Trig. Ratios for Angles Greater than 90 degrees.

• May 10th 2014, 09:59 AM
eleventhhour
Trig. Ratios for Angles Greater than 90 degrees.
1. State all the angles between 0 degrees and 360 degrees that make each equation true.
d) tan135 = -tan______________ (there's a blank space here)

So, I understand that you have to use the appropriate equation, but the textbook gives two answers in the book, while I only got one. Also, I don't really understand what this process is doing. Could someone explain what the equations mean for each quadrant?

I attached a photo from the textbook so you can see what I'm confused about. The textbook doesn't really explain well, and I can't find anything similar online. So, I'd really appreciate an explanation of this! Thanks.

And this was my solution (for one of the two possible answers according to the textbook):
tan(180-135) = -tanX
45 = X
• May 10th 2014, 10:17 AM
romsek
Re: Trig. Ratios for Angles Greater than 90 degrees.
Quote:

Originally Posted by eleventhhour
1. State all the angles between 0 degrees and 360 degrees that make each equation true.
d) tan135 = -tan______________ (there's a blank space here)

So, I understand that you have to use the appropriate equation, but the textbook gives two answers in the book, while I only got one. Also, I don't really understand what this process is doing. Could someone explain what the equations mean for each quadrant?

I attached a photo from the textbook so you can see what I'm confused about. The textbook doesn't really explain well, and I can't find anything similar online. So, I'd really appreciate an explanation of this! Thanks.

And this was my solution (for one of the two possible answers according to the textbook):
tan(180-135) = -tanX
45 = X

The best way to understand all this I think is to be able to visualize the geometry.

Imagine a circle with radius 1, centered at (0,0).

Now draw a line segment from (0,0) to somewhere on that circle. Note the angle that segment makes with the x-axis.

That point on the circle will have coordinates (x,y).

Let the angle the segment makes with the axis be $\theta$. Then

$\cos(\theta) = x$

$\sin(\theta) = y$

$\tan(\theta) = \dfrac y x$

and this is true for whatever angle/point on the circle you choose.

So for a given $\theta$ there are going to be 2 points on the circle that have a given value for $\cos(\theta)$ and $\sin(\theta)$ and $\tan(\theta)$

Suppose you have a point (x,y) in the first quadrant that makes angle $\theta$ with the x-axis.

From above you can see that the points

(x,y) and (x,-y) will have the same cosine

(x,y) and (-x,y) will have the same sine

(x, y) and (-x,y) will have cosines that are the negative of one another

(x,y) and (x,-y) will have sines that are the negative of one another

so the cosine is preserved when reflecting about the y axis and the sine is preserved when reflecting about the x axis.

the cosine is negated when reflecting about the x axis, and the sine is negated when reflecting about the y axis.

The tangent is thus preserved when reflecting about the origin, i.e. (x,y) -> (-x,-y) and is negated when reflecting about a single axis.

Now with a bit of thought you can turn these reflections into operations on $\theta$

for example a reflection about the y axis is $\theta \to (\pi - \theta)$

see if you can come up with the angular operation for reflection about the x-axis and about the origin.
• May 10th 2014, 03:57 PM
romsek
Re: Trig. Ratios for Angles Greater than 90 degrees.
Ugh, terrible mistake in here.

cosine is preserved reflecting about the x axis and negated reflecting about the y axis.

sine is preserved reflecting about the y axis and negated reflecting about the x axis.