Class 10:
Tri - 3 , Gono - sides , Metron - Measure
The study of relationship between sides and angles of a triangle .
Trigonometric ratios
sine = opp / hyp ,
cosecant = 1 / sin = hyp / opp ,
cosine = adj / hyp ,
secant = 1 / cos = hyp / adj ,
tangent = sin / cos = opp / adj ,
cotangent = 1 / tan = adj / opp.
The values of the trigonometric ratios of an angle do not vary with length of the sides of the triangles if the angles remains the same.
[!fact] sin or cos is always less than 1 because hyp is always the largest side in the triangle
Pythagoras theorem
o p p 2 opp^2 o p p 2 + a d j 2 adj^2 a d j 2 = h y p 2 hyp^2 h y p 2
Trigonometric ratios for specific angles
All angles which are integral multiples of π/2 are called quadrantal angles
Trigonometric Quadratic identities
s i n 2 sin^2 s i n 2 θ + c o s 2 cos^2 co s 2 θ = 1
s i n 2 sin^2 s i n 2 θ = 1 − c o s 2 cos^2 co s 2 θ
c o s 2 cos^2 co s 2 θ = 1 − s i n 2 sin^2 s i n 2 θ
t a n 2 tan^2 t a n 2 θ - s e c 2 sec^2 se c 2 θ= 1
t a n 2 tan^2 t a n 2 θ + 1 = s e c 2 sec^2 se c 2 θ
t a n 2 tan^2 t a n 2 θ = s e c 2 sec^2 se c 2 θ − 1
c o s e c 2 cosec^2 cose c 2 θ - c o t 2 cot^2 co t 2 θ= 1
c o t 2 cot^2 co t 2 θ + 1 = c o s e c 2 cosec^2 cose c 2 θ
c o t 2 cot^2 co t 2 θ = c o s e c 2 cosec^2 cose c 2 θ − 1
Opposite Angle Identities
s i n ( − θ ) = − s i n ( θ ) sin(−θ) = −sin(θ) s in ( − θ ) = − s in ( θ )
c o s ( − θ ) = c o s ( θ ) cos(−θ) = cos(θ) cos ( − θ ) = cos ( θ )
t a n ( − θ ) = − t a n ( θ ) tan(−θ) = −tan(θ) t an ( − θ ) = − t an ( θ )
Trigonometric Identities of Complementary Angles
S i n ( 90 – θ ) = C o s θ Sin (90 – θ) = Cos θ S in ( 90– θ ) = C os θ
C o s ( 90 – θ ) = S i n θ Cos (90 – θ) = Sin θ C os ( 90– θ ) = S in θ
T a n ( 90 – θ ) = C o t θ Tan (90 – θ) = Cot θ T an ( 90– θ ) = C o t θ
C o t ( 90 – θ ) = T a n θ Cot ( 90 – θ) = Tan θ C o t ( 90– θ ) = T an θ
S e c ( 90 – θ ) = C o s e c θ Sec (90 – θ) = Cosec θ S ec ( 90– θ ) = C osec θ
C o s e c ( 90 – θ ) = S e c θ Cosec (90 – θ) = Sec θ C osec ( 90– θ ) = S ec θ
Trigonometric Identities of Supplementary Angles
s i n ( 180 ° − θ ) = s i n θ sin (180°- θ) = sinθ s in ( 180° − θ ) = s in θ
c o s ( 180 ° − θ ) = − c o s θ cos (180°- θ) = -cos θ cos ( 180° − θ ) = − cos θ
c o s e c ( 180 ° − θ ) = c o s e c θ cosec (180°- θ) = cosec θ cosec ( 180° − θ ) = cosec θ
s e c ( 180 ° − θ ) = − s e c θ sec (180°- θ)= -sec θ sec ( 180° − θ ) = − sec θ
t a n ( 180 ° − θ ) = − t a n θ tan (180°- θ) = -tan θ t an ( 180° − θ ) = − t an θ
c o t ( 180 ° − θ ) = − c o t θ cot (180°- θ) = -cot θ co t ( 180° − θ ) = − co tθ
Angle Quadrant sin(…) cos(…) tan(…) Explanation − θ -\theta − θ or 360 ∘ − θ 360^\circ - \theta 36 0 ∘ − θ IV − sin θ -\sin \theta − sin θ cos θ \cos \theta cos θ − tan θ -\tan \theta − tan θ Cos is + in Q4. Function stays same.180 ∘ − θ 180^\circ - \theta 18 0 ∘ − θ II sin θ \sin \theta sin θ − cos θ -\cos \theta − cos θ − tan θ -\tan \theta − tan θ Sin is + in Q2. Function stays same.180 ∘ + θ 180^\circ + \theta 18 0 ∘ + θ III − sin θ -\sin \theta − sin θ − cos θ -\cos \theta − cos θ tan θ \tan \theta tan θ Tan is + in Q3. Function stays same.
Angle Quadrant sin(…) cos(…) tan(…) Explanation 90 ∘ − θ 90^\circ - \theta 9 0 ∘ − θ I cos θ \cos \theta cos θ sin θ \sin \theta sin θ cot θ \cot \theta cot θ All + in Q1. Change to co-function.90 ∘ + θ 90^\circ + \theta 9 0 ∘ + θ II cos θ \cos \theta cos θ − sin θ -\sin \theta − sin θ − cot θ -\cot \theta − cot θ Sin is + in Q2. Change to co-function.270 ∘ − θ 270^\circ - \theta 27 0 ∘ − θ III − cos θ -\cos \theta − cos θ − sin θ -\sin \theta − sin θ cot θ \cot \theta cot θ Tan is + in Q3. Change to co-function.270 ∘ + θ 270^\circ + \theta 27 0 ∘ + θ IV − cos θ -\cos \theta − cos θ sin θ \sin \theta sin θ − cot θ -\cot \theta − cot θ Cos is + in Q4. Change to co-function.
Double Angle Identities
sin 2 x = 2 sin x cos x = 2 tan x 1 + tan 2 x \sin 2x = 2\sin x \cos x = \frac{2\tan x}{1 + \tan^2 x} sin 2 x = 2 sin x cos x = 1 + t a n 2 x 2 t a n x
cos 2 x = cos 2 x − sin 2 x \cos 2x = \cos^2 x - \sin^2 x cos 2 x = cos 2 x − sin 2 x
cos 2 x = 2 cos 2 x − 1 \cos 2x = 2\cos^2 x - 1 cos 2 x = 2 cos 2 x − 1
cos 2 x = 1 − 2 sin 2 x \cos 2x = 1 - 2\sin^2 x cos 2 x = 1 − 2 sin 2 x
cos 2 x = 1 − tan 2 x 1 + tan 2 x \cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x} cos 2 x = 1 + t a n 2 x 1 − t a n 2 x
tan 2 x = 2 tan x 1 − tan 2 x \tan 2x = \frac{2\tan x}{1 - \tan^2 x} tan 2 x = 1 − t a n 2 x 2 t a n x
Triple Angles Identities
sin 3 x = 3 sin x − 4 sin 3 x \sin 3x = 3\sin x - 4\sin^3 x sin 3 x = 3 sin x − 4 sin 3 x
cos 3 x = 4 cos 3 x − 3 cos x \cos 3x = 4\cos^3 x - 3\cos x cos 3 x = 4 cos 3 x − 3 cos x
tan 3 x = 3 tan x − tan 3 x 1 − 3 tan 2 x \tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} tan 3 x = 1 − 3 t a n 2 x 3 t a n x − t a n 3 x
cosec 3 x = cosec 3 x 3 cosec 2 x − 4 \text{cosec } 3x = \frac{\text{cosec}^3 x}{3\text{cosec}^2 x - 4} cosec 3 x = 3 cosec 2 x − 4 cosec 3 x
sec 3 x = sec 3 x 4 − 3 sec 2 x \sec 3x = \frac{\sec^3 x}{4 - 3\sec^2 x} sec 3 x = 4 − 3 s e c 2 x s e c 3 x
cot 3 x = cot 3 x − 3 cot x 3 cot 2 x − 1 \cot 3x = \frac{\cot^3 x - 3\cot x}{3\cot^2 x - 1} cot 3 x = 3 c o t 2 x − 1 c o t 3 x − 3 c o t x
Geek for Geeks - Proof
Angle Sum and Difference Identities
Note that ± means you can use plus or minus, and the ∓ means to use the opposite sign.
s i n ( A ± B ) = s i n ( A ) c o s ( B ) ± c o s ( A ) s i n ( B ) sin(A ± B) = sin(A)cos(B) ± cos(A)sin(B) s in ( A ± B ) = s in ( A ) cos ( B ) ± cos ( A ) s in ( B )
c o s ( A ± B ) = c o s ( A ) c o s ( B ) ∓ s i n ( A ) s i n ( B ) cos(A ± B) = cos(A)cos(B) ∓ sin(A)sin(B) cos ( A ± B ) = cos ( A ) cos ( B ) ∓ s in ( A ) s in ( B )
t a n ( A ± B ) = t a n ( A ) ± t a n ( B ) / 1 ∓ t a n ( A ) t a n ( B ) tan(A ± B) = tan(A) ± tan(B) / 1 ∓ tan(A)tan(B) t an ( A ± B ) = t an ( A ) ± t an ( B ) /1 ∓ t an ( A ) t an ( B )
c o t ( A ± B ) = c o t ( A ) c o t ( B ) ∓ 1 / c o t ( B ) ± c o t ( A ) cot(A ± B) = cot(A)cot(B) ∓ 1 / cot(B) ± cot(A) co t ( A ± B ) = co t ( A ) co t ( B ) ∓ 1/ co t ( B ) ± co t ( A )
S i n A + S i n B = 2 S i n ( ( A + B ) / 2 ) . C o s ( ( A − B ) / 2 ) Sin A + Sin B = 2 {Sin((A+B)/2) . Cos((A-B)/2)} S in A + S in B = 2 S in (( A + B ) /2 ) . C os (( A − B ) /2 )
C o s A + C o s B = 2 C o s ( ( A + B ) / 2 ) . C o s ( ( A − B ) / 2 ) Cos A + Cos B = 2 {Cos((A+B)/2) . Cos((A-B)/2)} C os A + C os B = 2 C os (( A + B ) /2 ) . C os (( A − B ) /2 )
S i n A – S i n B = 2 C o s ( ( A + B ) / 2 ) . S i n ( ( A − B ) / 2 ) Sin A – Sin B = 2 {Cos((A+B)/2) . Sin((A-B)/2)} S in A – S in B = 2 C os (( A + B ) /2 ) . S in (( A − B ) /2 )
C o s A – C o s B = − 2 S i n ( ( A + B ) / 2 ) . S i n ( ( A − B ) / 2 ) Cos A – Cos B = -2 {Sin((A+B)/2) . Sin((A-B)/2)} C os A – C os B = − 2 S in (( A + B ) /2 ) . S in (( A − B ) /2 )
S i n A . S i n B = C o s ( A – B ) – C o s ( A + B ) / 2 Sin A. Sin B = {Cos (A – B) – Cos (A + B)}/2 S in A . S in B = C os ( A – B ) – C os ( A + B ) /2
C o s A . C o s B = C o s ( A + B ) + C o s ( A – B ) / 2 Cos A. Cos B = {Cos (A + B) + Cos (A – B)}/2 C os A . C os B = C os ( A + B ) + C os ( A – B ) /2
S i n A . C o s B = S i n ( A + B ) + S i n ( A – B ) / 2 Sin A. Cos B = {Sin (A + B) + Sin (A – B)}/2 S in A . C os B = S in ( A + B ) + S in ( A – B ) /2
C o s A . S i n B = S i n ( A + B ) – S i n ( A – B ) / 2 Cos A. Sin B = {Sin (A + B) – Sin (A – B)} / 2 C os A . S in B = S in ( A + B ) – S in ( A – B ) /2
Trigonometric Ratios Derivation
https://www.onlinemath4all.com/trigonometric-ratio-table.html
Class 10
8.1 find ratios from triangle 11 questions show that 8.2 use the table and solve 4 questions 8.3 use identities and other formulas 5 questions
RD Sharma new
Trigonometric ratios
find the values of other ratios - use the definition and pythagoras theorem
be sure where the theta is formed , which can be identified from the angle given
in case of prove that or show that questions, check whether any identity can be used, if so proceed ahead
if not then try to find the other side using the pythagoras theorem then substitute in the given LHS
if not then either divide by sin or cos, as depending upon the given, if tan is given divide by cos, if cot is give, divide by sin.
if needed, draw 2 right triangles, marking one angle at each diagram to avoid confusion
be sure of the table values, if the degree is given, then substitute and simplify
find the theta - get the value in the right side, then take inverse theta and get the angle
Trigonometric identities
always look the right hand side, before proceeding
rationalise if the identities cant be substituted
write the lhs in terms of rhs
remember the three square identities of sin, tan, cosec
if the rhs is one, try to substitute in its basic terms of sin and cos
if the value of the some function is asked, and one of the ratio is given, then plot it in the right triangle and find the other side. then from the definition, find the ratios, and subtitute in the function.
Class 12
Degrees, Minutes, Seconds
A degree is divided into 60 minutes, and a minute is divided into 60 seconds . One sixtieth of a degree is called a minute, written as 1′, and one sixtieth of a minute is called a second, written as 1″.
Angle diagrams for different degrees
Radian Measure
Another unit for measurement of an angle, called the radian measure.
radian measures and real numbers can be considered as one and the same.
Angle subtended at the centre by an arc of length 1 unit in a unit circle (circle of radius 1 unit) is said to have a measure of 1 radian.
The circumference of a circle of radius 1 unit is 2π. Thus, one complete revolution of the initial side subtends an angle of 2π radian.
If in a circle of radius r, an arc of length L subtends an angle θ radian at the centre, we have
Radian measure = {π / 180} ×Degree measure
Degree measure = {180 / π} ×Radian measure
Graphs
Sine function
Cosecant function
Cosine function
Secant graph
Tan function
Cot function
Periodicity of ratios
Trig Ratio (f(x)) Domain (Valid x) Range (Output Values) Period General Solution (if f(x)=f(y)) Value at x=nπ (n∈Z) sin x \sin x sin x R \mathbb{R} R [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] 2 π 2\pi 2 π x = n π + ( − 1 ) n y x = n\pi + (-1)^n y x = nπ + ( − 1 ) n y 0 0 0 cos x \cos x cos x R \mathbb{R} R [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] 2 π 2\pi 2 π x = 2 n π ± y x = 2n\pi \pm y x = 2 nπ ± y ( − 1 ) n (-1)^n ( − 1 ) n tan x \tan x tan x R − { ( 2 n + 1 ) π 2 } \mathbb{R} - \{ (2n+1)\frac{\pi}{2} \} R − {( 2 n + 1 ) 2 π } R \mathbb{R} R π \pi π x = n π + y x = n\pi + y x = nπ + y 0 0 0 cot x \cot x cot x R − { n π } \mathbb{R} - \{ n\pi \} R − { nπ } R \mathbb{R} R π \pi π x = n π + y x = n\pi + y x = nπ + y Not Defined (∞ \infty ∞ ) sec x \sec x sec x R − { ( 2 n + 1 ) π 2 } \mathbb{R} - \{ (2n+1)\frac{\pi}{2} \} R − {( 2 n + 1 ) 2 π } R − ( − 1 , 1 ) \mathbb{R} - (-1, 1) R − ( − 1 , 1 ) 2 π 2\pi 2 π x = 2 n π ± y x = 2n\pi \pm y x = 2 nπ ± y ( − 1 ) n (-1)^n ( − 1 ) n cosec x \text{cosec } x cosec x R − { n π } \mathbb{R} - \{ n\pi \} R − { nπ } R − ( − 1 , 1 ) \mathbb{R} - (-1, 1) R − ( − 1 , 1 ) 2 π 2\pi 2 π x = n π + ( − 1 ) n y x = n\pi + (-1)^n y x = nπ + ( − 1 ) n y Not Defined (∞ \infty ∞ )
Note:
For cos ( n π ) \cos(n\pi) cos ( nπ ) , the value oscillates:
it is 1 1 1 when n n n is even (e.g., 0 , 2 π 0, 2\pi 0 , 2 π ) and
− 1 -1 − 1 when n n n is odd (e.g., π , 3 π \pi, 3\pi π , 3 π ).
This is written as ( − 1 ) n (-1)^n ( − 1 ) n
Domain excluded points
Ratio Formula Denominator becomes 0 when… Domain (Excluded Points) sin x \sin x sin x Opp Hyp \frac{\text{Opp}}{\text{Hyp}} Hyp Opp Never R \mathbb{R} R cos x \cos x cos x Adj Hyp \frac{\text{Adj}}{\text{Hyp}} Hyp Adj Never R \mathbb{R} R tan x \tan x tan x sin x cos x \frac{\sin x}{\mathbf{\cos x}} c o s x s i n x cos x = 0 \cos x = 0 cos x = 0 (at 90 ∘ , 270 ∘ . . . 90^{\circ}, 270^{\circ}... 9 0 ∘ , 27 0 ∘ ... )R − { ( 2 n + 1 ) π 2 } \mathbb{R} - \{ (2n+1)\frac{\pi}{2} \} R − {( 2 n + 1 ) 2 π } sec x \sec x sec x 1 cos x \frac{1}{\mathbf{\cos x}} c o s x 1 cos x = 0 \cos x = 0 cos x = 0 (at 90 ∘ , 270 ∘ . . . 90^{\circ}, 270^{\circ}... 9 0 ∘ , 27 0 ∘ ... )R − { ( 2 n + 1 ) π 2 } \mathbb{R} - \{ (2n+1)\frac{\pi}{2} \} R − {( 2 n + 1 ) 2 π } cot x \cot x cot x cos x sin x \frac{\cos x}{\mathbf{\sin x}} s i n x c o s x sin x = 0 \sin x = 0 sin x = 0 (at 0 ∘ , 180 ∘ . . . 0^{\circ}, 180^{\circ}... 0 ∘ , 18 0 ∘ ... )R − { n π } \mathbb{R} - \{ n\pi \} R − { nπ } cosec x \text{cosec } x cosec x 1 sin x \frac{1}{\mathbf{\sin x}} s i n x 1 sin x = 0 \sin x = 0 sin x = 0 (at 0 ∘ , 180 ∘ . . . 0^{\circ}, 180^{\circ}... 0 ∘ , 18 0 ∘ ... )R − { n π } \mathbb{R} - \{ n\pi \} R − { nπ }
General solution
Ratio Quadrants where value matches How we get there General Solution (if f(x)=f(y)) Mathematical Trick Sin I & II Go Forward from even π \pi π , Backward from odd π \pi π x = n π + ( − 1 ) n y x = n\pi + (-1)^n y x = nπ + ( − 1 ) n y ( − 1 ) n (-1)^n ( − 1 ) n handles the sign flip.Cos I & IV Go Up (+ + + ) or Down (− - − ) from the start line (2 n π 2n\pi 2 nπ ) x = 2 n π ± y x = 2n\pi \pm y x = 2 nπ ± y ± \pm ± handles the direction.Tan I & III Just keep adding half-circles (π \pi π ) x = n π + y x = n\pi + y x = nπ + y Simple addition (n π n\pi nπ ).