Half-Angle Trig Identity...how can it end up in a different quadrant if angle < 90?

[math_knight]

Half-Angle Trig Identity...how can it end up in a different quadrant if angle < 90?

The book doesn't give an angle but states that tan T = 8/15 and T lies in Quadrant III ...180 < T < 270

The Pythagorean works out so that the hypotenuse = 17, cos = 15, sin = 8, an acute angle. (The calc says angle T is about 28 degrees by the way.)

The book then asks to find sin (T/2). The book then draws a triangle in Quadrant III. OK, so far so good. Quadrants must be kept in mind since we need to choose signs of square roots, fair enough.

However the books states that you must watch the domain and that it must be halved. 180/2 < T < 270/2. I understand that in certain situations that would be true, like say half of 120 would change the quadrant, but visually, if you have an acute angle in Quad III , and halve that, it will still be in Quad III. In fact you could keep halving it and at no point would it leave Quadrant III! The Initial side of an angle is by definition is the x-axis right? So to halve the domain doesn't make sense to me in the case of acute angles.

However, the book says that after halving the domain you end up with 90 < T < 135, and that the terminal side of the angle therefore lies in Quadrant II and picks a positive value for sin for the new angle? The book doesn't draw the new triangle with half the original angle by the way. What am I missing? Any help would be appreciated.

edit: pretty sure it's just a mistake in the book, right? Wouldn't be the first one I've spotted, just little things where the sign of something is off.

 

[mmm4444bot]

… The book then draws a triangle in Quadrant III …

That is a reference triangle. It does not contain your angle T in QIII. Reference triangles contain a copy of their QI counterpart. We use reference triangles as models because trig values in other quadrants match the corresponding values from QI, except possibly for sign.

You know symbol T represents only one angle, the terminal side of which is located in QIII. Yet, we're discussing two particular angles in this exercise, located in QI and QIII. I will use symbols T1 and T3, respectively.

(In other words, symbols T and T3 each represent the same angle.)

Think of angle 0° in standard position. Now visualize the terminal ray rotating counterclockwise. It reaches angle T1.

The angle continues increasing; the terminal ray passes through the remainder of QI, crosses the positive vertical axis, passes through QII, crosses the negative horizontal axis, and rotates within QIII until it reaches T3.

Again, T1 is in QI and T3 is in QIII.

Now, where do you think the terminal ray will be, if it rotates from 0° only halfway to T3?

… The calc says angle T is about 28 degrees by the way …

We always need to interpret what machines tell us. Don't be fooled!

You're told that angle T is located in QIII. It cannot be "about 28 degrees".

… However, the book … picks a positive value for sin for the new angle?

The "new angle" means T/2, yes? The sine function is positive, for all QII angles.

Has your class seen the graph of y = sin(x) yet?

 

[math_knight]

Hmm, I see your points, ty.

 

[mmm4444bot]

You're welcome. I notice that I forgot to stress the following.

When we say an angle is located in quadrant III (or we designate it as a QIII angle), we are not saying that the angle itself lives inside the third quadrant. Statements like those are a shortened way of stating that the angle's "terminal ray" is located in QIII.

Also, I mentioned the graph of y = sin(x) because it's a simple visual aid, to quickly recall the sign of sin(x) in each quadrant.

The graph of y = sin(x) plotted from x=0° to x=360° shows the behavior of sine over one period (i.e., the graph shows one complete "wave" of oscillating behavior, as the terminal ray of angle x rotates through all four quadrants -- making one complete revolution). You can remember the picture of this graph: the first half of the wave is above the x-axis and the second half is below the x-axis. Hence, the first half of the wave shows sin(x) is positive in QI and QII, and the second half of the wave shows sin(x) is negative in QIII and QIV.

You can use a similar aid, for the sign of cos(x). One period (from 0° to 360°) shows the graph below the x-axis in the middle of the wave, and it shows the wave above the x-axis at each end. In other words, cos(x) is negative in QII and QIII, and cos(x) is positive in QI and QIV.

One just needs to remember the shape of each wave and realize that 0° to 360° includes all four quadrants.