course Mth 164
Some of these problems and the book problems I left unanswered because I spent so much time doing and redoing some of the identity problems. Some of those problems I probably spent 3-4 hrs on and still didn't get them right.
Section 6.3 (7.3 starting with 8th edition)Establish each identity.
6. Sin `theta ( cot `theta + tan `theta ) = sec `theta
Sin(theta)cot(theta) + sin(theta)tan(theta) = sec(theta)
Sin(theta)(cos(theta)/sin(theta)+)sin(theta)(sin(theta)/cos(theta) = 1/cos(theta)
Cos(theta)+sin^2(theta)/cos(theta) = 1/cos(theta)
Cos^2(theta)+1-cos^2(theta)/cos(theta) = 1/cos(theta) = sec(theta)
10. ( csc `theta - 1 ) ( csc `theta + 1 ) = cot^2(`theta)
Csc^2-1 = (1+cot^2(theta)) - 1 = cot^2(theta)
12. ( csc `theta + cot `theta ) ( csc `theta - cot `theta ) = 1
Csc^2(theta)-cot^2(theta) =1
(1+cot^2(theta)-cot^2(theta) ) = 1
18. csc ˆ4 `theta- csc^2(`theta) = cotˆ4 `theta + cot^2(`theta)
(1+cot^2(theta))(1+cot^2(theta)-(1+cot^2(theta) =
1+2cot^2(theta)+cot^4(theta)-1+cot^2(theta) = cot^4(theta)+cot^2(theta)
20. csc `theta - cot `theta =( sin `theta / ( 1 + cos `theta )
1/sin(theta) - cos(theta)sin(theta) = (sin(theta)(1-cos(theta)/(sin(theta)(sin(theta) =
(sin)(1-cos(theta)/sin^2(theta)= (sin(theta)(1-cos(theta)/(1-cos^2(theta) = sin(theta)/(1+cos(theta)
24. 1 - ( sin² `theta / ( 1- cos `theta ) ) = -cos `theta
1-(1+cos(theta)(1-cos(theta)/(1-cos(theta) = cos(theta)
30. ( cos `theta + 1/ ( cos `theta - 1) ) = ( 1+ sec `theta) /( 1- sec `theta )
Cos^2(theta)+2cos(theta)+1/cos^2-1 = (1+2(1/cos(theta)+(1/cos(theta))(1/cos(theta)))/(1-(1(1/cos(theta))(1/cos(theta))**couldn’t get this one any farther
36. ( 1 - cos `theta ) / ( 1 + cos `theta) = ( csc `theta - cot `theta )^2
(1-cos(theta))*(1-cos(theta))/(1+cos(theta))*(1-cos(theta)) =
(1-2cos(theta+cos^2(theta))/(1-cos^2(theta)) = 1-2cos(theta)+cos^2(theta)/(1-sin^2(theta)
Cos^2 - 2(cos(theta)/sin(theta)*cos(theta)+cos^2(theta)/sin^2(theta) ?
40. sin `theta cos `theta / ( cos^2 (`theta) - sin^2( `theta) ) = tan `theta / ( 1- tan^2( `theta ))
Sin(theta)cos(theta)/(cos(theta)-(1-cos^2(theta)) =
sin(theta)cos(theta)/(cos^2(theta)-1+cos^2(theta)) =
sin(theta)cos(theta)/(cos^2(theta)-sin^2(theta)) ?
42. (sin `theta — cos `theta +1) /( sin `theta + cos `theta – 1 ) = ( sin `theta +1) / cos `theta
(sin(theta)-cos(theta)+1)/(sin(theta)+cos(theta)-1)*(sin(theta)+cos(theta)+1)/(sin(theta)+cos(theta)+1=
Sin^2(theta)-cos^2(theta)+sin(theta)+1/sin^2(theta)+2sin(theta)cos(theta)+sin(theta)+cos^2(theta)-1=
Sin(theta)(sin(theta)+sin(theta)+1)/sin(theta)(2cos(theta)+1)*****could not get it past this - I think it has something to do with what I multiplied by - I tried (sin(theta)-cos(theta)+1) and that didn’t work either
Starting with
(sin `theta — cos `theta +1) /( sin `theta + cos `theta – 1 ) = ( sin `theta +1) / cos `theta,
I would first multiply both sides by the common denominator ( sin `theta + cos `theta – 1 ) * cos(theta) to get
(sin `theta — cos `theta +1) * cos(theta) = ( sin `theta +1) * ( sin `theta + cos `theta – 1 ),
then multiply everything out to get
sin(theta) cos(theta) - cos^2(theta) + cos(theta) = sin^2(theta) + sin(theta) cos(theta) - sin(theta) + sin(theta) + cos(theta) - 1
sin(theta) cos(theta) and cos(theta) both appear on both sides and - sin(theta) + sin(theta) on the right-hand side is zero, so we get
- cos^2(theta) = sin^2(theta) - 1, which quickly rearranges to
1 = sin^2(theta) + cos^2(theta).
48. sec`theta / (1 + sec `theta ) = ( 1 – cos `theta ) /sin^2( `theta )
Cos(theta)cos(theta)/(cos(theta)+1) = (cos^2(theta))/(1+cos(theta))*(1-cos(theta))/(1-cos(theta)) = 1-sin^2(theta)-cos^2(theta)cos(theta)/(1-cos^2(theta) =
Cos^2(theta)?(1-cos(theta))/1-cos^2(theta) = (1-cos(theta)/sin^2(theta)
50. (1- cot^2( `theta ) ) / ( 1+ cot^2(`theta) )+ 2 cos^2(`theta) = 0
(1+cot(theta))(1-cot(theta))/(1+cot(theta))(1+cot(theta)) = -2cot(theta) =
-2(cos(theta)/sin(theta)) = -cos(theta)-cos(theta)/sin(theta) = 0
54. tan `theta + cot `theta –sec `theta csc `theta = 0
Sin(theta)/cos(theta)+cos(theta)/cos(theta)-(1/cos(theta)(1(sin(theta)) =
Sin^2+cos^2/(cos(theta)sin(theta)-(1/sin(theta))(1/sin(theta)) - 1/cos(theta)sin(theta) = 0
60. (sec^2(`theta) – tan^2(`theta) + tan `theta) / sec `theta =sin `theta + cos `theta
1+tan^2(theta)-tan^2(theta)+tan(theta) = 1+tan(theta)/sec(theta) = (1+sin(theta)/cos(theta))/(1/cos(theta)) = cos(theta)+sin(theta)
66. ( cos `theta + sin `theta–sin³`theta) /sin`theta = cot `theta + cos`theta
Cos(theta)+sin(theta)-(sin(theta)sin(theta)sin(theta)/sin(theta) = cos(theta)+sin(theta)-(1-cos^2(theta)(sin(theta)) = cos(theta)+sin(theta)(cos(theta)/sin(theta)
70. ( 1 + cos `theta + sin `theta) / ( 1+ cos `theta–sin `theta )= sec `theta+ tan `theta
(1+cos(theta)+sin(theta)/(1+cos(theta)-sin(theta))*(1+cos(theta)+sin(theta))/(1+cos(theta)+sin(theta)) =
1/cos(theta)+sin(theta)/cos(theta)
(2+2cos(theta)+2sin(theta)+2cos(theta)sin(theta)/cos^2(theta)+2cos(theta)
=sin(theta)(cos(theta)+1)/cos(theta) ?
72. ( 2a sin `theta cos`theta)^2 + a^2(cos^2(theta) – sin^2 (`theta))^2=a^2
78. ln |tan`theta |=ln |sin`theta| –ln| cos`theta |
80. ln |sec`theta+tan`theta |+ln |sec`theta–tan`theta |=0
Not sure what to do with these 2 - is ln referring to logarithms
Section 6.4 (7.4 starting with 8th edition)
Find the exact value of each trigonometric function.
6. sin 105 `deg = 7pi/12 = sin(15pi/12-8pi/12) = .9659
10. tan(19 `pi /12) = tan(21pi/12-2pi/12) = -3.732
12. cot(-5 `pi /12) = did not know what to do w/ negative
Find the exact value of each expression.
18. (tan 40 `deg –tan 10 `deg) / (1+tan 40 `deg tan 10 `deg) = tan(4pi/18-pi/18) = .5774
20. cos (5 `pi /12) cos (7 `pi /12) – sin (5 `pi /12) sin (7 `pi /12)=cos(a-B) = cos(5pi/12-7pi/12) = .8660 = sq. rt. Of 3/2
Find the exact value of each of the following under the given conditions.
24. cos `alpha = `sqrt( 5 )/5, 0<`alpha < `pi /2; sin `beta = -4/5, - `pi /2<`beta<0 ?
30. If cos `theta=¼,with `theta in quadrant IV, find the exact value of
(a) sin `theta = y/r = -sq. rt. Of 15/4 y = -sq. rt. Of 15
(b) sin ( `theta – `pi / 6) = sin(-sq. rt. Of 15/4-pi/6 = -.9969
(c) cos (`theta + `pi / 3 ) = (¼_+pi/3) = .2702
(d) tan ( `theta – `pi / 4) = tan(sq. rt. Of 15 - pi/4) = -18.5
Establish each identity.
36. Cos ( `pi + `theta ) = -cos `theta (-1)(cos(theta) - (0)(sin(theta) = -sin(theta)
40. cos (3 `pi / 2 + `theta ) = sin `theta (0)cos(theta)-(-1)sin(theta) = sin(theta)
42. cos ( `alpha + `beta ) + cos (`alpha – `beta ) = 2 cos `alpha cos `beta
Cos a cos B-sin a sin B+ cos a cos B+ sin a sin B = 2cos a cos B
48. cos (`alpha + `beta)/cos (`alpha–`beta) = (1- tan `alpha tan `beta) / (1+ tan `alpha tan `beta)
Cos a cos B-sin a sin B/ cos a cos B+ cos a cos B = (1-(sin a/cos a)*(sin B/cos B)/(1+( sin a /cos a)*(sin B/cos B) = cos a cos b- sin a sin B/(cos a cos B) ?
50.cot(`alpha–`beta) = (cot`alpha cot`beta+1)/ (cot`beta–cot`beta)
Does not exist because denom. = 0 ?
54.cos(`alpha–`beta)cos(`alpha+`beta) = cos^2(`alpha) –^ sin^2(`beta)
Cos^2a cos^2B - sin^2a sin^2B could not get rid of cos^2B, sin^2a in middle
60.If tan `alpha = x+1and tan`beta = x–1,show that 2 cot(`alpha–`beta) = x^2
Simplified to (x^2-2)/(x^2-) = (2/(x^2-2) , could not get it any farther
I've answered the one question I could find (marked with ****) and noted that your work generally looks quite good. I think you're in good shape here.
You should be submitting the 'open' qa's and queries, which is a lot less work than typing in all the problems and solutions. You're doing well at it, but you would get more in-depth feedback from the 'open' documents.
&#Let me know if you have questions. &#