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## Precalculus

### Course: Precalculus>Unit 2

Lesson 1: 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?

• 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 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).
• How do you find the trig ratios(sine, cosine and tangent) for other angles like pi/5, pi/10 or any other arbitrary angle.
• 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).
• Why is tangent = Sine over cosine?
• sine = opposite/hypotenuse

• why are pi/3 60degrees and pi/6 30degrees?
• Because unit cirkel is 360 degrees, equal to 2 pi r. 1 pi r 180, and 1/3 pi r therefore 60 degrees.
• Can we say that for all θ measured in radians, where 0≤θ≤π/2, that sinθ=cos(π/2-θ) and cosθ=sin(π/2-θ)?
• For all theta that are part of the real numbers
• : How does Sal get b is the sqrt of 3 over 2? He doesn't seem to show his work there...
• He starts from b^2 = 3/4. The opposite of squaring is square rooting (sort of, but Sal mentions we are taking the primary -positive - root only). So taking square root of everything, you get √b^2 = √3/√4. On the left, the square and root cancel, on the right √4=2, so you end up with b = √3/2.
• Does that mean if hypotenues = A then adjecent side is A.cos(theta) and opposite side is A.sin(theta)?
(1 vote)
• Yes, since sin(<)=opp/hyp, you could multiply by hyp to get opp=hyp sin(<), same for cos(<).