Auxillary angle method (help please) (1 Viewer)

[idling fire]

I always get this wrong because I get confused between the + or - in:
Rsin(x+θ) , Rsin(x-θ) , Rcos(x+θ) , Rcos(x-θ)

Here's one example:
a) Change y= 3cos(x) - 4sin(x) into the form y=Rcos(x-θ)
and sketch.
b) hence solve 3sinx - 4cosx = 2

Thanks

 

[ssglain]

a sin(x) + b cos(x) = R sin(x + theta)
a sin(x) - b cos(x) = R sin(x - theta)
a cos(x) + b sin(x) = R cos(x + theta)
a cos(x) - b sin(x) = R cos(x - theta)

For you question, you needn't worry about which one of these to use because the form in which you should express y= 3cos(x) - 4sin(x) has been given.

Expanding y = R cos(x - theta) = R cos(x)cos(theta) + R sin(x)sin(theta)

Now you can compare and equate the cos(x) and sin(x) terms in this expression:
3cos(x) - 4sin(x) = R cos(x)cos(theta) + R sin(x)sin(theta)

Clearly, R cos(theta) = 3 and R sin(theta) = -4

Now square both expressions and add them:
R^2*{cos^2(theta) + sin^2(theta)} = 25
R^2*{1} = 25
R = 5

Substitute R=5 back into either expression for cos(theta) or sin(theta) to solve for theta.

For part two, solve R cos(x - theta) = 2 instead.

That particular question doesn't seem to like handing out nice numbers though. Please correct me if I perhaps did something wrong there.

 

[idling fire]

I'm sure you're right, because I think the numbers are meant to be icky.

So is theta the acute angle? (When my teacher did it, she just wrote the obtuse one, but I can't tell if her answers are messed up. Then again, she always took shortcuts and didn't explain them >.<)

How about the graphs...? Each time I end up with one looking completely different...

 

[ssglain]

Theta is supposed to be the related (auxiliary) angle, which means it should be positive acute. For your question theta seems to be negative acute, from what I have. This makes me a bit worried about these two:
a cos(x) + b sin(x) = R cos(x + theta)
a cos(x) - b sin(x) = R cos(x - theta)

I am more and more inclined to think that they should have been the other way around so:
a cos(x) + b sin(x) = R cos(x - theta)
a cos(x) - b sin(x) = R cos(x + theta)

Which graphs are you referring to?

 

[idling fire]

Just when I tried to draw the graph of y= 3cos(x) - 4sin(x) using the result from part a, it would be displaced incorrectly along the x-axis.

The best I got was to draw y= -5cos(x+ 36'52') after rearranging everything and getting confused...

 

[ssglain]

Oh right. Forgot about the sketch bit. The value for theta that you got doesn't seem to be correct. I'm somehow getting theta = 0.92729 radians which is closer to 53 degs.

I think the question would prefer you to sketch y = 5 cos(x - theta) because to sketch y = 3cos(x) - 4sin(x) by the subtraction of functions is walking along the edge between Ext-1 and Ext-2. For y = 5 cos(x - theta), you should graph a general y = 5 cos(x) curve and shift the curve right by theta units (alternatively, shift the y-axis theta units to the left).

 

[idling fire]

Okay, thanks heaps.

 

[elseany]

an old bump sorry but i saw u say sub in the value for r to get theta.

since u know Rsin(*)=-4 and Rcos(*)=3 you can just divide them to get an expression for tan(*) and then use inverse trig to get your value for theta.

 

[fishy89sg]

the proof for those formulae is pretty cool