Help with trig graphs... (1 Viewer)

[wccchick]

just another problem - could someone give me a hand?

this is the question:

sketch the graph y=cosx-sinx for 0/<x>/2pie

(/< & >/ are greater than or equal to)

i have no idea how to start!

 

[Riviet]

The y-coordinates that you should investigate are when y=0, 1, -1 on either graph, because these are the easiest to add.

As an alternative you can subtract the graph [y-coordinates] of y=sinx from the graph of y=cosx.

 

[insert-username]

The easiest way to do it is probably to transform y = cosx - sinx into the form rcos(x+@), because then you have just the one function to graph and no ordinates to add or subtract. So:

y = cosx - sinx

a = 1 (The coefficients of cos and sin, ignoring sign)
b = 1

r² = a² + b²

r² = 1 + 1 = 2

Therefore r = √2

tan @ = 1/1

tan @ = 1

Therefore @ = π/4

Therefore cosx - sinx = √2cos(x + π/4)

So the graph of cosx - sinx is a cos curve with amplitude √2, period 2π, and shifted π/4 units to the left of the y-axis.

I hope that helps,


I_F