RE:
These are some good problems. You should show us what you have before we solve them...
1. use manual steps to solve Cos(x) Tan(x) + (3)^1/2 Cos(x) = 0
2. use calculator methods: consider y=-3sin(x-3)+2 on .5radians < x < 5.5r
find: max, min points, inflection, and give calculus reasons
3. Antiderive:
A. 3x/4+x^2 dx
B. 3/(4-X^2)^1/2 dx
C. 3/4+x^2 dx
D. 3+2x/(1-x^2)^1/2
4. if sin (x) = -3/7, x is in quad 3 and cos (y)= -2/5, y is in quad 2. calculate exact values for :
a) sin (x-y) =
b) tan (2x) =
5. show steps to prove that:
csc^2(x)tan(x)-cot(x) is equal to tan(x)csc^2(x)tan(x)-cot(x) = tan(x)
thanks guys
First of all, the functions are cos(x) and tan(x), not Cos(x) and Tan(x). Yes, capital letters are important!
You need to factor:
cos(x)[tan(x) + sqrt{3}] = 0
So either
cos(x) = 0 ==> x = (pi)/2, 3(pi)/2 rad
or
tan(x) + sqrt{3} = 0
tan(x) = -sqrt{3}
x = 2(pi)/3, 5(pi)/3 rad
So
x = (pi)/2, 2(pi)/3, 5(pi)/3, 3(pi)/2 rad
-Dan
A. Int[3x/4+x^2 dx] = (3/4)*Int[x dx] + Int[x^2 dx]
= (3/4)*(1/2)x^2 + (1/3)x^3 + C
B. Int[3/(4-X^2)^1/2 dx] = 3*Int[1/sqrt{4 - x^2} dx]
Let y = 2x ==> dy = 2 dx
Int[3/(4-X^2)^1/2 dx] = 3*Int[1/sqrt{4 - x^2} dx] = 3*Int[1/sqrt{4 - 4y^2} (dy/2)]
= (3/2)*Int[(1/2)*1/sqrt{1 - y^2} dy]
= (3/4)*Int[1/sqrt{1 - y^2} dy]
Now let y = sin(t) ==> dy = cos(t) dt
Int[3/(4-X^2)^1/2 dx] = (3/4)*Int[1/sqrt{1 - y^2} dy]
= (3/4)*Int[1/sqrt{1 - sin^2(t)} * cos(t) dt]
= (3/4)*Int[1/cos(t) * cos(t) dt]
= (3/4)*Int[dt] = (3/4)*t + C = (3/4)*asn(y) + C
= (3/4)*asn(2x) + C
C. Int[3/4+x^2 dx] = 3/4*Int[dx] + Int[x^2 dx]
= (3/4)*x + (1/2)x^3 + C
D. Int[3+2x/(1-x^2)^1/2 dx] = 3*Int[dx] + Int[2x/sqrt{1 - x^2} dx]
For the second integral let y = 1 - x^2 ==> dy = 2x dx
Int[3+2x/(1-x^2)^1/2 dx] = 3*Int[dx] + Int[2x/sqrt{1 - x^2} dx]
= 3*x + Int[1/sqrt{y} (-dy)] = 3x - (2)*sqrt{y} + C
= 3x - 2*sqrt{1 - x^2} + C
-Dan
a) sin(x - y) = sin(x)cos(y) - sin(y)cos(x)
Now, sin(y) = -sqrt{1 - cos^2(y)} <-- Negative since y is in QIII
sin(y) = -sqrt{1 - (-2/5)^2} = -sqrt{1 - 4/25} = -sqrt{21/25}
cos(x) = -sqrt{1 - sin^2(x)} <-- Negative since x is in QII
cos(x) = -sqrt{1 - (-3/7)^2} = -sqrt{1 - 9/49} = -sqrt{40/49}
So
sin(x - y) = sin(x)cos(y) - sin(y)cos(x)
= (-3/7)(-2/5) - (-sqrt{21/25})(-sqrt{40/49})
= 21/35 - sqrt{(21*40)/(25*49)}
= 21/35 - (2/35)*sqrt{210}
= (21 - 2*sqrt{210})/35
b. tan(2x) = sin(2x)/cos(2x) = 2*sin(x)*cos(x)/(2*cos^2(x) - 1)
= 2*(-3/7)*(-sqrt{40/49})/(2*(-sqrt{40/49})^2 - 1)
= (6*sqrt{40/49)/7)/(2*40/49 - 1)
= (6*sqrt{40}/49)/(80/49 - 1)
= (12*sqrt{10}/49)/(31/49) <-- Cancel the 49s
= 12*sqrt{10}/31
-Dan
This is not possible, which is why Jhevon was asking about it. You have multiplied the first term by a tan(x) and said it is somehow equal to the original. A simpler version of what you are saying:
Show that x + 2 = 2x + 2 for all x.
This simply isn't true. (For x = 0, yes, but not for all x.)
-Dan
Convert everything into sines and cosines:
csc(x) = 1/sin(x)
tan(x) = sin(x)/cos(x)
cot(x) = cos(x)/sin(x)
So we have to show:
1/[sin(x)cos(x)] - cos(x)/sin(x) = sin(x)/cos(x)
Add the fractions on the LHS:
1/[sin(x)cos(x)] - cos^2(x)/[sin(x)cos(x)]
= [1 - cos^2(x)]/[sin(x)cos(x)]
= sin^2(x)/[sin(x)cos(x)]
So as long as sin(x) is not equal to 0 (which we require anyway for csc(x) and cot(x) to exist)
= sin(x)/cos(x)
which is the RHS.
-Dan
Here:
Tips:
1) start on the most complicated side, it gives you more options to change things
2) it is USUALLY better to change everything to sines and cosines. these are the ones students are more familiar with, so it's easier to see connections
3) always consider working on both sides one at a time to bring them to the same thing. sometimes working on one side only to get the answer is too hard or even impossible in some sense
EDIT: You're too quick for me Dan!...can i call you Dan?