Some basic mathematic for physic CLASS 11 CBSE NEET ALL parts viral

part 1 

part 2 

Basic Mathematics 

TRIGONOMETRY Terminal line It is one of the important branch of mathematics which deal with relations of sides and angles of triangle and also with the relevant functions of any angle. Consider a ray OA. If this ray revolves about its end point O in anti-clockwise direction and takes position OB, then we say that the angle ∠AOB has been generated as shown in the following figure.

 B O Initial line (i) Positive angle Initial line O A A B (ii) Negative angle Fig. 0.1 Or simply say that angle is a measure of an amount of revolution of a given ray about its initial point. The angle is positive (or negative), if the initial line revolves in anti-clockwise (or clockwise) direction to get the terminal line. System of measurement of angles (i) Sexagesimal system In this system, a right angle is divided into 90 equal parts, called degree. 

Terminal line Thus, 

1 right angle = ° 90 (degrees) 

1 60 ° = ′ (minutes) 

1 60 ′ = ′′ (seconds)

(ii)CircularsystemInthissystem, theunitof measurement isradian. θ= l r radianorrad 1rightangle= π 2 rad 1straightangle=πrad and1completeangle=2πrad Oneradian(denotedby1c) isthemeasureofan anglesubtendedatthecentreofacirclebyanarcof lengthequal totheradiusofthecircle. 1 rad (1c)= ° ≈ ° ′ ′′≈ ° 180 571745 π 57.3 Note (i) π= = 22 7 314. (ii)360 2 °= πradian, π 2 radian= ° 90 Example0.1 Find the radian measures corresponding to the following degree measures. (i)75° (ii)− ° ′ 3730 (iii)53730 ° ′ ′′ Sol. We have,1 180 c= ° π ⇒1 180 °=       π c (i)75 75 180 5 12 °= ×       =       π π c c (ii)− ° ′=−       ° =−       ° =− ×       = 3730 37 1 2 75 2 75 2 180 π c −       5 24 π c (iii)53730 5 5 8 45 8 45 8 180 ° ′ ′′=       ° =       ° = ×       π c =      π 32 c Example0.2 Find the degree measures corresponding to the following radian measures. (i) 2 15 π       c (ii) π 8       c (iii) ( ) −2c Sol.We have, (i) 2 15 2 15 180 24 π π π       = ×       ° = ° c (ii) π π π 8 8 180 45 2       = ×       ° =       ° c =       ° = ° ×       ′ 22 1 2 22 1 2 60 = ° ′ 2230 (iii) ( ) − = − ×       ° =− × ×       ° 2 2 180 2 180 22 7 c π =−       ° 114 6 11 =− ° ×       =− °       ′ ′ 114 6 11 60 114 32 8 11 =− ° ′ ×       ″ − ° ′ ′′ 11432 8 11 60 114 32 44 • [ ] Trigonometricalratios(orT-ratios) Consider the two fixed linesXOX ′ andYOY′intersecting each other at right angle at pointOas shown in the following figure Then, (i) IntersectionpointOiscalledorigin. (ii)XOX ′ andYOY′arecalledX-axisandY-axis, respectively. (iii)TheportionXOY,YOX′,XOY ′ ′andYOX ′ are knownasI, II, IIIandIVquadrants,respectively. Now, consider that the revolving lineOAhas traced out an angleθin anti-clockwise direction (in I quadrant). From pointA, drawAB OX ⊥ which results a right angled ∆ABO, whereAB=perpendicular,OA=hypotenuse and OB=base. The three sides of right angled triangle are related to each other through side having different ratios, called trigonometrical ratios orT-ratios, which are given as (i) sinθ= = Perpendicular Hypotenuse AB OA (FromFig.0.3) (ii)cosθ= = Base Hypotenuse OB OA (iii) tanθ= = Perpendicular Base AB OB (iv)cotθ= = Base Perpendicular OB AB (v) secθ= = Hypotenuse Base OA OB (vi)cosec Hypotenuse Perpendicular θ= =OA AB FundamentalofT-ratiosor trigonometricfunctions Foranyacuteanglesayθ( ) < ° 90 ,thefunctionsaregivenas (i)cosecθ θ = 1 sin (ii) sec cos θ θ = 1 2 OBJECTIVEPhysicsVol.1 X′ X Y′ Y θ A B O Fig. 0.3Trigonometrical ratios θ O r l Fig. 0.2Circular system (iii)cot tan θ θ = 1 (iv) sin cos 2 2 1 θ θ + = (v)1 2 2 + = tan sec θ θ (vi)1 2 + = cot θ θ cosec2 Signs of trigonometric ratios orT-ratios in various quadrants (i) InIquadrant,allT-ratiosarepositive. (ii) InIIquadrant,sinθis+ve,cosθandtanθare−ve. (iii) InIIIquadrant,tanθis +ve,sinθandcosθare−ve. (iv) InIVquadrant,cosθis+ve,sinθandtanθare−ve. Example0.3 Ifsinθ= 4 5, whereθlies in the first quadrant, then find all the other T-ratios. Sol. Let∆PQRberightangledtriangle,rightangledatQ. sinθ= 4 5 (Given) ∴ PR=5andPQ=4 On applying Pythagoras theorem in∆PQR, we have ( ) ( ) ( ) PR PQ QR 2 2 2 = + ⇒ ( ) ( ) ( ) 5 4 2 2 2 = +QR ⇒ QR= − = = 25 16 9 3 [Taking positive value of square root] Now, cosθ= = QR PR 3 5, tanθ= = PQ QR 4 3, cotθ= = QR PQ 3 4, secθ= = PR QR 5 3 and cosecθ= = PR PQ 5 4 T-ratiosofalliedangles In trigonometry two angles are said to be allied angles when their sum or difference is a multiple of 90°. TheT-ratios of the following allied angles are as (i)Whenangle(say)θisnegative, then (a)sin( ) sin − =− θ θ (b)cos( ) cos − = θ θ (c) tan( ) tan − =− θ θ (ii)Whenangleθislessthan90° (i.e., liesinIquadrant), then (a)sin( ) cos 90°− = θ θ (b)cos( ) sin 90°− = θ θ (c) tan( ) cot 90°− = θ θ (iii)Whenangleθliesbetween90°and180° (i.e., liesinIIquadrant), then (a) sin( ) cos 90°+ = θ θ cos(90 ) sin °+ =− θ θ tan( ) cot 90°+ =− θ θ (b) sin( ) sin 180°− = θ θ cos(180 cos °− =− θ) θ tan( ) tan 180°− =− θ θ (iv)Whenangleθliesbetween180°and270° (i.e., liesinIIIquadrant), then (a) sin( ) sin 180°+ =− θ θ cos( ) cos 180°+ =− θ θ tan( ) tan 180°+ = θ θ (b) sin( ) cos 270°− =− θ θ cos( ) sin 270°− =− θ θ tan( ) cot 270°− = θ θ (v)Whenangleθliesbetween270°and360° (i.e., liesinIVquadrant), then (a) sin( ) cos 270°+ =− θ θ cos( ) sin 270°+ = θ θ tan( ) cot 270°+ =− θ θ (b) sin( ) sin 360°− =− θ θ cos( ) cos 360°− = θ θ tan( tan 360°− =− θ) θ Values ofT-ratios of some standard angles Angle (θ) 0° 30° =      π 6 45° =      π 4 60° =      π 3 90° =      π 2 120° =       2 3 π 135° =       3 4 π 150° =       5 6 π 180° (=π) sinθ 0 1 2 1 2 3 2 1 3 2 1 2 1 2 0 cosθ 1 3 2 1 2 1 2 0 − 1 2 − 1 2 − 3 2 −1 tanθ 0 1 3 1 3 ∞ − 3 −1 − 1 3 0 BasicMathematics 3 I All+ve IV onlycosand secare+ve II onlysinandcosec are+ve III onlytanand cotare+ve Fig. 0.4Sign of T-ratios Q P R θ 4 5 Example0.4 Find the value of (i)sin( ) − ° 45 (ii) tan225° (iii)cos300° (iv)sec120° Sol. (i) sin( ) sin − ° =− ° 45 45 [Qsin( sin − =− θ) θ] =− 1 2 [Qsin45 1 2 °= ] (ii) tan tan( ) cot 225 270 45 45 °= °− ° = ° [Qtan( cot 270°− = θ) θ] =1 [Qcot45 1 °= ] (iii) cos cos( ) sin 300 270 30 30 °= °+ ° = ° [Qcos( sin 270°+ = θ) θ] = 1 2 [Qsin30 1 2 °= ] (iv) sec sec( ) sec 120 180 60 60 °= °− ° =− ° [Qsec( sec 180°− =− θ) θ] =−2 Someimportantformulaeof trigonometry sin( ) sin cos cos sin A B A B A B + = + sin( ) sin cos cos sin A B A B A B − = − cos( ) cos cos sin sin A B A B A B + = − cos( ) cos cos sin sin A B A B A B − = + tan( ) tan tan tan tan A B A B A B + = + −1 tan( ) tan tan tan tan A B A B A B − = − +1 sin sin cos tan tan 2 2 2 1 2 A A A A A = = + cos cos sin cos 2 2 1 2 2 2 A A A A = − = − = − = − + 1 2 1 1 2 2 2 sin tan tan A A A tan tan tan 2 2 1 2 A A A = − sin sin sin 3 3 4 3 A A A = − cos cos cos 3 4 3 3 A A A = − tan tan tan tan 3 3 1 3 3 2 A A A A = − − sin( ) sin( ) sin cos A B A B A B + + − =2 sin( ) sin( ) cos sin A B A B A B + − − =2 cos( ) cos( ) cos cos A B A B A B + + − =2 cos( ) cos( ) sin sin A B A B A B + − − =−2 sin sin sin cos C D C D C D + = + 2 2 2 cos cos cos cos C D C D C D + = +       ⋅ −       2 2 2 cos cos sin sin C D C D C D − =− +  2 2 2 Example0.5 Find the value of (i)sin15°(ii) tan75° Sol. (i)We have, sin sin( ) 15 45 30 °= °− ° = ° °− ° ° sin cos cos sin 45 30 45 30 [Qsin( ) sin cos A B A B − = −cos sin A B] = ⋅ − ⋅ 1 2 3 2 1 2 1 2 (Qsin cos 45 45 1 2 °= °= ,cos30 3 2 °= andsin30 1 2 °= ) = −3 1 2 2 (ii)We have, tan tan( ) 75 45 30 °= °+ ° = °+ ° − °⋅ ° tan tan tan tan 45 30 1 45 30 Qtan( ) tan tan tan tan A B A B A B + = + − ⋅       1 = + − ⋅ 1 1 3 1 1 1 3 Qtan tan 45 1 30 1 3 °= °=       and = + − 3 1 3 1 4 OBJECTIVEPhysicsVol.1 1.Find the radian measures corresponding to the following degree measures. (i)25° (ii)− ° ′ 4730 (iii)392230 ° ′ ′′ Ans. (i) 5 36 π (ii)− 19 72 π (iii) 7 32 π 2.Find the degree measures corresponding to the following radian measures. (i) 18 5 π       c (ii)( ) −3c (iii) −      5 6 π c (iv) 9 5 π       c Ans. (i)648° (ii) − ° ′ ′′ 171495 (iii)− ° 150 (iv)324° 3.Findsinθandtanθ, ifcosθ=− 12 13andθlies in thethird quadrant. Ans.sinθ=− 5 13 andtanθ= 5 12 4.Find the values of other fiveT-ratios, iftanθ=− 3 4andθ lies in II quadrant. Ans.sinθ= 3 5 ,cosθ=− 4 5 ,cosecθ= 5 3 ,secθ=− 5 4 andcotθ=− 4 3 5.Find the values of the followingT-ratios (i) cosec315° (ii) cos210° (iii)sin( ) − ° 330 Ans.(i)− 2 (ii)− 3 2 (iii) 1 2 6.Find the value of (i)sec165° (ii) cot105° Ans.(i)( ) 2 6 − (ii) 1 3 1 3