{x + y}{1 - xy}

. The formula without the

\pi

gives a negative angle for a sum of two positive ones — a sign that the condition was ignored.

Cancelling an inverse outside its range

\tan^{-1}(\tan 2\theta) = 2\theta

needs

-\frac{\pi}{4} < \theta < \frac{\pi}{4}

. If

x = \tan\theta > 1

, the cancellation gives the wrong branch; read the stated range of

x

before cancelling.

Inverse Trigonometric Equations — Solve, Then Check Every Root

Learn this subtopic in the notes

Equations Solved by a Complementary Pair

Replace one of the pair

\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x \qquad \cot^{-1}x=\frac{\pi}{2}-\tan^{-1}x

Equations Solved by the Addition Formula — and the Roots It Adds

Combine, then take tangents

\tan^{-1}u+\tan^{-1}v=\theta \;\Rightarrow\; \frac{u+v}{1-uv}=\tan\theta\quad(\text{then check }uv<1)

Converting Both Sides to One Function, Substituting, and Checking the Domain

Two conversions and a substitution

\sin(\cot^{-1}u)=\cos(\tan^{-1}u)=\frac{1}{\sqrt{1+u^2}} \qquad x=\tan\theta:\ \sin^{-1}\frac{2x}{1+x^2}=2\theta\ (|x|\le1)

Common traps

Keeping the root outside the range

2t^2 - \pi t - \frac{3\pi^2}{8} = 0

gives

t = -\frac{\pi}{4}

or

\frac{3\pi}{4}

, but

t = \tan^{-1}x

lies in

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

. Only

-\frac{\pi}{4}

is allowed, so

x = -1

.

Answering with the number of roots of the quadratic

6x^2 + 5x - 1 = 0

has two roots,

\frac16

and

-1

, but the set was defined with

x \ge 0

. 'Contains two elements' is the planted option; the set is a singleton.

Reporting both signs after squaring

Squaring

\sin^{-1}4x + \sin^{-1}4\sqrt3 x = -\frac{\pi}{2}

gives

x = \pm\frac18

, but

x = \frac18

makes the left side

+\frac{\pi}{2}

. Substitute each candidate. (The paper's options are all written with

\pm

, and its key is

\pm\frac18

.)

Trigonometric Functions formulas

Trigonometric Equations — General Solutions, Counting Roots and a cos x + b sin x

General and Principal Solutions of sin θ = k, cos θ = k, tan θ = k

Equations That Reduce to a Quadratic in One Ratio — and the Roots to Reject

a cos x + b sin x = c — the R Form, When a Solution Exists, and Roots as a Pair

Solving by Factorisation — Sum-to-Product and Multiple Angles

Range Arguments — No Solution, Exponential Forms and Trigonometric Inequalities

Common traps

Solution of a Triangle — the Sine, Cosine and Projection Rules

The Sine Rule and the Circumradius

The Cosine Rule — an Angle from Three Sides, a Side from Two and the Included Angle

Reading an Angle off a Relation Among the Sides

The Projection Rule

Common traps

Solution of a Triangle — Half-Angle Formulas, Napier's Analogy and Area

Half-Angle Formulas in Terms of s

Napier's Analogy — the Difference of Two Angles

Right Triangle: tan of the Two Half-Angles as Roots of a Quadratic

Area — Heron's Formula and Its Consequences

Common traps

Inverse Trigonometric Functions — Principal Values and Evaluating Expressions

A Trigonometric Ratio of an Inverse Value — the Right-Triangle Conversion

Principal Ranges, Negative Arguments and f⁻¹(f(x))

Common traps

Inverse Trigonometric Identities — Complementary Pairs, the Addition Formula, Substitution and Telescoping

Complementary Pairs — sin⁻¹x + cos⁻¹x = π/2

The Addition Formula — tan⁻¹x ± tan⁻¹y, 2 tan⁻¹x and Three-Term Sums

Simplifying by Substitution — x = tan θ, cos θ or cos 2θ

Telescoping Sums of Inverse Tangents

Common traps

Inverse Trigonometric Equations — Solve, Then Check Every Root

Equations Solved by a Complementary Pair

Equations Solved by the Addition Formula — and the Roots It Adds

Converting Both Sides to One Function, Substituting, and Checking the Domain

Common traps

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