SHORT NOTES: Trigonometry

Trigonometric functions :

$\begin{array}{l} \sin x = \dfrac{p}{h} \,\,\,\,\,\,\,\, {\rm{cosec}}\;x = \dfrac{h}{p}\\ \cos x = \dfrac{b}{h} \,\,\,\,\,\,\,\, {\rm{sec}}\;x = \dfrac{h}{b}\\ \tan x = \dfrac{p}{b} \,\,\,\,\,\,\,\,\cot \;x = \dfrac{b}{p} \end{array}$

Aid to memory: AFTER SCHOOL TO CINEMA

Steps of finding the value when ${\color{Blue} x> 90}$ (e.g $\sin 480$ ) :

Express x as $x=\left ( 90n\pm \theta \right )$ where $0\leq \theta\leq 90$ (e.g $480=90\times5+30$ )
If n is even function remains same.If n is odd Sin becomes cos and vice versa tan becomes cot & vice versa , sec becomes cosec & vice versa (e.g n is odd so $\cos 30$ )
Find whether function is positive or negative in above fig showing four regions(e.g $\sin 480$ is in 2nd quadrant(360+120) so + )
find the value of function in step-2 and use step-3 to get answer(e.g $\sin 480=+\cos 30=\frac{\sqrt{3}}{2}$

Basic relations between these functions:
(A )Sum

${\sin ^2}x + {\cos ^2}x = 1$

$1 + {\tan ^2}x = {\sec ^2}x$

$1 + {\cot ^2}x = {\rm{cose}}{{\rm{c}}^2}x$

(B)Product

$cosec x=\frac{1}{\sin x}\\$

$\sec x=\frac{1}{\\cos x}$

$\tan x=\frac{1}{\cot x}=\frac{\sin x}{\cos x}$

Periodicity and graphs:

Formulae involving multiple and sub-multiple angles:
1.

$\sin \left( {A \pm B} \right) = \sin A\cos B \pm \cos A\sin B$

$\cos (A \pm B) = \cos A\cos B \mp \sin A\sin B$ NOTE: ${\color{Magenta} \mp }$ SIGN

$\tan (A \pm B) = \dfrac{{\tan A \pm \tan B}}{{1 \mp \tan A\tan B}}$

2.By putting A in place of B in eq(1)

$\sin 2A = 2\sin A\cos A = \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}$

$\cos 2A = {\cos ^2}A - {\sin ^2}A = \dfrac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = 2{\cos ^2}A - 1 = 1 - 2{\sin ^2}A$

$\tan 2A=\frac{2\tan A}{1-\tan^2 A}$

3.By putting 2A and A in place of A & B in eq(1) and applying eq(2)

$\sin 3A = 3\sin A - 4{\sin ^3}A$

$\cos 3A = 4{\cos ^3}A - 3\cos A$

$\tan 3A = \dfrac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}$

4.By putting A/2 in place A in eq (2)

$\sin \dfrac{A}{2} = \sqrt {\dfrac{{1 - \cos A}}{2}}$

$\cos \dfrac{A}{2} = \sqrt {\dfrac{{1 + \cos A}}{2}}$

$\tan \dfrac{A}{2} = \sqrt {\dfrac{{1 - \cos A}}{{1 + \cos A}}} = \dfrac{{1 - \cos A}}{{\sin A}} = \dfrac{{\sin A}}{{1 + \cos A}}$

5.Taking A+B=C & A-B=D in eq(1)

$\sin C + \sin D = 2\sin \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$

$\sin C - \sin D = 2\cos \dfrac{{C + D}}{2}\;\sin \dfrac{{C - D}}{2}$

$\cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\;\cos \dfrac{{C - D}}{2}$

$\cos C - \cos D = 2\sin \dfrac{{C + D}}{2}\;\sin \dfrac{{D - C}}{2}$

General solution of trigonometric equations:

$\alpha \in \left [ -\frac{\pi }{2},{\frac{\pi }{2}} \right ]$

$\alpha\in [0,\pi ]$

$\alpha \in \left [ -\frac{\pi }{2},{\frac{\pi }{2}} \right ]$

Relations between sides and angles of a triangle:

Sine Rule:

$\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}$

${\Delta = \dfrac{1}{2}ab\sin C = \dfrac{1}{2}bc\sin A = \dfrac{1}{2}ca\sin B\;}$

Cosine Rule:

${a^2} = {b^2} + {c^2} - 2bc\;\cos A$

${b^2} = {c^2} + {a^2} - 2ac\;\cos B$

${c^2} = {a^2} + {b^2} - 2ab\;\cos C$

Projection Rule:

$a=b\cos C+c\cos B$

$b=c\cos A+a\cos C$

$c=a\cos B+b\cos A$

half-angle formula:

$s=\frac{a+b+c}{2}$

$\sin\frac{A}{2}=\sqrt{\frac{\left ( s-b \right )\left ( s-c \right )}{bc}}\ \sin\frac{B}{2}=\sqrt{\frac{\left ( s-c \right )\left ( s-a \right )}{ca}} \ \sin\frac{C}{2}=\sqrt{\frac{\left ( s-a \right )\left ( s-b \right )}{ab}}$

$\cos\frac{A}{2}=\sqrt{\frac{\left ( s \right )\left ( s-a \right )}{bc}},\ \cos\frac{B}{2}=\sqrt{\frac{\left ( s \right )\left ( s-b \right )}{ca}}, \ \cos\frac{C}{2}=\sqrt{\frac{\left ( s \right )\left ( s-c \right )}{ab}}$

$\tan\frac{A}{2}=\sqrt{\frac{\left ( s-b \right )\left ( s-c \right )}{s\left ( s-a \right )}},\ \tan\frac{B}{2}=\sqrt{\frac{\left ( s -c\right )\left ( s-a \right )}{s\left ( s-b \right )}}, \ \tan\frac{C}{2}=\sqrt{\frac{\left ( s-a \right )\left ( s-b \right )}{s\left ( s-c \right )}}$

Area of a triangle:

Area $=\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}=\frac{abc}{4R}=rs$

R = radius of circumcircle

r = radius of incircle

Inverse trigonometric functions (principal value only):

$f(x) = {\sin ^{ - 1}}x,\;[ - 1,\;1] \to \left[ {\dfrac{{ - \pi }}{2},\;\dfrac{\pi }{2}} \right]$

$f(x) = {\cos ^{ - 1}}x,\;[ - 1,\;1] \to [0,\;\pi ]$

$f(x) = {\tan ^{ - 1}}x, \, \mathbb{R}\left( {\dfrac{{ - \pi }}{2},\;\dfrac{\pi }{2}} \right)$

$f(x) = {\cot ^{ - 1}}x,\; \mathbb{R}\to (0,\;\pi )$

$f(x) = {\sec ^{ - 1}}x,\;\left( { - \infty ,\; - 1} \right] \cup \left[ {1,\;\infty } \right) \to [0,\;\pi ]\backslash \left\{ {\pi /2} \right\}$

$f(x) = {\rm{cose}}{{\rm{c}}^{ - 1}}\,x,\;\left( { - \infty ,\; - 1} \right]\; \cup \left[ {1,\;\infty } \right) \to \left[ {\dfrac{{ - \pi }}{2},\;\dfrac{\pi }{2}} \right]\;\backslash \;\left\{ 0 \right\}$

Properties:

${{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x = \dfrac{\pi }{2}}$

${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$ for all $x \in\mathbb{R}$

${\sec ^{ - 1}}x + {\rm{cose}}{{\rm{c}}^{ - 1}}x = \dfrac{\pi }{2}$ for all $x \in \;\left( { - \infty , - 1} \right] \cup \left[ {1,\;\infty } \right)$

1 comment:

Chuck BushDecember 5, 2024 at 3:55 PM
Trigonometry often feels complex, but your approach makes it much more digestible. I believe understanding these relationships is foundational to mastering higher-level math. For anyone struggling with similar topics, I'd highly recommend seeking math tuition to ensure a deeper grasp of the subject. Personalized guidance can make a huge difference in improving one's confidence and skills in mathematics.

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