Trigonometry - Associated Acute Angles

[BOAS]

Hello,

i'm doing some revision and working through the textbook my course follows and have a small problem.

The question is an incomplete table of values for sinθ, cosθ, tanθ, [itex]\alpha[/itex] and θ. I have to work out the blanks using what is given.

I thought I knew the correct method of doing this, and I got all the answers correct except for the last two lines where I got the value correct, but did not see why I needed to include a ± sign.

I'll show what I did for one of the lines, and hopefully someone can see why I'm not realising the values can be negative or positive.

Homework Statement

sinθ =
cosθ =
tanθ =
[itex]\alpha[/itex] = 45°
θ =

The Attempt at a Solution

I worked out sin, cos and tan for this value of alpha and they were all positive, so I thought I could assume, based on the CAST diagram, that the associated acute angle [itex]\alpha[/itex], must lie in the 1st quadrant. Therefore θ = 45°

However, my textbook says all the trig values I obtained can be ± and θ = 45° or 135°

What am I missing?

 

[Mark44]

Hello,

i'm doing some revision and working through the textbook my course follows and have a small problem.

The question is an incomplete table of values for sinθ, cosθ, tanθ, [itex]\alpha[/itex] and θ. I have to work out the blanks using what is given.

I thought I knew the correct method of doing this, and I got all the answers correct except for the last two lines where I got the value correct, but did not see why I needed to include a ± sign.

I'll show what I did for one of the lines, and hopefully someone can see why I'm not realising the values can be negative or positive.

Homework Statement

sinθ =
cosθ =
tanθ =
[itex]\alpha[/itex] = 45°
θ =

The Attempt at a Solution

I worked out sin, cos and tan for this value of alpha and they were all positive, so I thought I could assume, based on the CAST diagram, that the associated acute angle [itex]\alpha[/itex], must lie in the 1st quadrant. Therefore θ = 45°

However, my textbook says all the trig values I obtained can be ± and θ = 45° or 135°

What am I missing?

How are α and θ related?
Are you given a picture of the triangle?
You haven't provided enough information for us to be able to help you.

 

[BOAS]

How are α and θ related?
Are you given a picture of the triangle?
You haven't provided enough information for us to be able to help you.

Sorry, I didn't stop to think whether my explanation involving α would make sense.

I'll do my best to explain it, but it's easier with a set of axes in front of you.

Take o to be the origin, and draw a line op at say 45° to the x axis. Directly below p, on the x axis, we label the point q to construct a right angled triangle. The anticlockwise direction is taken as +ve. We can say that α is the associated acute angle for θ here because it lies in the 'first quadrant'. They are numbered from 1-4 anticlockwise.

Now suppose you rotate the line op to a position of 135°, p lies in the 'second' quadrant and θ = 135°, but the associated acute angle α is 45° (the triangle opq)

I hope that makes enough sense to see what my question is getting at.

 

[tiny-tim]

i don't get it …

if OP is 120°, what is α?​

 

[BOAS]

i don't get it …

if OP is 120°, what is α?​

If the line op is rotated 120° anticlockwise, then θ is 120° and α is the acute angle made with the x axis, so it's 60°.

EDIT:

This is in relation to the CAST diagram, that tells us which trig ratios are positive in which quadrant.

 

[tiny-tim]

If the line op is rotated 120° anticlockwise, then θ is 120° and α is the acute angle made with the x axis, so it's 60°.

I see.

And what about 240° and 300° … are they 60° or -60° ?

 

[BOAS]

In the first quadrant α = θ - 360°
second quadrant α = 180° - θ
Third quadrant α = θ - 180°
Fourth quadrant α = 360° - θ

So, if θ = 240°, the line op lies in the third quadrant and α = 60°.

If θ = 300°, the line op lies in the fourth quadrant and α = 60°

I'm pretty sure it's setup so that α is always positive.

 

[tiny-tim]

Homework Statement

sinθ =
cosθ =
tanθ =
[itex]\alpha[/itex] = 45°
θ =

The Attempt at a Solution

I worked out sin, cos and tan for this value of alpha and they were all positive, so I thought I could assume, based on the CAST diagram, that the associated acute angle [itex]\alpha[/itex], must lie in the 1st quadrant. Therefore θ = 45°

However, my textbook says all the trig values I obtained can be ± and θ = 45° or 135°

What am I missing?

ok, then if θ = 45° or 135°, then both cosθ and tanθ can be ± (though sinθ can only be +)

however, on your explanation of α, i don't see why θ can't be 225° or 315°

 

[BOAS]

ok, then if θ = 45° or 135°, then both cosθ and tanθ can be ± (though sinθ can only be +)

however, on your explanation of α, i don't see why θ can't be 225° or 315°

I'm sorry, this whole question is due to me mis-reading the answer at the back of the book.

It was written as θ = ±45° or ±135° and I didn't see the signs in front of the values it gave for θ. I have it all making sense now.

Thanks for your patience.