MH Board · Chapter

Trigonometric Functions

The concepts MH Board needs, one connected idea at a time.

0 / 15

concepts mastered

Trigonometric equations0/5
Polar coordinates0/1
Solution of a triangle0/5
Inverse trigonometric functions0/4

Ready for you

Up nextPrincipal solutionsThe solutions with $0 \le \theta < 2\pi$ : find the reference angle, then the quadrants where the ratio has the given signStart
Polar coordinatesStart $x = r\cos\theta$ , $y = r\sin\theta$ ; back again with $r = \sqrt{x^2 + y^2}$
The sine ruleStart $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ ;
The cosine ruleStart $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$ ;
The projection ruleStart $a = b\cos C + c\cos B$ : a side is the sum of the other two sides' projections on it
Principal values of inverse functionsStart $\sin^{-1}$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ ,

Whole chapter

How the ideas build on each other. Each one opens once you've mastered what it needs.

Trigonometric equations0 / 5

Principal solutionsThe solutions with $0 \le \theta < 2\pi$ : find the reference angle, then the quadrants where the ratio has the given signStart
General solution of sin θ = sin α

and

\theta

in the point's quadrant

the sides are in the ratio of the sines, not of the angles

the largest angle is obtuse when

a^2 > b^2 + c^2

\cos^{-1}

in

[0, \pi]

,

\tan^{-1}

in

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

:

one angle, in that range

\theta = n\pi + (-1)^n\alpha

,

n \in \mathbb{Z}

; no solution when

|\sin\theta|

would exceed

1

Locked

General solutions of cos θ and tan θ

\cos\theta = \cos\alpha

:

\theta = 2n\pi \pm \alpha

;

\tan\theta = \tan\alpha

:

\theta = n\pi + \alpha

Locked

Equations in sin²θ, cos²θ, tan²θ

\sin^2\theta = \sin^2\alpha

(likewise

\cos^2

,

\tan^2

) gives

\theta = n\pi \pm \alpha

; both signs of the root countLocked

Equations that reduce to the standard formsFactorise, use identities or divide out a common ratio to reach

\sin\theta = k

,

\cos\theta = k

or

\tan\theta = k

; don't divide by a factor that can be

0

Locked

Polar coordinates0 / 1

Polar coordinates $x = r\cos\theta$ , $y = r\sin\theta$ ; back again with $r = \sqrt{x^2 + y^2}$ and $\theta$ in the point's quadrantStart

Solution of a triangle0 / 5

The sine rule $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ ; the sides are in the ratio of the sines, not of the anglesStart
The cosine rule $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$ ;
The projection rule $a = b\cos C + c\cos B$ : a side is the sum of the other two sides' projections on itStart
Half-angle formulae and areaWith $s = \dfrac{a + b + c}{2}$ : $\sin\dfrac{A}{2} = \sqrt{\dfrac{(s - b)(s - c)}{bc}}$
Proving triangle identitiesTurn every side into a sine (or every cosine into sides), then use the angle sum A + B + C = πLocked

Inverse trigonometric functions0 / 4

Principal values of inverse functions $\sin^{-1}$ in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ , $\cos^{-1}$ in $[0, \pi]$ , $\tan^{-1}$ in $(-\frac{\pi}{2}, \frac{\pi}{2})$ : one angle, in that rangeStart
sin⁻¹(sin x) and its relatives $\sin^{-1}(\sin x) = x$ only for $x$ in the principal range; otherwise find the angle in the range with the same sineLocked
Properties of inverse functions $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ ;
Equations in inverse functionsCombine the inverse terms with the properties, take the ratio of both sides, then keep only roots in the principal rangesLocked

the largest angle is obtuse when

a^2 > b^2 + c^2

Start

and area

= \sqrt{s(s - a)(s - b)(s - c)}

Locked

\tan^{-1}x + \tan^{-1}y = \tan^{-1}\dfrac{x + y}{1 - xy}

when

xy < 1

Locked