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## Class 11 Physics (India)

### Course: Class 11 Physics (India)>Unit 3

Lesson 6: The graphs of sine, cosine, and tangent

# Graph of y=tan(x)

Sal draws the graph of the tangent function based on the unit circle definition of the function. Created by Sal Khan.

## Want to join the conversation?

• Where is Sal getting square root of 2 over 2 at ?
• At that point, Sal is dealing with pi/4 or 45 degrees, which is an angle that you can know sin, cos, and tan values without having to calculate anything. 45 degrees is one of those "special" angles that it would be good to memorize or at least know how to very easily find its sin/cos/tan value without a calculator.

If you have a right triangle where the other two angles are both 45 degrees, that means your two non-hypotenuse side lengths are going to be the same (it's an isosceles right triangle). For this "special" triangle, you can remember easily the values 1, 1, and √2. The hypotenuse will be the larger value (√2) and the other two sides will both be 1. This makes sense, especially when we plug these values into the Pythagorean theorem:

a^2 + b^2 = c^2
1^2 + 1^2 = (√2)^2
1 + 1 = 2
2 = 2

So now we have our sides, so we can very easily find sin/cos/tan values.
sin = O/H = 1/√2
cos = A/H = 1/√2
tan = O/A = 1/1 = 1

I personally don't know why they don't like irrational numbers in the denominator of fractions, but they don't. So they usually convert that fraction (in both sin and cos) by multiplying by √2/√2:

sin = O/H = 1/√2 = 1/√2 * √2/√2 = (1*√2) / (√2*√2) = √2/2
cos = √2/2

Because 45 degrees is so easy to remember, Sal just quickly wrote in that its sin and cos are √2/2. Just remember that when you're working with negative angles, the quadrant changes, and so your sign my change. sin(-45) = -√2/2
• Why do you have to draw the dotted vertical lines at ?
• TAN to 90 degrees (PI/2 Radians) is 1/0, which is undefined, so you can't graph a result that's not there. You can get as close as you want to 90 degrees, as long as you don't land on it.
Example:
TAN (89.9999999999) ≈ 572,957,795,131
TAN (90) = 1/0 = UNDEFINED
TAN (90.0000000001) ≈ -572,957,795,131
• Why do we take the angle in radians and not in degrees?
• Once you get to higher math, like calculus, it becomes more clear. One simple example, though, is finding arc length. If you have a 30º arc with a radius of 1, to find the arc length you convert 30º to a fraction of 180º (30/180, or 1/6) and multiply that by πr. In radians, all you have to do is multiply the radius by the arc measure. π/6 * 1 = π/6, which you would get both ways. 30º is obviously a simple example, but it still is easier to receive the angle in radians instead.

Another way (involving calculus) is the derivatives of trigonometric functions. The derivative of a function is the function's slope at a given point, and (in radians) the derivative of sin(x) = cos(x). When you put it in degrees, however, the derivative of sin(x) is π/180 * cos(x).

Hope this helps!
• At several times in the video, Sal refers to "vertical asymptotes". Could anyone please explain their definition?
• A vertical asymptote is a line that the graph would approach but never reach.
It occurs at values where the function is undefined, in this case where its denominator is zero.
For tangent, that would be at values of x that make cos(x) = 0 --- in other words, at x = 90 degrees and at x = 270 degrees for 0 <= x <=360.
• What would be the range and the domain then?
• The range is the entire set of real numbers, and the domain is all real numbers except for (k+0.5)π.
• so what is the domain on the tangent function since there are gaps in the line?
• Domain: all real numbers except pi/2 + k pi, k is an integer.

It cannot be pi/2 or any multiple of pi/2 because that creates an instance where tangent is divided by 0.
In the graph of tangent, pi/2 or any multiple of pi/2 is represented by an asymptote.
• How do you find the vertical asymptotes?
• 𝑓(𝑥) = 𝑎 tan(𝑏𝑥 + 𝑐) + 𝑑 = 𝑎 sin(𝑏𝑥 + 𝑐)∕cos(𝑏𝑥 + 𝑐) + 𝑑

The vertical asymptotes of 𝑓(𝑥) occur when
cos(𝑏𝑥 + 𝑐) = 0

This happens when
𝑏𝑥 + 𝑐 = 𝜋∕2 + 𝑛𝜋 for any integer 𝑛

⇒ 𝑥 = (𝜋∕2 + 𝑛𝜋 − 𝑐)∕𝑏
= (𝜋∕2 − 𝑐)∕𝑏 + 𝑛𝜋∕𝑏
• Why is the period of a tangent graph pi and not 2pi like the sine and cosine graphs?
• Because tangent is sine/cosine. Sine and cosine start positive, cosine goes negative, then sine goes negative before cosine goes back positive and sine goes back positive. negative/negative = positive/positive and negative/positive = positive/negative.
• What is the Domain of the function y=tanx ? As it is evident that every multiple of pi/2 is not defined.
• Every odd multiple of 𝜋/2 is not defined. Thus, we can write the domain as:
𝑥 ∈ ℝ/{(2𝑘 – 1)𝜋/2} : 𝑘 ∈ ℤ
Which is basically saying that the input can be any real number other than odd multiples of 𝜋/2.
• OK so the the very important question is
what is the domain and range of tangent function??
• mushuwu has the range right, but domain not so much. if you graph it you will notice vertical asymptotes along a regular interval. now, the key to finding this interval is knowing that tangent is sine divided by cosine. What can you never do in a division problem? divide by 0. so cosine can never be 0, and that is what the vertical asymptotes mean.

When is cosine 0? at pi/2 and 3pi/2. plus 2pi * n of course, where n is any integer. so pi/2 plus every 2pi rotation of that after, and the same for 3pi/2

Let me know if that doesn't make sense.