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 20o tan 30o tan 40o tan 80o = 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


Chapter 8 - Transformation Formulae Exercise Ex. 8VSAQ

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

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