Need help with trig cofunctions

[max]

Hi,

Why does cos(pi/2 -x) = sin(x)?

When plotting points, I see that it is true, but I get a different graph using transformations.
I reflect the graph over the y axis (the graph looks identical to the original), and I move the grapth to the left since pi/2 is positive.

Also, if the graph of cos(x) = cos(-x), why doesn't cos(pi/2+x),which equals sin(x), = cos(-(pi/2)-x)?

Thanks

 

[Denis]

Why does cos(pi/2 -x) = sin(x)?

Do you realise that circumference of unit circle = 2pi ?
Which means arc pi/2 subtends a 90 degrees angle.

 

[Deleted member 4993]

Hi,

Why does cos(pi/2 -x) = sin(x)?

When plotting points, I see that it is true, but I get a different graph using transformations.
I reflect the graph over the y axis (the graph looks identical to the original), and I move the grapth to the left since pi/2 is positive.

Also, if the graph of cos(x) = cos(-x), why doesn't cos(pi/2+x),which equals sin(x) no... cos(pi/2+x) = - sin(x), watch that negative sign after "equal to" sign = cos(-(pi/2)-x)?

Thanks

One of the ways to show that would be to use:

cos(A-B) = cos(A) * cos(B) + sin(A)*sin(B)

If you understand (or take for granted) the equation above then

\(\displaystyle cos(\frac{\pi}{2} - \theta) = \, cos(\frac{\pi}{2}) \cdot cos(\theta)\, + \, sin(\frac{\pi}{2})\cdot sin(\theta) = \, sin(\theta)\)

 

[max]

Thanks for helping see why it is true. But how can I picture the transformations correctly? I take the graph and flip it over the y axis then move it to the left. What am I doing wrong?