trignometry basic questions

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some basic questions to enhance understanding of trignometry

Transcript of trignometry basic questions

  • L EC TURE

    22 Trigonometric Functions

    1. TABLE REPRESENTING ALL TRIGONAL RATIOS (TRs) IN TERMS OF ONE TRIGONAL RATIO

    sin cos tan cot sec cosec

    sin sin 21 cos 2tan

    1 tan

    + 21

    1 cot+ 2sec 1

    sec

    1cosec

    cos 21 sin cos

    2

    11 tan+ 2

    cot1 cot

    +

    1sec

    2cosec 1cosec

    tan 2sin

    1 sin

    21 cos

    cos

    tan

    1cot 2sec 1 2

    1cosec 1

    cot21 sin

    sin

    2cos

    1 cos

    1tan cot 2

    1sec 1

    2cosec 1

    sec 21

    1 sin 1

    cos 21 tan+ 21 cot

    cot+

    sec 2

    coseccosec 1

    1. TABLE REPRESENTING ALL TRIGONAL RATIOS (TR

    2. T-RATIOS (OR TRIGONOMETRIC FUNCTIONS)

    sin ,cos , tan ,p b p

    h h b = = =

    cosec ,sec ,cot ;h h b

    p b p = = =

    2. T-RATIOS (OR TRIGONOMETRIC

    O

    p

    B

    A

    h

    bp + b = h2 2 2

    p perpendicular, b base and h stands for hypotenuse.

  • A.18 Trigonometric Functions

    3. FOLLOWING ARE SOME OF THE FUNDAMENTAL TRIGONOMETRIC IDENTITIES

    (i) 1sincosec

    =

    or 1cosecsin

    =

    (ii) 1cossec

    =

    or 1seccos

    =

    (iii) 1cottan

    =

    or 1tancot

    =

    (iv) sintancos

    =

    or coscotsin

    =

    (v) sin2+cos2=1orsin2 =1cos2orcos2 =1sin2

    (vi) sec2tan2=1 or1+tan2sin2

    or 1sec tansec tan

    =+

    (vii) cosec2cot2=1or1+cot2=cosec2

    or 1cosec cotcosec cot

    =+

    4. MAXIMUM AND MINIMUM VALUES OF TRIGONOMETRICAL FUNCTIONS

    (i) 1 sin x 1 (ii) 1 cos x 1 (iii) < tan x < (iv) < cot x < (v) |sec x | 1 i.e., sec x 1orsecx 1 (vi) |cosec x| 1 i.e., cosec x 1orcosecx 1

    3. FOLLOWING ARE SOME OF THE

    4. MAXIMUM AND MINIMUM VALUES OF

    (vii) sin2 x +cosec2 x 2, x R (viii) cos2 x +sec2 x 2, x R (ix) tan2 x +cot2 x 2, x R (x) |sin x+cosecx | 2 (xi) |cos x+secx | 2 (xii) |tan x+cotx | 2 (xiii) If a sin x+b cos x=c, then (a cos xb sin x)2

    =a2 + b2 c2

    5. SOME USEFUL RESULTS

    (i) sin2+cos4=sin4+cos2=1sin2cos2 (ii) sin4+cos4=12sin2cos2 (iii) sin6+cos6 =13sin2cos2 (iv) sin4cos4 =12cos2 (v) sin8cos8=(sin2cos2)(12sin2

    cos2) (vi) sec2+cosec2 =sec2cosec2=tan2+

    cot2+2

    (vii) 1 1 sinsec tansec tan 1 sin

    + = + =

    (viii) 1 1 sinsec tansec tan 1 sin

    = =+ +

    (ix) 1 1 coscosec cotcosec cot 1 cos

    + = + =

    (x) 1 1 coscosec cotcosec cot 1 cos

    = =+ +

    (xi) 1 sin cos sec tancos 1 sin = =

    +

    (xii) 1 cos sin cosec cotsin 1 cos = =

    +

    (xiii) 1 sin cos sec tancos 1 sin = =

    +

    (xiv) 1 cos sin cosec cotsin 1 cos = =

    +

    5. SOME USEFUL RESULTS

    (i) sin

  • Trigonometric Functions A.19

    1. Show that 3 3

    (1 tan cot )(sin cos )sec cosec

    + +

    Solution

    L.H.S.=3 3

    3 3

    sin cos1 (sin cos )(1 tan cot )(sin cos ) cos sin

    1 1sec coseccos sin

    + + + + =

    3 3

    3 3

    sin cos1 (sin cos )(1 tan cot )(sin cos ) cos sin

    1 1sec coseccos sin

    + + + + =

    2 2

    3 3

    3 3

    (sin cos sin cos )(sin cos )sin cossin cossin cos

    + + =

    3 3 3 3

    3 3

    (sin cos ) sin cossin cos (sin cos )

    = 2 2 3 3( ( )( ) )a b ab a b a b+ + =

    2 2sin cos= = R.H.S.

    2. If cos2sin2=tan2,thenshowthattan2 =cos2sin2.

    Solution

    Given, cos2sin2=tan22 2 2

    2 2 2

    cos sin sincos sin cos

    = +

    2 2 2

    2 2 2

    cos sin coscos sin sin

    + =

    2 2( cos sin 1) + =Applying componendo and dividendo, we get

    2 2 2 2 2 2

    2 2 2 2 2 2

    cos sin cos sin cos sincos sin cos sin cos sin

    + + + = + +

    2

    2 2 2

    2cos 12sin cos sin

    =

    22 2

    2

    sin cos sincos

    =

    2 2 2tan cos sin , = as desired.

    3. If sinsin

    A mB= and cos

    cosA nB= ,findthevalueof

    tan B; n2 < 1 < m2.

    Solution

    Given sin sin sinsin

    A m A m BB= = .........(1)

    and cos cos coscos

    A n A n BB= = ..........(2)

    By squaring (1) and (2) and adding, we get

    2 2 2 21 sin cosm B n B= +2

    2 22 2

    1 sincos cos

    Bm nB B

    = + (Dividing by cos2 B)

    2 2 2 2sec tan = +B m B n

    2 2 2 21 tan tan + = +B m B n 2

    2 2 22

    11 ( 1) tan tan1

    nn m B Bm

    2 = =

    2

    2

    1tan1

    nBm =

    4. Eliminatebetween cosec sin , = a sec cos = b.

    Solution

    Given cosec sin a = .............(1)and sec cos b = .............(2)

    From (1), 21 cossin

    sin sina a = =

    .............(3)

    From (2), 21 sincos

    cos cosb b = =

    ..............(4)Squaring(3)andmultiplyingby(4),weget

    3 2 2 1/3cos cos ( )a b a b = = ............(5)

    Squaring(4)andmultiplyingby(3),weget3 2 2 1/3sin sin ( )ab ab = = ............(6)

  • A.20 Trigonometric Functions

    5. Eliminateand between cos cos , cos sin , sin .x r y r z r= = =

    Solution

    Given cos cos , cos sinx r y r= = andsin .= z r

    Squaring and adding, we get2 2 2+ +x y z

    2 2 2 2 2 2 2 2cos cos cos sin sinr r r= + + 2 2 2 2 2 2cos (cos sin ) sinr r= + + 2 2 2 2(sin cos )r r= + =

    2 2[ sin cos 1] + =

    6. If sin cosandsin cos

    m n = =

    , than prove that

    22 2

    2

    1tan ; 1 .1

    m n n mn m

    = < .

    [IIT-JEE-1981]

    Solution

    sin cos 22+ =

  • A.26 Trigonometric Functions

    1. Ifsin= 22

    1tt+

    ,thencosisequalto

    (a) 22

    1tt

    (b) 22

    1tt+

    (c) 2

    2

    11

    tt

    +

    (d) 2

    2

    11

    tt

    +

    2. sin cos1 cot 1 tan

    +

    is equal to

    [Karnataka CET-1998]

    (a) 0 (b) 1(c) cossin (d) cos+sin

    3. If for real values of x,cos=x+ 1x

    , then

    [MPPET-1996](a) isanacuteangle(b) isarightangle(c) isanobtuseangle(d) Novaluesofispossible

    4. The equation sec2 = 24

    ( )xy

    x y+ is only

    possible when [MPPET-1986; IIT-1996](a) x=y (b) x < y(c) x > y (d) None of these

    5. Which of the following relations is correct? [WBJEE-91](a) sin 1< sin 1o

    (b) sin 1 > sin 1o

    (c) sin1=sin1o

    (d) sin1 sin1180

    =

    6. Ifsin+cosec=2,thensin2+cosec2is equal to [MPPET-1992; MNR-1990; UPSEAT-2002](a) 1 (b) 4(c) 2 (d) None of these

    7. Iftan= 2021

    ,coswillbe [MPPET-1994]

    (a) 2041

    (b) 121

    (c) 2129

    (d) 20

    21

    8. If x=a cos3,y=b sin3,then(a) (a/x)2/3+(b/y)2/3=1(b) (b/x)2/3+(a/y)2/3=1(c) (x/a)2/3+(y/b)2/3=1(d) (x/b)2/3+(y/a)2/3=1

    9. If x=sec+tan,thenx+ 1x

    is equal to

    [MPPET-1986](a) 1 (b) 2sec(c) 2 (d) 2tan

    10. If(1+sinA)(1+sinB)(1+sinC)=(1sinA) (1sinB)(1sinC), then each side is equal to(a) sin A sin B sin C(b) cosA cosB cos C(c) sin A cos B cos C(d) cos A sin B sin C

    11. If sin1+ sin2+ sin3=3, thencos1+cos2+cos3 is equal to [EAMCET-1994](a) 3 (b) 2(c) 1 (d) 0

    12. Iftancot=aandsin+cos=b, then (b21) 2 (a2+4)isequalto [WB JEE-1979](a) 2 (b) 4(c) 4 (d) 4

    13. Iftan=sin

    1 cosx

    x

    and tan =

    sin1 cos

    yy

    ,

    then xy

    is equal to [MPPET-1991]

    (a) sinsin

    (b)

    sinsin

    (c) sin1 cos

    (d) sin1 cos

  • Trigonometric Functions A.27

    14. Ifsec+tan=p,thentanisequalto [MPPET-1994]

    (a) 22

    1p

    p (b)

    2 12

    pp

    (c) 2 12

    pp+ (d) 2

    21

    pp +

    15. If 2sin cos

    and =1 cos sin 1 sin

    p q =+ + + , then

    [MPPET-2001]

    (a) pq=1 (b) 1qp=

    (c) qp=1 (d) q+p=1

    16. Theminimumvalueof9tan2+4cot2 is(a) 13 (b) 9

    (c) 6 (d) 12

    17. The maximum value of 4 sin2 x+3cos2 x is [Karnataka CET-2003](a) 3 (b) 4

    (c) 5 (d) 7

    18. Which value of k, (cosx+sinx) 2 +k sin x cos x1=0isidentity? [Kerala (Engg.)-2001](a) 1 (b) 2

    (c) 0 (d) 1

    20. If sinx+sin2x=1,thencos8x+2cos6 x+cos4x is equal to(a) 0 (b) 1(c) 2 (d) 1

    21. The least value of 2 sin2 + 3 cos2 is [MPPET-2010]

    (a) 1 (b) 2 (c)3 (d)5

    22. If 1sec ,4

    A xx

    = + then the value of sec A+tan A is [MPPET-2010](a) 3x (b)

    3

    x

    (c)

    2x (d) 2x

    1. (c) Given 22sin

    1tt

    =+

    By Pythagoras theoremAC2=AB2+BC2

    (1+t2)2=(2t)2+BC2

    (1+t2)2=4t2+BC2

    BC2=(1+t2)24t2=1+t4+2t24t2=1+t42t2

    BC2=(1t2) BC=1t2

    2

    2

    1cos1

    BC tAC t

    = =+

    Second Method

    22

    2 tan2 2sin .1 1 tan

    2

    tt

    = = + + If tan

    2t =

    %WW

    &

    $

    22

    22

    1 tan 12cos11 tan

    2

    tt

    = = ++

    2. (d) Given sin cos1 cot 1 tan

    +

    sin coscos sin1 1sin cos

    = +

    sin cossin cos cos sin

    sin cos

    = +

    2 2sin cos

    sin cos cos sin +

  • A.28 Trigonometric Functions

    2 2sin cos

    sin cos sin cos

    2 2sin cos

    sin cos

    (sin cos )(sin cos )(sin cos )

    +

    sin+cos

    3. (d)Givenequationiscos=x+1/xor x2xcos+1=0for real value of x By B2 > 4AC

    A=1,B=cos,C=1

    cos2>4(1)(1)or|cos|2

    Whichisimpossiblebecause|cos|