Question

area of a curve

Answer 1

how do i find the points of intersection betwen the curves y=cosx and y=sin2x?

Answer 2

you cant determine the area of a curve. you can define the area between 2 curves

Answer 3

the intersection of 2 functions is where they are equal

Answer 4

sorry, thats what i meant. i know i equate them but ow do i solve for x?

Answer 5

with trig, you usually have to remember the trig identities to simplify stuff

Answer 6

sin(2x) = sin(x+x)

recall the sin(a+b) indentity

Answer 7

sin(a+b) = sin(a)cos(b)+sin(b)cos(a)

sin(x+x) = sin(x)cos(x)+sin(x)cos(x)
= 2 sin(x) cos(x)

so equate:

cos(x) = 2sin(x) cos(x)

Answer 8

i'll try to work it out quickly

Answer 9

x=1 or x=0?

Answer 10

cos(0) not equal, 2(0) cos(0)

Answer 11

in fact, sin(x) would have to equal 1/2 to give us\[cos(x)=2(\frac12)cos(x)\]

Answer 12

okay so i dont do this:
cosx-2sinxcosx=0
cosx(1-2sinx)=0
arccos(0)=x or arcsin(1/2)=x
x=1 or x=30

Answer 13

the x=30 sounds best

Answer 14

and of course cos(x) = 0 will fit fine

Answer 15

cos(odd pi/2) = 0

sin(pi/6+n 2pi) = 1/2
sin(5pi/6+n 2pi) = 1/2

Answer 16

thanks and how about y=sinx and y=e^x?

Answer 17

that might have to compare their power series equivalents

Answer 18

let say:\[sin(x)=\sum_0a_nx^n\]\[e^x=\sum_0b_nx^n\]
\[\sum_0a_nx^n=\sum_0b_nx^n\]
\[\sum_0a_nx^n-\sum_0b_nx^n=0\]
\[\sum_0(a_n-b_n)x^n=0\]
\[a_n-b_n=0\]

Answer 19

in other words, we turn them into polynomials and life gets simpler, with any luck

Answer 20

do i have to check the points of intersection if x is between 0 and pi/2?