Imagine an engaging and educational image, free of any written words. At the center, you can see a huge diagram of a unit circle on a Cartesian plane. Half of the circle is lightly shaded, representing an angle which measures 13pi/12 radians. The Cartesian plane axes are marked with 'cosine' and 'sine' to signify respective relations. On the right side, there are visually appealing mathematical equations and illustrations demonstrating the application of sin and cos sum and difference formulas, each step in the calculations clearly demonstrated. The overall ambiance of the scene is that of a crucial math study session.

How could you evaluate tan (13pi/12) if you did not know the sum and difference formula for tangent? Would you use the sin and cos sum and difference formulas, and if so, can someone walk me through it? Thank you!!!

OR

so we can't use tan expansions, ok
tan(13pi/12) = sin(13pi/12) / cos(13pi/12)

When I was learning this stuff many eons ago, I personally found that
I could work it out better in degrees than all that radian stuff with π's
(13pi/12) = 195°
several ways to get 195 with basic angles, oobleck did 60 + 135
I am going to try 180+15
we need sin15 and cos15.
I know cos30 = √3/2 and cos30 = 1 - 2sin^2 15
2sin^2 15 = 1 - √3/2 = (2 - √3)/2
sin15= √((2-√3)/4)
similarly, cos15 = √((2+√3)/4)
so tan15 = √((2-√3)/4) / √((2+√3)/4) = √(2-√3) / √(2+√3)
but tan 195° = - tan15° = -√(2-√3) / √(2+√3)

which does reduce to 2 - √3, but that's another story.

Don't claim this is easier, just another way. What else do I have to do
after 4 weeks of quarantine lock-down ??

OR

13 π / 12 = 12 π / 12 + π / 12 = π + π / 12

tan ( A + B ) = ( tan A + tan B ) / ( 1 - tan A ∙ tan B )

In this case:

tan ( 13 π / 12 ) = tan ( π + π / 12 ) = ( tan π + tan π / 12 ) / ( 1 - tan π ∙ tan π / 12 )

Since tan π = 0

tan ( 13 π / 12 ) = ( 0 + tan π / 12 ) / ( 1 - 0 ∙ tan π / 12 )

tan ( 13 π / 12 ) = ( tan π / 12 ) / 1

tan ( 13 π / 12 ) = tan π / 12

tan ( θ / 2 ) = ( 1 - cos θ ) / sin θ

In this case π / 12 = ( π / 6 ) / 2

tan ( π / 12 ) = tan ( π / 6 ) / 2 = [ 1 - cos ( π / 6 ) ] / sin ( π / 6 ) =

( 1 - √3 / 2 ) / ( 1 / 2 ) =

( 2 / 2 - √3 / 2 ) ∙ 2 =

( 2 - √3 ) / 2 ∙ 2 = 2 - √3

_____________________

Remark:

1 / ( 1 / 2 ) = 2

so

( 1 - √3 / 2 ) / ( 1 / 2 ) =

( 1 - √3 / 2 ) ∙ 1 / ( 1 / 2 ) =

( 2 / 2 - √3 / 2 ) ∙ 2
_____________________

Solution:

tan 13 π / 12 = 2 - √3

Haha you're right Reiny. Thank you very much and I hope you continue to entertain yourself during quarantine.

well, that is certainly one way. Since 13π/12 = 3π/4 + π/3

sin(3π/4 + π/3) = (1/√2)(1/2) + (-1/√2)(√3/2)
cos(3π/4 + π/3) = (-1/√2)(1/√2) - (-1/2)(√3/2)
Now simplify, divide and simplify the result. You should ned up with 2-√3

Thank you both sooo much! It was the last problem on my hw and I've been stuck for hours.

Oh, tan(13π/12), you're trying to take me back to my algebra days! Well, fear not because your friendly neighborhood Clown Bot is here to help you out with a twist of humor.

Now, tan(13π/12) is not the friendliest angle to deal with, but don't worry, we can handle it without the sum and difference formula for tangent. Instead, we'll use the sin and cos sum and difference formulas in a clever way.

First, let's break down the angle 13π/12 into two familiar angles. To do that, we'll subtract a more manageable angle from it. Let's subtract π/6, because hey, π/6 is a pie everyone loves!

So, 13π/12 - π/6 = (26π - π)/12 = 25π/12. Great! Now we have a simpler angle to work with.

Next, we want to express 25π/12 as the sum or difference of two angles. Let's go for the sum: π/3 + 8π/12 = 4π/3 + 2π/12.

Now, we can apply the sin and cos sum formulas:

sin(25π/12) = sin(4π/3 + 2π/12)
= sin(4π/3)cos(2π/12) + cos(4π/3)sin(2π/12)

cos(25π/12) = cos(4π/3 + 2π/12)
= cos(4π/3)cos(2π/12) - sin(4π/3)sin(2π/12)

Using the values of sin(π/3), sin(2π/12), cos(4π/3), and cos(2π/12), which you hopefully know, we can evaluate this expression bit by bit.

And voila! By plugging in these values and doing some calculations, you'll ultimately find the value of tan(13π/12) without needing the specific sum and difference formula for tangent.

Hope this humor-filled explanation helped! Remember, math can be a joke - you just have to find the punchline!

To evaluate tan (13π/12) without using the sum and difference formula for tangent, you can indeed use the sum and difference formulas for sine and cosine. Here's how you can go about it:

1. Recall the formula for tangent: tan(x) = sin(x) / cos(x).

2. Divide the numerator and denominator by cos(13π/12):
tan(13π/12) = [sin(13π/12) / cos(13π/12)] / [cos(13π/12) / cos(13π/12)].

3. Apply the angle sum formula for sin:
sin(a + b) = sin(a)cos(b) + cos(a)sin(b).

4. Using the angle sum formula for sin, rewrite the numerator:
sin(13π/12) = sin(π/3 + 3π/4) = sin(π/3)cos(3π/4) + cos(π/3)sin(3π/4).

5. Recall that sin(π/3) = √3/2 and cos(π/3) = 1/2 from the unit circle.
Also, cos(3π/4) = -sin(π/4) = -1/√2 and sin(3π/4) = cos(π/4) = 1/√2.

6. Plug in the values found in step 5 into the numerator and simplify:
sin(13π/12) = (√3/2)(-1/√2) + (1/2)(1/√2) = -√3/2√2 + 1/2√2 = (-√3 + 1) / (2√2).

7. Now simplify the denominator, cos(13π/12):
cos(13π/12) = cos(π - π/12) = -cos(π/12).

8. Apply the angle difference formula for cosine:
cos(a - b) = cos(a)cos(b) + sin(a)sin(b).

9. Using the angle difference formula for cosine, rewrite the denominator:
cos(π/12) = cos(π/4 - π/6) = cos(π/4)cos(π/6) + sin(π/4)sin(π/6).

10. Recall that cos(π/4) = sin(π/4) = 1/√2 and sin(π/6) = 1/2.
Plug in the values found in step 10 into the denominator and simplify:
cos(13π/12) = (1/√2)(1/2) + (1/√2)(1/√2)
= 1 / (2√2) + 1 / (2√2)
= 2 / (4√2)
= 1 / (2√2).

11. Now that we have the numerator and denominator simplified, the final result is:
tan(13π/12) = [(-√3 + 1) / (2√2)] / [1 / (2√2)]
= (-√3 + 1) / 1
= -√3 + 1.

So, tan(13π/12) simplifies to -√3 + 1.

Remember that this process involves applying the sum and difference formulas for sine and cosine to find the values of sine and cosine for the given angle and then substituting them into the formula for tangent.