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### Course: Precalculus>Unit 2

Lesson 10: Using trigonometric identities

# Using the tangent angle addition identity

Find the tangent of 13pi/12 without a calculator using the tangent angle addition identity. Created by Sal Khan.

## Want to join the conversation?

• why are the practice problems so much harder than what we learn in the video?
i feel like there are not enough videos here to go ahead to the parctice problem because he does not show how to solve so many of them
• You could read the article at the end of this section, this covers equations ranging from the inverse of sin to the reciprocal of tangent and the equations to solve the problems.
• It should be noted that it's possible to simplify the answer even further. Sal came up with (sqrt(3)-1)/(sqrt(3)+1), which is not wrong; however, if you multiply both numerator and denominator by (sqrt(3)-1), the numerator becomes (sqrt(3)-1)^2, which simplifies to 4-2sqrt(3), while the denominator simplifies to the difference of two squares, 3-1 = 2. Thus the entire expression can be simplified to 2-sqrt(3).
• whoa whoa whoa, where the heck did he get 15pi/12 - 2pi/12? is he just picking randomn fractions(angles or whatver)? By that reasoning, why can't I just use 3pi/12 plus 20pi/12?
• He picked them because 15pi/12-2pi/12=13pi/12, which is the angle we are trying to evaluate. Why did he pick those two angles to find the difference of (when there are many others that would also equal 13pi/12)? Because of what they simplify down to. The 15pi/12 simplifies to 5pi/4. We know how to find tan(5pi/4), which is 1. The same thing goes for 2pi/12. It simplifies to pi/6, and tan(pi/6)=1/sqrt(3).

Hope this helps!
• Isn't the slope change in y over change in x? Why does he use one point and just does y over x = slope?
• The line goes through the origin (0,0) and the point (x,y). So, the complete formula would be (y-0)/(x-0)=y/x.

So, if you are finding the slope of a line that goes through the origin, all you have to do is take y/x.

Hope this helps!
• When Sal says the slope of the tangent on the unit circle is just the radius for 5pi/4 = 1, why is it different for pi/6?
• Since tangent is opposite divided by adjacent, the values in the 45-45-90 triangle are both the same so it'll just be 1.
• Why did Sal rationalize the variables last? I mean, I can see why that was a good idea since it was faster than just rationalizing them from the start, but is there a rule on when to rationalize or is it just common sense on when it's a good idea to rationalize now than later?

Edit: I notice he didn't have to rationalize this way.
• It's always easier to work with smaller numbers, so personally I lean to rationalizing denominators afterwards unless I see some obvious benefit of doing it beforehand.
• In , Sal said we've proven tangent identities in another video. Where can I find that video?

I was trying to search the tangent identities but couldn't find a video with the sum and difference identities of the tangent.
• How would you solve an equation such as (cos(x))/(1+csc(x))*(1-csc(x))/(1-csc(x))?
I end up always getting these wrong, and they always answer (such as with this one) with something like tan(x)-tan(x)sin(x). why and how did they get to that answer??
• Find cos(13pi/12) exactly using an angle addition or subtraction formla.
I broke up (13pi/12) to (4pi/12 + 9pi/12). This futher simplifies to (pi/3 + 3pi/4). pi/3 is angle 60 degrees in the first quadrant and 3pi/4 is angle 45 degrees in the 2nd quadrant.
I used the addition formula cos(x+y) = cos(x)cos(y)-sin(x)sin(y). Assigned x to pi/3 and y to 3pi/4.
Cos(pi/3) = cos(60) = adjacent/hypotenuse = +x/1 = .5/1 = .5 (1st quadrant)
Cos(3pi/4) = cos(45) = adjacent/hypotenuse = +y/1 = sqroot2/2 divided by 1 = sqroot2/2 (2nd quadrant)
Sin(pi/3) = sin(60) = opposite/hypotenuse = +y/1 = sqroot3/2 divided by 1 = sqroot3/2 (1st quadrant)
Sin(3pi/4) = sin(45) = opposite / hypotenuse = -x/1 =
-sqroot2/2 divide by 1 = -sqroot2/2 (2nd quadrant)
Appling the sin and cosine values to the formula cos(x+Y) i got the answer sqroot2/4 + sqroot6/4.
The correct answer to the problem given is -sqroot2/4 -sqroot6/4.
• So there two definitions of the trignometric functions one based on SOHCAHTOA and the other the definitions of the unit circle.

For this question there is no need to use SOHCAHTOA so I would avoid it.

There are two mistakes.
1) cos(3pi/4) = sqrt(2)/2 should be cos(3pi/4) = -sqrt(2)/2
2) sin(3pi/4) = -sqrt(2)/2 should be sin(3pi/4) = sqrt(2)/2

This will make sense when you think about it. By definition cos(theta) = x and sin(theta) = y. But in the second quadrant sin(theta) is positive and cos(theta) is negative.

Fixing that you should arrive at the correct answer.

As a side note I would avoid writing sqroot instead write sqrt as it is more commonly used. Also use parenthesis when using sqrt as it not clear from your post what argument that is the input to the sqrt function is meant to be.