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### Course: Trigonometry>Unit 2

Lesson 4: Special trigonometric values in the first quadrant

# Cosine, sine and tangent of π/6 and π/3

With the unit circle and the Pythagorean theorem, we can find the exact sine, cosine, and tangent of the angles π/6 and π/3. Created by Sal Khan.

## Want to join the conversation?

• trig is scary
• just... get better, lol jk try here we go. embracing the challenge of trigonometry with a positive and determined mindset. Visualize concepts through diagrams and real-world applications to make the subject more engaging and relevant. Set achievable goals and celebrate each milestone, no matter how small. Seek support when needed, and remember that patience and persistence are key to mastering trigonometry. Stay positive, avoiding self-doubt, and view difficulties as opportunities to learn and grow. Draw inspiration from the achievements of mathematicians and scientists who have utilized trigonometry to solve fascinating problems throughout history. With dedication and motivation, you can unravel the beauty and practicality of trigonometry, and its applications will inspire your mathematical journey.

I really hope this help mate :)) stay strong!
• Hello

I just completed composite functions in Pre-calculus and the videos were really informative. However, the next section, which is this (trigonometry), is a little confusing. I don't understand the video above that I just watched, am I lacking some sort of foundational math that someone can refer me to before coming on to this section? I took elementary math in high school and not additional math which covers pre-calculus to calculus and that's why I am learning pre-calculus now before moving on to calculus but I don't recall learning this before.

Thank you :)
• So you are still in the pre-calculus section but in the trignometry unit.

Instead go to trigonometry section and start at the first unit. There you will learn trigonometry from the start.
• How does Sal determine that the point of which π/3 intersects the actual unit circle in in ? I still don't understand that especially the part of which he says that it is (cos π/3,sin π/3)? Please let me know, thank you.
• So π/3 is 60 degrees (π/3*180/π) which is how he estimates about where π/3 is. He then uses trig functions to get the points. By drawing a right triangle, the hypotenuse is 1 (radius of unit circle), the adjacent part along the x axis is defined by the function cos(π/3) = adj/hyp, but since the hyp=1, you get adj = cos(π/3) and the opposite part of the triangle would be sin(π/3) = opp/hyp, so the opp =sin(π/3). So staying in the first quadrant, the point on the unit circle defined by 0<x<π/2 (or 0<x<90) would always be (cos(x),sin(x)).
• I think this video also needs to appear in the Algebra 2 curriculum. Right now Unit 11 Lesson 4 of Algebra 2 only has the video "Trig values of π/4" and the exercise "Trig values of special angles."
• why can you just assume the length of the hypotenuse is 1?
• That's the idea behind using the "unit circle" i.e. a circle with a radius of 1. So a triangle drawn from the center of the circle will always have a hypotenuse of 1.
• I found using radians stranger than degrees. BTW, why is radian used here rather than degrees ?
• Radians are used because they're a more "natural" way to measure angles, especially in higher-level math. They're based on the radius of a circle, which makes many formulas simpler and neater. For example, when we use radians, we can describe the length of a part of the circumference of a circle directly in terms of the radius and the angle.
• I recently saw a question on someone else's test that I wasn't sure about.
`csc 158° 10'`

I assume that the problem was saying 10 * csc(158), where the tick mark mean distance or perhaps feet? The only question was to calculate the answer.

I am not aware of the notation with the little tickmark as shown.

• A degree of arc is subdivided into 60 'minutes of arc', or just 'minutes'. An arcminute is further divided into 60 arcseconds. So there are 60^2=3600 arcseconds in a degree.

We denote an arcminute with a ', and an arcsecond with a ".

So 158º 10' is 158 degrees, 10 minutes, or 158 and one-sixth degrees (since 10/60=1/6).
• Does that mean if hypotenues = A then adjecent side is A.cos(theta) and opposite side is A.sin(theta)?
• Yes, since sin(<)=opp/hyp, you could multiply by hyp to get opp=hyp sin(<), same for cos(<).
• How do you find the trig ratios(sine, cosine and tangent) for other angles like pi/5, pi/10 or any other arbitrary angle.
(1 vote)
• The two angles you mention, 𝜋∕5 = 36° and 𝜋∕10 = 18°, we can actually find exact expressions for that aren't too complicated.

We let the base of this isosceles triangle have length 1, and the legs have length 𝑥.

Then we draw the bisector to one of the 72° angles.
This splits the triangle into two new isosceles triangles, one of which is a 72-72-36 triangle.

We can figure out that this triangle has a base of length 𝑥 − 1, while the legs are of length 1.

Since this triangle is similar to the original triangle, we can set up the equation
𝑥 − 1 = 1∕𝑥.

Multiplying both sides by 𝑥 and then subtracting 1 from both sides, we get
𝑥² − 𝑥 − 1 = 0,
which has the positive solution
𝑥 = (1 + √5)∕2.

Now we go back to the original triangle.
Because it is isosceles, we know that its height bisects both the 36° angle and the base, thus splitting the triangle into two congruent right triangles –
either of which having one angle being 18° with the opposite side having length 1∕2
and a hypotenuse of length 𝑥 = (1 + √5)∕2.

Thus, sin(18°) = (1∕2)∕((1 + √5)∕2),
which simplifies to (√5 − 1)∕4.

Because 18° lies in the first quadrant, we know that cos(18°) is positive:
cos(18°) = +√(1 − sin²(18°)),
which we calculate to √(10 + 2√5)∕4.

To get a somewhat neat expression for tan(18°) we begin by finding
tan²(18°) = sin²(18°)∕cos²(18°) = (1 − cos²(18°))∕cos²(18°) = 1∕cos²(18°) − 1
= 16∕(10 + 2√5) − 1,
which can be simplified to (25 − 10√5)∕5².

Again, 18° lies in the first quadrant, so tan(18°) is positive, and we get
tan(18°) = √(25 − 10√5)∕5.

– – –

To find sin(36°) we can use the double angle formula
sin(2𝑥) = 2 sin(𝑥) cos(𝑥)
⇒ sin(36°) = 2 sin(18°) cos(18°) = 2(√5 − 1)∕4⋅√(10 + 2√5)∕4
= (√5 − 1)√(10 + 2√5)∕8.

Again, squaring both sides allows us to get a neater result:
sin²(36°) = (6 − 2√5)(10 + 2√5)∕64,
simplifying to (10 − 2√5)∕16.
36° lies in the first quadrant, so
sin(36°) = √(10 − 2√5)∕4.

cos(36°) = √(1 − sin²(36°)),
which we calculate to √(6 + 2√5)∕4.

Finally,
tan²(36°) = 1∕cos²(36°) − 1 = 16∕(6 + 2√5) − 1,
which simplifies to 5 − 2√5
⇒ tan(36°) = √(5 − 2√5).