what is y=sin3x+cos4x, 0<=x<=2pi ?

Question

y=sin3x+cos4x, 0<=x<=2pi

Answer 1

y = sin(3x) + cos(4x)
y = sin(2x + x) + cos(2x + 2x)

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

sin(2x) = 2sinxcosx
cos(2x) = cos(x)^2 - sin(x)^2 = 2cos(x)^2 - 1 = 1 - 2sin(x)^2

(sin(2x)cos(x) + sin(x)cos(2x)) + (cos(2x)^2 - sin(2x)^2)

(2sin(x)cos(x)cos(x) + sin(x)(2cos(x)^2 - 1)) + ((2cos(x)^2 - 1)^2 - (2sin(x)cos(x))^2)

(2sin(x)cos(x)^2 + 2sin(x)cos(x)^2 - sin(x)) + (((2cos(x)^2 - 1)(2cos(x)^2 - 1)) - 4sin(x)^2cos(x)^2)

(4sin(x)cos(x)^2 - sin(x)) + ((4cos(x)^4 - 2cos(x)^2 - 2cos(x)^2 + 1) - 4sin(x)^2cos(x)^2)

4sin(x)cos(x)^2 - sin(x) + (4cos(x)^4 - 4cos(x)^2 + 1 - 4sin(x)^2cos(x)^2)

4sin(x)cos(x)^2 - sin(x) + 4cos(x)^4 - 4cos(x)^2 + 1 - 4sin(x)^2cos(x)^2

cos(x)^2 = 1 - sin(x)^2

4sin(x)(1 - sin(x)^2) - sin(x) + 4(1 - sin(x)^2)^2 - 4(1 - sin(x)^2) + 1 - 4sin(x)^2(1 - sin(x)^2)

4sin(x) - 4sin(x)^3 - sin(x) + 4((1 - sin(x)^2)(1 - sin(x)^2)) - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4

4sin(x) - 4sin(x)^3 - sin(x) + 4(1 - sin(x)^2 - sin(x)^2 + sin(x)^4) - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4

4sin(x) - 4sin(x)^3 - sin(x) + 4(1 - 2sin(x)^2 + sin(x)^4) - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4

4sin(x) - 4sin(x)^3 - sin(x) + 4 - 8sin(x)^2 + 4sin(x)^4 - 4 + 4sin(x)^2 + 1 - 4sin(x)^2 + 4sin(x)^4

(4 + 4)sin(x)^4 - 4sin(x)^3 + (-8 + 4 - 4)sin(x)^2 + (4 - 1)sin(x) + (4 - 4 + 1)

sin(3x) + cos(4x) = 8sin(x)^4 - 4sin(x)^3 - 8sin(x)^2 + 3sin(x) + 1

for a graph, go to http://www.calculator.com/calcs/GCalc.html

just type in sin(3x) + cos(4x), click enter

Answer 2

x lies in the first two quadrants and the equation is valid only between these limits

http://rajkiranpro.googlepages.com/graph1.gif

Answer 3

It varies. You're going to have to graph it.