RD SHARMA Solutions for Class 11-science Maths Chapter 8 - Transformation Formulae

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Chapter 8 - Transformation Formulae Exercise Ex. 8.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

prove that ten 20^o tan 30^o tan 40^o tan 80^o = 1

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Chapter 8 - Transformation Formulae Exercise Ex. 8.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

$text LHS = end text sin 105 to the power of ring operator plus cos 105 to the power of ring operator equals sin 105 to the power of ring operator plus cos left parenthesis 90 to the power of ring operator plus 15 to the power of ring operator right parenthesis equals sin 105 to the power of ring operator minus sin 15 to the power of ring operator equals 2 sin open parentheses fraction numerator 105 to the power of ring operator minus 15 to the power of ring operator over denominator 2 end fraction close parentheses cos open parentheses fraction numerator 105 to the power of ring operator plus 15 to the power of ring operator over denominator 2 end fraction close parentheses equals 2 sin 45 to the power of ring operator cos 60 to the power of ring operator equals 2 fraction numerator 1 over denominator square root of 2 end fraction 1 half equals fraction numerator 1 over denominator square root of 2 end fraction equals cos 45 to the power of ring operator$

Question 7

Solution 7

Question 8

$cos 55 to the power of 0 plus cos 65 to the power of 0 plus cos 175 to the power of 0 space equals space 0$

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Prove that:

$cosθcos straight theta over 2 minus cos 3 θcos fraction numerator 9 straight theta over denominator 2 end fraction equals sin 7 θsin 8 straight theta$

Solution 26

$table attributes columnalign left end attributes row cell Consider text the left hand side of the given expression : end text end cell row cell straight L. straight H. straight S. equals cosθcos straight theta over 2 minus cos 3 θcos fraction numerator 9 straight theta over denominator 2 end fraction end cell row cell We text know that 2 cosAcosB = cos end text left parenthesis text A + B end text right parenthesis plus cos left parenthesis straight A minus straight B right parenthesis end cell row cell Thus comma end cell row cell straight L. straight H. straight S. equals 1 half open square brackets text cos end text open parentheses straight theta text + end text straight theta over 2 close parentheses plus cos open parentheses straight theta minus straight theta over 2 close parentheses close square brackets minus 1 half open square brackets text cos end text open parentheses 3 straight theta text + end text fraction numerator 9 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses 3 straight theta minus fraction numerator 9 straight theta over denominator 2 end fraction close parentheses close square brackets end cell row cell text end text equals 1 half open square brackets text cos end text open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses straight theta over 2 close parentheses close square brackets minus 1 half open square brackets text cos end text open parentheses fraction numerator 15 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses negative fraction numerator 3 straight theta over denominator 2 end fraction close parentheses close square brackets end cell row cell text end text equals 1 half open square brackets text cos end text open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses straight theta over 2 close parentheses close square brackets minus 1 half open square brackets text cos end text open parentheses fraction numerator 15 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses close square brackets text end text text end text left square bracket because cos left parenthesis negative straight theta right parenthesis equals cosθ right square bracket end cell row cell text end text equals 1 half open square brackets text cos end text open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses plus cos open parentheses straight theta over 2 close parentheses minus text cos end text open parentheses fraction numerator 15 straight theta over denominator 2 end fraction close parentheses minus cos open parentheses fraction numerator 3 straight theta over denominator 2 end fraction close parentheses close square brackets end cell row cell text end text equals 1 half open square brackets cos open parentheses straight theta over 2 close parentheses minus text cos end text open parentheses fraction numerator 15 straight theta over denominator 2 end fraction close parentheses close square brackets end cell row cell Also text we know that , end text end cell row cell text cosD end text minus cosC equals 2 sin fraction numerator straight C plus straight D over denominator 2 end fraction sin fraction numerator straight C minus straight D over denominator 2 end fraction end cell row cell Therefore comma end cell row cell straight L. straight H. straight S. equals 1 half cross times 2 sin fraction numerator fraction numerator 15 straight theta over denominator 2 end fraction plus straight theta over 2 over denominator 2 end fraction sin fraction numerator fraction numerator 15 straight theta over denominator 2 end fraction minus straight theta over 2 over denominator 2 end fraction end cell row cell text end text equals sin fraction numerator fraction numerator 16 straight theta over denominator 2 end fraction over denominator 2 end fraction sin fraction numerator fraction numerator 14 straight theta over denominator 2 end fraction over denominator 2 end fraction end cell row cell text end text equals sin fraction numerator 8 straight theta over denominator 2 end fraction sin fraction numerator 7 straight theta over denominator 2 end fraction end cell row cell text end text equals straight R. straight H. straight S. end cell row blank row cell asterisk times Note colon text Question given in the book is incorrect. end text end cell row cell text R. H. S . should be equal to end text sin fraction numerator 8 straight theta over denominator 2 end fraction sin fraction numerator 7 straight theta over denominator 2 end fraction. end cell end table$

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

$text LHS end text equals fraction numerator s i n 3 A cos 4 A minus sin A cos 2 A over denominator sin 4 A sin A plus cos 6 A cos A end fraction equals fraction numerator 2 left parenthesis s i n 3 A cos 4 A minus sin A cos 2 A right parenthesis over denominator 2 left parenthesis sin 4 A sin A plus cos 6 A cos A right parenthesis end fraction equals fraction numerator 2 s i n 3 A cos 4 A minus 2 sin A cos 2 A over denominator 2 sin 4 A sin A plus 2 cos 6 A cos A end fraction equals fraction numerator s i n left parenthesis 4 A plus 3 A right parenthesis minus s i n left parenthesis 4 A minus 3 A right parenthesis minus open square brackets s i n left parenthesis 2 A plus A right parenthesis minus s i n left parenthesis 2 A minus A right parenthesis close square brackets over denominator cos left parenthesis 4 A minus A right parenthesis minus cos left parenthesis 4 A plus A right parenthesis plus cos left parenthesis 6 A plus A right parenthesis plus cos left parenthesis 6 A minus A right parenthesis end fraction equals fraction numerator s i n left parenthesis 7 A right parenthesis minus s i n left parenthesis A right parenthesis minus s i n left parenthesis 3 A right parenthesis plus s i n left parenthesis A right parenthesis over denominator cos left parenthesis 3 A right parenthesis minus cos left parenthesis 5 A right parenthesis plus cos left parenthesis 7 A right parenthesis plus cos left parenthesis 5 A right parenthesis end fraction equals fraction numerator s i n left parenthesis 7 A right parenthesis minus s i n left parenthesis 3 A right parenthesis over denominator cos left parenthesis 3 A right parenthesis plus cos left parenthesis 7 A right parenthesis end fraction equals fraction numerator 2 s i n open parentheses fraction numerator 7 A minus 3 A over denominator 2 end fraction close parentheses cos open parentheses fraction numerator 7 A plus 3 A over denominator 2 end fraction close parentheses over denominator 2 cos open parentheses fraction numerator 7 A plus 3 A over denominator 2 end fraction close parentheses cos open parentheses fraction numerator 7 A minus 3 A over denominator 2 end fraction close parentheses end fraction equals fraction numerator s i n 2 A over denominator cos 2 A end fraction equals tan 2 A equals R H S$

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

$I f space x cos theta equals y cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses equals z cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses comma space P r o v e space t h a t space x y plus y z plus z x equals 0$

Solution 54

$G i v e n space x cos theta equals y cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses equals z cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses equals k left parenthesis s a y right parenthesis x equals fraction numerator k over denominator cos theta end fraction y equals fraction numerator k over denominator cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses end fraction z equals fraction numerator k over denominator cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction x y plus y z plus z x equals k squared open square brackets fraction numerator 1 over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses end fraction plus fraction numerator 1 over denominator cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction plus fraction numerator 1 over denominator cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses cos theta end fraction close square brackets equals k squared open square brackets fraction numerator cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses plus cos theta plus cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction close square brackets equals k squared open square brackets fraction numerator cos theta cos fraction numerator 4 straight pi over denominator 3 end fraction minus sin theta sin fraction numerator 4 straight pi over denominator 3 end fraction plus cos theta plus cos theta cos fraction numerator 2 straight pi over denominator 3 end fraction minus sin theta sin fraction numerator 2 straight pi over denominator 3 end fraction over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction close square brackets equals k squared open square brackets fraction numerator cos theta open parentheses begin display style fraction numerator minus 1 over denominator 2 end fraction end style close parentheses minus sin theta open parentheses begin display style fraction numerator minus square root of 3 over denominator 2 end fraction end style close parentheses plus cos theta plus cos theta open parentheses begin display style fraction numerator minus 1 over denominator 2 end fraction end style close parentheses minus sin theta open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction close square brackets equals k squared open square brackets fraction numerator minus cos theta plus sin theta open parentheses begin display style fraction numerator square root of 3 over denominator 2 end fraction end style close parentheses plus cos theta plus minus sin theta open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses over denominator cos theta cos open parentheses theta plus fraction numerator 2 straight pi over denominator 3 end fraction close parentheses cos open parentheses theta plus fraction numerator 4 straight pi over denominator 3 end fraction close parentheses end fraction close square brackets equals 0 H e n c e space P r o v e d$

Question 55

$If text end text straight m text end text sinθ equals text end text straight n text end text sin left parenthesis straight theta plus 2 straight a right parenthesis comma text end text prove text end text that text end text tan left parenthesis straight theta plus straight a right parenthesis equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction tana$

Solution 55

$table attributes columnalign left end attributes row cell Given text that end text straight m text end text sinθ equals text end text straight n text end text sin left parenthesis straight theta plus 2 straight a right parenthesis comma text end text end cell row cell text We need to end text prove text end text that text end text tan left parenthesis straight theta plus straight a right parenthesis equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction tana end cell row cell straight m text end text sinθ equals text end text straight n text end text sin left parenthesis straight theta plus 2 straight a right parenthesis end cell row cell rightwards double arrow fraction numerator sin left parenthesis straight theta plus 2 straight a right parenthesis over denominator text end text sinθ end fraction equals straight m over straight n end cell row cell text Using Componendo-Dividendo , we have , end text end cell row cell rightwards double arrow fraction numerator sin left parenthesis straight theta plus 2 straight a right parenthesis plus sinθ over denominator sin left parenthesis straight theta plus 2 straight a right parenthesis minus sinθ end fraction equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction.... left parenthesis 1 right parenthesis end cell row cell We text know that , end text end cell row cell text sinC + sinD = 2 sin end text fraction numerator text C + D end text over denominator text 2 end text end fraction text cos end text fraction numerator text C end text minus text D end text over denominator text 2 end text end fraction end cell row and row cell text sinC end text minus text sinD = 2 cos end text fraction numerator text C + D end text over denominator text 2 end text end fraction text sin end text fraction numerator text C end text minus text D end text over denominator text 2 end text end fraction end cell row cell text Applying the above formulae in equation ( 1 ), we have , end text end cell row cell fraction numerator text 2 sin end text fraction numerator straight theta text + 2 a + end text straight theta over denominator text 2 end text end fraction text cos end text fraction numerator straight theta text + 2 a end text minus straight theta over denominator text 2 end text end fraction over denominator text 2 cos end text fraction numerator straight theta text + 2 a + end text straight theta over denominator text 2 end text end fraction text sin end text fraction numerator straight theta text + 2 a end text minus straight theta over denominator text 2 end text end fraction end fraction equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction end cell row cell rightwards double arrow fraction numerator text 2 sin end text left parenthesis straight theta plus straight a right parenthesis cosa over denominator text 2 cos end text left parenthesis straight theta plus straight a right parenthesis sina end fraction equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction end cell row cell rightwards double arrow fraction numerator text tan end text left parenthesis straight theta plus straight a right parenthesis over denominator tana end fraction equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction end cell row cell rightwards double arrow text tan end text left parenthesis straight theta plus straight a right parenthesis equals fraction numerator straight m plus straight n over denominator straight m minus straight n end fraction cross times tana end cell row cell text Hence proved. end text end cell end table$