# Tan (a) = -1

Gold Member
The problem
I want to solve ##tan(a)=-1## where a is in degrees.

The attempt

## tan(a)= \frac{sin(a)}{cos(a)}=-1 \ \ cos(a) \neq 0 \\ \frac{sin(a)}{cos(a)}=-1 \\ sin(a)=-cos(a) \\ -sin(a)=cos(a) \\ sin(-a)=cos(a) ##

I have no idea how to continue from here.

Related Precalculus Mathematics Homework Help News on Phys.org
Ray Vickson
Homework Helper
Dearly Missed
The problem
I want to solve tan(a)=-1 where a is in degrees.

The attempt

## tan(a)= \frac{sin(a)}{cos(a)}=-1 \ \ cos(a) \neq 0 \\ \frac{sin(a)}{cos(a)}=-1 \\ sin(a)=-cos(a) \\ -sin(a)=cos(a) \\ sin(-a)=cos(a) ##

I have no idea how to continue from here.
First solve ##\tan(a) = +1##, then use sign-change properties of the tangent function to get the solution of the equation you want.

Rectifier
Gold Member
First solve ##\tan(a) = +1##, then use sign-change properties of the tangent function to get the solution of the equation you want.
To be honest, ##tan(a)=1## doesnt make my life easier :(. I still don know how to solve it. I know that ##sin(a)## and ##cos(a)## are equal when ##a=45## but I dont know how I can come up with that from the equation ##sin(a)=cos(a)##.

Ray Vickson
Homework Helper
Dearly Missed
To be honest, ##tan(a)=1## doesnt make my life easier :(. I still don know how to solve it. I know that ##sin(a)## and ##cos(a)## are equal when ##a=45## but I dont know how I can come up with that from the equation ##sin(a)=cos(a)##.
I don't see the problem. You say you know that ##\sin(a) = \cos(a)## when ##a = 45 \deg##. You don't "come up with that" from any equation at all; it is just a fact about geometry, translated into trigonometric terms.

Of course, there are infinitely many other solutions of the equation ##\tan(a) = 1##, but they will all differ from ##a = 45 \deg## in a simple way. just look at the graph of ##y = \tan(x)## over a few "periods" to see what is happening.

Mark44
Mentor
To be honest, ##tan(a)=1## doesnt make my life easier :(. I still don know how to solve it. I know that ##sin(a)## and ##cos(a)## are equal when ##a=45## but I dont know how I can come up with that from the equation ##sin(a)=cos(a)##.
There are a small number of angles whose trig functions (especially sine, cosine, and tangent) you should know by heart. They are 0°, 30°, 45°, 60°, and 90°. In radians, these are ##0, \pi/6, \pi/4, \pi/4##, and ##\pi/2##, respectively. From basic geometry and the unit circle, you can easily derive all six of the trig functions for angles in the 2nd, 3rd, and 4th quadrants.

The problem
I want to solve tan(a)=−1tan(a)=-1 where a is in degrees.

The attempt

tan(a)=sin(a)cos(a)=−1 cos(a)≠0sin(a)cos(a)=−1sin(a)=−cos(a)−sin(a)=cos(a)sin(−a)=cos(a) tan(a)= \frac{sin(a)}{cos(a)}=-1 \ \ cos(a) \neq 0 \\ \frac{sin(a)}{cos(a)}=-1 \\ sin(a)=-cos(a) \\ -sin(a)=cos(a) \\ sin(-a)=cos(a)

I have no idea how to continue from here.
Just out of curiosity , if you were to solve sin(x) = 0.5 , how would you ( rectifier ) go about it ?

SteamKing
Staff Emeritus
Homework Helper
To be honest, ##tan(a)=1## doesnt make my life easier :(. I still don know how to solve it. I know that ##sin(a)## and ##cos(a)## are equal when ##a=45## but I dont know how I can come up with that from the equation ##sin(a)=cos(a)##.
If tan (a) = sin (a) / cos (a) , by definition

and

tan (a) = 1

then what does the first statement say about the second?

Since it hasn't been specified where the angle is being taken from, let's just assume it's in relation to the positive x-axis.

Now, to solve this problem you just need to know the table of exact values, which I'll attach to this post.
From the table, we can see that tan(45)= 1

Therefore; tan(-45) = -1 | This is true because -45 degrees from the positive x-axis puts you in the fourth quadrant, where tan values are negative.
So, if you're given tan(a) = -1, you should know that 'a' must equal -45 degrees (if we are dealing with values only within 360 degrees)

Generally, you wont be asked to prove this mathematically because it's sort of a pre-established rule. Kind of like how you don't get asked to prove Pythagoras' Theorem.

#### Attachments

• 47.9 KB Views: 438
Student100
Gold Member
Generally, you wont be asked to prove this mathematically because it's sort of a pre-established rule. Kind of like how you don't get asked to prove Pythagoras' Theorem.

It is also generally bad form to post solutions here.

It is also generally bad form to post solutions here.
Firstly, sorry for posting the solution then.

As for the first part, aren't values from the table of exact values just assumed knowledge, and you don't have to prove them to be true?

Student100
Gold Member
Firstly, sorry for posting the solution then.

As for the first part, aren't values from the table of exact values just assumed knowledge, and you don't have to prove them to be true?
It isn't just assumed knowledge, it's quite easy to derive these exact values using the Pythagorean theorem or unit circle.

To suggest it's "pre-established" is akin to saying it's magic and not care at all where it comes from.

Qwertywerty
Does not the concept of "arctan X" or "tan-1 X" fit in here somewhere?

BruceW
Homework Helper
yes. The problem statement is essentially to find the solutions of arctan(-1)

You could try to solve it graphically (seems to be just a suitable a method)

If you use
desmos.com

you can plot both
y = tan x
y = -1

and note the value of where the lines meet (you may need to convert from radians to degrees).

You will find there are lots of points, so from there, you can either take one common sense answer (the one nearerst the origin) of make a simple formula to show that the answer is periodic.

Does not the concept of "arctan X" or "tan-1 X" fit in here somewhere?
Not exactly . Firstly , the OP's question is a basic one and he may therefore not have knowledge of this .

Also , the arctan only gives you but one solution to the question , when in fact there are ∞ .

Draw a unit circle. Break it up into 4 quadrants. Where is tan negative? We can restrict the solution to 0 to 2 pie or for all possible solutions. Let's start from the restricted case. We know that tanx=1 when x is 45 degrees. We can multiply 45 by at most 8 in our restricted case. It gives us 45 degrees in each respective quadrant. 45*1= 1. 45×2=90,45×3= 135, 45×4=180 etc. Notice that the 45 degree in every Quadrant perfectly divides it into 2?.

Just where you reached is perfect...
sin (-a) = cos (a) only and only if (iff) a = 135 degrees

Rectifier
The problem
I want to solve ##tan(a)=-1## where a is in degrees.

The attempt

## tan(a)= \frac{sin(a)}{cos(a)}=-1 \ \ cos(a) \neq 0 \\ \frac{sin(a)}{cos(a)}=-1 \\ sin(a)=-cos(a) \\ -sin(a)=cos(a) \\ sin(-a)=cos(a) ##

I have no idea how to continue from here.
If ## tan(a) = -1##, it means that

## sin(a) = - cos(a) ##

Since the domain of ## tan(x) ## is between ## -\frac{\pi}{2} ## to ## \frac{\pi}{2} ##,

## a = -\frac{\pi}{4} ##, or -45 degrees.

Kind of like how you don't get asked to prove Pythagoras' Theorem.
What's wrong with proving the Pythagorean theorem? Aside from being well know, there is nothing obvious to it. It is in fact a statement which is very difficult to believe, even with proof, since it gives an amazing and unexpected relationship between algebra and geometry.

Just where you reached is perfect...
sin (-a) = cos (a) only and only if (iff) a = 135 degrees
not entirely true. Since in my restricted case. we are looking at ALL the quadrants from 0 to 2pie.

so yes, at 135 degrees is an answer but not the only answer. Can you see another quadrant where tan values are negative besides the second?

Figure this out, then we can find tanx=-1 for all possible combinations.

It's pretty easy tan(-a)= -tan a
So, Tan(-45) = -1
Now tan is negative in the 2nd and 4rth quadrant.
So tan(180-45)= -1, tan(360-45)=-1,tan(540-45)=-1 It has Infinite solutions.
Could be a=135,315 etc.