Without using a calculator, determine all values of A in the interval [ 0 , π 2 ) \left[0,\frac{\pi}{2}\right) with the following trigonometric function value. cos A = 3 2 \cos A=\frac{\sqrt3}{2} cos A = 2 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

Let's give this problem a try. So this problem says without using a calculator, determine all the values of a in the interval from 0 to pi over 2 with the following trigonometric function value. And we're given the cosine of a is equal to the square root of 3 over 2. Now, this problem is actually quite tricky because we can't use a calculator, so we're going to have to think outside the box to solve this. But let's see if using the skills that we've learned from trigonometry so far, we can go about figuring out what this angle a is. So let's try this. Now what I'm first going to do is draw a right triangle so we can visualize this better. And I'm going to go ahead and say that our angle a is right about there. Now to figure out how this relates to the sides of the triangle, we're going to use SOHCAHTOA. SOHCAHTOA is this memory tool that we use to remember how the trigonometric functions relate to the sides of a right triangle. And since I can see that we're dealing with cosine, I'm going to use this middle term here that says that cosine is adjacent over hypotenuse. So if I go ahead and go to our angle, the adjacent side is going to be this, which is square root of 3, and then the hypotenuse is going to be this bottom piece, which is 2. Again, because cosine is adjacent over hypotenuse. Now from here, what I'm going to do is solve for the missing side. And to do this, I can use the pythagorean theorem a squared plus b squared is equal to c squared. Now a and b are always going to be the two legs of the triangle. So I'm gonna go ahead and say that our a is going to be the square root of 3. So we're gonna have the square root of 3 squared, and then b is going to be this side of the triangle, which we do not know. And then this is all going to equal c squared, which is always the hypotenuse. So we'll have the hypotenuse squared, and now we just need to solve for b. Now the square root of 3 squared is just gonna give you 3, and this is plus b squared is equal to 2 squared, which is 4. Now what I can do is subtract 3 on both sides of this equation, and that's going to give us that b squared is equal to 4 minuteus 3 which is 1, and then I can take the square root on both sides of this equation to get that b is equal to just 1. So that means that our missing side b is equal to 1. So now that we found all the sides of the right triangle, there's something that we can observe about the behavior of this triangle. Recall that when dealing with a 30, 60, 90 triangle, the hypotenuse of your triangle is always the short leg multiplied by 2. And recall that also the long leg of a 30, 60, 90 triangle is equal to the short leg multiplied by the square root of 3. Now we can see in this triangle that the short leg is 1. So So what that means is that our hypotenuse would just be 1 times 2 or 2, and that checks out. Now what we also can see is that the long leg is gonna be the short leg multiplied by the square root of 3, and one times the square root of 3 would just be the square root of 3, and that also checks out for the long leg of this triangle. So that means we are dealing with a 30, 60, 90 right triangle. And if this angle would be 60 degrees up here, then that means that our angle a must be 30 degrees. Although we're not asked to give our answer in degrees, because notice the interval is from 0 to pi over 2. Pi means we're dealing with radians. So what we actually want to do is convert this angle to radians by taking this and multiplying it by pi over 180 degrees. We've discussed how to do these conversions in previous videos. Now what that means is that the degrees are going to go away, and we can also cancel one of these zeros. Now 3 is going to be right there, and then 18, that's the same thing as 3 times 6. Because 3 times 6 gives you 18. So we can cancel the 3 with that 3 down there, giving us just pi over 6. So the solution for our angle a is going to be pi over 6 radians, and this is the only solution within the interval from 0 to pi over 2. So that's how you can go about solving this problem. As you can see, it's pretty tricky and there's a lot of steps. But hopefully, you can understand the general intuition as to how we were able to find what this angle is. So hope you found this video helpful. Thanks for watching, and let me know if you have any questions.

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0. Review4h 36m

Algebraic Expressions
37m

Exponents
32m

Polynomials Intro
19m

Multiplying Polynomials
36m

Factoring Polynomials
1h 2m

Radical Expressions
15m

Simplifying Radical Expressions
35m

Rationalize Denominator
15m

Rational Exponents
4m

Pythagorean Theorem & Basics of Triangles
17m

1. Equations & Inequalities3h 26m

Linear Equations
31m

Rational Equations
21m

The Imaginary Unit
6m

Powers of i
11m

Complex Numbers
41m

Intro to Quadratic Equations
18m

The Square Root Property
11m

Completing the Square
12m

The Quadratic Formula
19m

Choosing a Method to Solve Quadratics
9m

Linear Inequalities
22m

2. Graphs & Functions4h 0m

Graphs and Coordinates
7m

Two-Variable Equations
23m

Lines
1h 12m

Intro To Functions And Graphs
35m

Common Functions
5m

Transformations
45m

Function Operations
21m

Function Composition
29m

3. Polynomial and Rational Functions3h 18m

Quadratic Functions
49m

Understanding Polynomial Functions
34m

Graphing Polynomial Functions
30m

Intro to Rational Functions
9m

Asymptotes
35m

Graphing Rational Functions
38m

4. Inverse, Exponential, and Logarithmic Functions2h 28m

Intro to Exponential Functions
9m

Graphing Exponential Functions
25m

The Number e
8m

Intro to Logarithms
22m

Graphing Logarithmic Functions
18m

Properties of Logarithms
27m

Solving Exponential & Logarithmic Equations
35m

5. Trigonometric Functions5h 25m

Angles in Standard Position
12m

Coterminal Angles
8m

Complementary & Supplementary Angles
10m

Converting between Degrees & Radians
8m

Trigonometric Functions on Right Triangles
45m

Special Right Triangles
30m

Cofunctions of Complementary Angles
26m

Solving Right Triangles
23m

Defining the Unit Circle
14m

Trig Functions on the Unit Circle
9m

Common Values of Sine, Cosine, & Tangent
11m

Reference Angles
38m

Reciprocal Trig Functions on the Unit Circle
6m

Graphs of the Sine and Cosine Functions
32m

Phase Shifts
14m

Graphs of Secant and Cosecant Functions
10m

Graphs of Tangent and Cotangent Functions
21m

6. Analytic Trigonometry4h 15m

Inverse Sine, Cosine, & Tangent
28m

Evaluate Composite Trig Functions
44m

Linear Trigonometric Equations
28m

Introduction to Trig Identities
1h 4m

Sum and Difference Identities
1h 3m

Double Angle Identities
16m

Solving Trig Equations Using Identities
9m

7. Applications of Trigonometry1h 38m

The Law of Sines
49m

The Law of Cosines
30m

Area of SAS & ASA Triangles
19m

8. Polar Coordinates and Complex Numbers4h 20m

Polar Coordinate System
29m

Convert Points Between Polar & Rectangular Coordinates
26m

Convert Equations Between Polar & Rectangular Forms
29m

Graphing Other Common Polar Equations
39m

Graphing Parametric Equations
12m

Eliminate the Parameter
32m

Writing Parametric Equations
21m

Graphing Complex Numbers
6m

Polar Form of Complex Numbers
22m

Products and Quotients of Complex Numbers
14m

Powers of Complex Numbers (DeMoivre's Theorem)
23m

9. Vectors2h 25m

Geometric Vectors
28m

Vectors in Component Form
32m

Direction of a Vector
23m

Unit Vectors and i & j Notation
27m

Dot Product
22m

Cross Product
11m

10. Systems of Equations & Matrices4h 5m

Two Variable Systems of Linear EquationsBonus
1h 7m

Introduction to Matrices
1h 11m

Determinants and Cramer's Rule
1h 3m

Graphing Systems of Inequalities
42m

11. Analytic Geometry2h 36m

Introduction to Conic Sections
6m

Circles
26m

Ellipses: Standard Form
33m

Parabolas
36m

Hyperbolas at the Origin
40m

Hyperbolas NOT at the Origin
12m

12. Sequences, Series, and Induction1h 22m

Sequences
40m

Arithmetic Sequences
25m

Geometric Sequences
15m

13. Combinatorics and Probability1h 45m

Factorials
11m

Combinatorics
46m

Probability
47m

5. Trigonometric Functions

Special Right Triangles

Algebra and Trigonometry

5. Trigonometric Functions

Special Right Triangles

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Multiple Choice

Without using a calculator, determine all values of A in the interval $open bracket 0 comma pi over 2 close paren$ with the following trigonometric function value.
$\cos\) A=\(\frac{\sqrt3}{2}$

$0$ only

$\frac{\pi}{4}$ only

$\frac{\pi}{6}$ only

$\frac{\pi}{3}$ only

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Struggling with Algebra and Trigonometry?

Without using a calculator, determine all values of A in the interval [0,π2)\(\left\)[0,\(\frac{\pi}{2}\]\right\)) with the following trigonometric function value.cosA=32\(\cos\) A=\(\frac{\sqrt3}{2}\)cosA=23

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Without using a calculator, determine all values of A in the interval $open bracket 0 comma pi over 2 close paren$ with the following trigonometric function value.
$\cos\) A=\(\frac{\sqrt3}{2}$