course Mth 164
I did the online problems while at home with 2 sick kids the past 2 days. I will submit those separately tonight. I did that so I could get the assignment done faster - won't make it a habit.
Assignment #4Section 7.3
3. Suppose that f and g are two functions with the same domain. If f(x) = g(x) for every x in the domain, the equation is called an identity. Otherwise it is called a conditional equation.
9. Rewrite in terms of sine and cosine: tan(theta)*csc(theta)=
Tan = y/x = sin(theta)/cos(theta) sin(theta)/cos(theta)*(1/sin(theta) = 1/cos(theta)
11. Multiply cos(theta)/1-sin(theta) by 1+sin(theta)/1+sin(theta) =
Cos(theta)*(1+sin(theta))/(1-sin^2(theta)) = cos(theta)(1+sin(theta))/(cos(theta))*(cos(theta)) = 1+sin(theta)/cos(theta)
13. Rewrite over a common denominator: (sin(theta)+cos(theta)/cos(theta))+(cos(theta)-sin(theta)/sin(theta)): mult. Left by sin(theta)/sin(theta), mult. Rt. by cos(theta)/cos(theta)= sin^2(theta)+cos^2(theta)/sin(theta)cos(theta) =
1/sin(theta)cos(theta)
15. Multiply and simplify:
(sin(theta)+cos(theta)+sin(theta)+cos(theta)-1)/sin(theta)cos(theta)=
(Sin^2(theta)+2sin(theta)cos(theta)+cos^2(theta) -1)/sin(theta)cos(theta)=
(sin^2(theta)+1-sin^2(theta)-1) +2sin(theta)cos(theta))/sin(theta)cos(theta) = 2
Establish each identity:
19. csc(theta)*cos(theta) = cot(theta) 1/(sin(theta))*cos(theta) = cos(theta)/sin(theta) = cot (theta)
21. 1 + tan^2(-theta) = sec^2(theta) 1+(-tan(theta)*(-tan(theta)) = 1+ tan^2(theta) = sec^2(theta)
23. cos(theta)*(tan(theta)+cot(theta)) = csc(theta)
Cos(theta)*(sin(theta)/cos(theta))+(cos(theta)+sin(theta)) = sin(theta) + cos^2(theta)/sin(theta) = (sin(theta))/(sin(theta))*(sin(theta)/1+(1-sin^2(theta)/sin(theta) =
1/sin(theta)
27. (sec(theta)-1)(sec(theta)+1) = tan^2(theta) = sec^2(theta) = tan^2(theta)
Tan^2(theta) +1-1 = tan^2(theta) tan^2(theta) = tan^2(theta)
33. (sin(theta)+cos(theta))^2 +(sin(theta)-cos(theta))^2 = 2
2sin^2(theta)+2cos^2(theta) = 2
Sin^2(theta)+cos^2(theta) = 1
Sin^2(theta)+cos^2(theta) = 1 = 2
45. (sec(theta)+csc(theta))+(sin(theta)+cos(theta)) = 2tan(theta)
((1/cos(theta))/(1/sin(theta))+(sin(theta))/(cos(theta)) = (sin(theta)/cos(theta))/(sin(theta)/cos(theta)) = 2(sin(theta)/cos(theta)) = 2tan(theta)
49. (1-sin v)/cos v + cos v/(1-sin v ) = 2 sec v
(1-sin v)/cos v + (cos v/(1-sin v)*(cos v/cos v) = 2 sec v
(1-sin v)^2 + cos^2/((1-sin v)(cos v)) = (1/cos v)*(1/cos v ) = 2/cos v =
(1/cos v) +(1/cos v) = 2/cos v
51. (sin(theta))/(sin(theta)-cos(theta)) = 1/(1-cot(theta))=
****could not figure this one out:divide the whole term by 1/sin(theta) ? 1/sin(theta)-cos(theta) does not equal 1/(1-(cos(theta)/sin(theta))
(sin(theta))/(sin(theta)-cos(theta)) * 1 / sin(theta) = 1 / (sin(theta) - cos(theta))
1/(1-cot(theta)) * 1 / sin(theta) = 1 / (sin(theta) - sin(theta) * cot(theta) ) = 1 / (sin(theta) - sin(theta) * cos(theta) / sin(theta) ) = 1 / (sin(theta) - cos(theta).
So the identity is valid.
53. (1-sin(theta))/(1+(sin(theta)) = (sec(theta) tan(theta))^2
(1-2sin(theta) + sin^2(theta))/(1-sin^2(theta)) = (1-sin(theta))*(1-sin(theta))/(1+sin(theta)*(1-sin(theta)) = (1-sin(theta))/(1+sin(theta))
57. tan(theta) + (cos(theta)+(1+sin(theta)) = sec(theta)
(1+sin(theta)*sin(theta))/((1+sin(theta)*cos(theta))+(cos(theta)*cos(theta))/(1+sin(theta))*cos(theta) = 1/cos(theta)
(1+sin(theta)*sin(theta) + cos^2(theta))/(1+sin(theta)*cos(theta)) =
(Sin+sin^2(theta) +cos^2(theta))/(1+sin(theta)*(cos(theta))
= (1+sin(theta))/((1+sin(theta))cos(theta)) = 1/cos(theta)
63. (tan u cot u)/(tan u + cot u) + 1 = 2 sin^2 u
69. (sec(theta) csc(theta))/(sec(theta)*csc(theta)) = sin(theta) cos(theta)
((1/cos(theta))-(1/sin(theta)))/((1/cos(theta))*(1/sin(theta))) =
(1/cos(theta))-(1/sin(theta))*(sin(theta)*(cos(theta)) =
Sin(theta) cos(theta) = sin(theta) cos(theta)
75. (sec(theta))/(1-sin(theta)) = (1+sin(theta)/cos^3(theta)) =
Sec(theta)/(1-sin(theta))*((1+sin(theta)/(1+sin(theta)) =
(1+sin(theta))/(1+sin(theta))(1-sin(theta))(cos(theta)) =
(1+sin(theta))/(cos^3(theta))
81. (sin^3(theta)+cos^3(theta))/(sin(theta)+cos(theta)) = 1-sin(theta)cos(theta) =
Sin^2(theta)+cos^2(theta) = 1*****
105. One thing that I do when trying to establish identities is to rewrite the identity using the letters to see if that gives me any insight. For example, I would represent sec(theta) as 1/x instead of 1/cos(theta), and cot(theta) as x/y. Sometimes this helps me to see more clearly the algebra that I need to do to establish the identity(but sometimes it doesnt !)
Sec. 7.4
3.a) sin(pi/4)*cos(pi/3) = (sq. rt. of 2)/2*(1/2) = (sq. rt. of 2)/4
b) tan(pi/4)- sin(pi/6) = ½
Find the exact value of:
9. sin(5pi/12) = 75 deg. 225 deg. 150 deg. = sin(5pi/4-5pi/6) =
(sq. rt. of 6)/4+(sq. rt. of 2)/4
11. cos(7pi/12) = cos(9pi/12 2pi/12) = (-sq. rt. of 2)/4 (sq. rt. of 6)/4 = -1/4)(sq. rt. of 2 + sq. rt. of 6)
15. tan 15 deg. = tan(21pi/12-20pi/12) = .2679 same as calculator gives in deg. mode = pi/12 = app. .2618
17. sin(17pi/12) = 3pi/4+2pi/3 = sin(a +B) = sin(3pi/4)cos(2pi/3)+cos(3pi/4)sin(2pi/3) = -.9659
21. sin20deg.cos10deg +.cos20deg.sindeg. = sin(12pi/6-11pi/6) = 0-(-1/2) = 1/2
23. cos70deg.cos20deg. sin70deg.sin20 = cos(a+B) = cos(70+20) = cos(90) = 0
27. sin(pi/12)cos(7pi/12) cos(pi/12)sin(7pi/12) = sin(15) sin(75) =
Sin(a+B)+(cos(a+B)-cos(a+B) = 1
31. sin a = 3/5, 0 less than a is less than pi/2; cos B = (2(sq. rt. of 5)/5), -pi/2 is less than B is less than 0 cos a = 4/5 sin B = (sq. rt. of 5)/5
A) sin(a +B) = sin(a)cos(B)+cos(a)sin(B) = (6*(sq. rt. of 5)/25 + 4*(sq. rt. of 5)/25) = 2(sq. rt. of 5)/5 ( I got this which is the decimal equivalent, .894427191, of the radical terms added together. The decimal equivalent of the answer in the book, 2*(sq. rt. of 5)/25, is.1788854382. Either I did something wrong or 10(sq. rt. of 5)/25 should reduce to 2(sq. rt. of 5)/5.
33. A)sin(a+B) =sin a cosB + cos a sinB = (4/5)(1/2) + (-3/5)(sq. rt. of 3)/2) =
4-3*(sq. rt. 33. tan a = -4/3, pi/2 is less than a is less than pi; cos B = ½, 0 is less than B is less than pi/2
B) Cos(a+B) = cos a cosB sin a sin B = (-3/5)(1/2) (4/5)(sq. rt. of 3/2) =
-3-4(sq. rt. of 3)/10
C) sin( a-B) sin a cos B cos a sin B = (4/5)(1/2)-(-3/5)((sq. rt. of 3)/ 2) =
4+3*(sq. rt. of 3)/10 = (4+3*(sq. rt. of 3))/10
D) tan(a-B) = (tan a tan B)/(1+tan a tan B) = (-4/3)-sq. rt. of 3)/(1+(-4/3)(sq. rt. of 3) = -2.34( same as( 25(sq. rt.of 3)+48)/39 given in book
Use the figures to evaluate each function if f(x) = sin x, g(x)= cos x, and h(x) = tan x
39. f(a+B) sin(a+B) = sinacosB+cos a sinB sin a = 1, cos a = sq. of 3, tan y/x = (sq. rt. of 3)/3 cos B = 1 sin B = sq. rt. of 2(sq. rt.of 2)/3
Sin(1)(1/3)+(sq. rt. of 3)(2 *sq rt. of 2)/3) = 2.04632
Establish each identity:
45. sin(pi/2 + (theta)) = cos(theta) sin(pi/2+(theta)) = cos(theta)
Sin(pi/2)cos(theta)+cos(pi/2)sin(theta) = (1)(cos(theta))+(0)(sin(theta)) = cos(theta)
51. tan(pi theta) = -tan tan(a B) = (tan a tanB)/1+tan a tanB =
(tan pi theta)/(1+tan pi*(theta)) = -tan(theta)/1 = -tan(theta)
57. sin( a+B)/sin a cos B = 1 + cot a tan B = (sin a cos B)+(cos a sin B)/sin a cos B +sin a cosB = (sin a cos B)(sin a Cos B)+ (cos a/sin a )*(sin B/cos B) ?
63. cot( a + B) = (cot a cot B 1)/(cot B + cot a)= ((cos a/sin a)(cos B/sin B) (sin a sinB)/(sin a sin B))/((sin a/sin a)(cos B/sin B)+(cos a)/(sin a)(sin B/sin B) = (sin a cos B+ cos a sin B)/(sin a sin B)
Establish the identity:
69. sin((theta)+ k pi) = (-1) ^k, sin(theta) k any integer
(sin(theta))(cos 2pi)+cos(theta)(sin 2pi) = (sin(theta))(1) + (cos(theta))(0) = (-1)^2 = 1
(sin(theta)(-1)+cos(theta)(0) = (-1)^1 = -1
Works for 1, -1
73. sin[sin ^-1 (3/5) cos^-1(-4/5)] = sin(a B) = sin a cos B cos a sin B =
(3/5)(-4/5) (4/5)(3/5) = -24/25
79. tan(sin^-1(3/5) + pi/6) = (tan a + tan B)/(1-tan a tanB) = ((5-(sq. rt. of 3)/3-(54/20(sq. rt. of 3)
Write as an algebraic expression
83. cos(cos^-1 u +sin^-1 v)
85. sin(tan^-1 u sin^-1 ) -pi/2 is less than or equal to a is less than or equal to pi/2
-1 is less than or equal to u is less than or equal to 1
Tan a = sq. rt. of 1-(cos a/sin a) ^2 = sq. rt. of 1- (v/u)^2
Sin B = sq. rt. of 1- cos^2 v = sq.rt. of 1-v^2
Tan a tan B + sina sin B =
(1/u)+ sq rt. of 1-(v/u)^2 sq. rt. of (v/u)^2)(sq. rt. of 1-v^2)
I'm sorry for the long delay in posting your work.
See my answers to your questions, and let me know if you have additional questions.
Note that you should be submitting the 'open' qa's and queries, according to instructions. You don't need to submit problems in this format, except problems on which you want to ask questions.