Trigonometry Problems: Solving For Cosine, Tangent, And Angles

Hey guys! Today, we're diving into some awesome trigonometry problems. We'll be tackling questions about finding cosine, tangent, and even constructing angles. So, grab your calculators and let's get started!

1. Finding $\cos\alpha$ when $\sin \alpha = \frac{1}{2}$

Okay, so the first problem asks us to find the value of $\cos\alpha$ given that $\sin \alpha$ is equal to $\frac{1}{2}$. To solve this, we'll need to use one of the most fundamental trigonometric identities: the Pythagorean identity. You might remember it as:

$\sin^2(\alpha) + \cos^2(\alpha) = 1$

This identity is super important in trigonometry, and it basically says that for any angle $\alpha$, the square of its sine plus the square of its cosine always equals 1. It's like a cornerstone of trigonometry, guys! Now, let's see how we can use it to solve our problem.

We already know that $\sin \alpha = \frac{1}{2}$, so we can plug that into our identity:

$(\frac{1}{2})^2 + \cos^2(\alpha) = 1$

This simplifies to:

$\frac{1}{4} + \cos^2(\alpha) = 1$

Now, we want to isolate $\cos^2(\alpha)$, so we subtract $\frac{1}{4}$ from both sides:

$\cos^2(\alpha) = 1 - \frac{1}{4}$

$\cos^2(\alpha) = \frac{3}{4}$

Great! We've got $\cos^2(\alpha)$, but we need $\cos(\alpha)$. To get that, we take the square root of both sides. Now, this is a crucial step because when you take the square root, you need to consider both positive and negative solutions:

$\cos(\alpha) = \pm\sqrt{\frac{3}{4}}$

$\cos(\alpha) = \pm\frac{\sqrt{3}}{2}$

So, we have two possible values for $\cos(\alpha)$: $\frac{\sqrt{3}}{2}$ and $-\frac{\sqrt{3}}{2}$. Which one is the correct answer? Well, it depends on the quadrant in which the angle $\alpha$ lies. Remember that sine is positive in the first and second quadrants. If $\alpha$ is in the first quadrant, cosine is also positive. If $\alpha$ is in the second quadrant, cosine is negative. Without more information about the angle $\alpha$, we can't definitively say which value is correct. Both $\frac{\sqrt{3}}{2}$ and $-\frac{\sqrt{3}}{2}$ are possible solutions. This is a super important thing to keep in mind when solving trig problems, guys! Always consider the possible quadrants.

2. Finding $tg\alpha$ when $\cos \alpha = -\frac{4}{5}$

Alright, let's move on to the next problem. This time, we need to find the tangent of $\alpha$ ($tg\alpha$) given that the cosine of $\alpha$ is $-\frac{4}{5}$. To tackle this, we'll need to remember the relationship between tangent, sine, and cosine. The tangent function is defined as the ratio of sine to cosine:

$tg(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$

So, we already know $\cos(\alpha)$, but we need to find $\sin(\alpha)$. Guess what? We can use the Pythagorean identity again! This identity is like our Swiss Army knife for trig problems. We know:

$\sin^2(\alpha) + \cos^2(\alpha) = 1$

We can plug in the value of $\cos(\alpha)$:

$\sin^2(\alpha) + (-\frac{4}{5})^2 = 1$

$\sin^2(\alpha) + \frac{16}{25} = 1$

Now, let's isolate $\sin^2(\alpha)$ by subtracting $\frac{16}{25}$ from both sides:

$\sin^2(\alpha) = 1 - \frac{16}{25}$

$\sin^2(\alpha) = \frac{9}{25}$

Taking the square root of both sides (remembering both positive and negative solutions!):

$\sin(\alpha) = \pm\sqrt{\frac{9}{25}}$

$\sin(\alpha) = \pm\frac{3}{5}$

Okay, now we have two possible values for $\sin(\alpha)$: $\frac{3}{5}$ and $-\frac{3}{5}$. To figure out which one is correct, we need to consider the quadrant where $\alpha$ lies. We know that $\cos(\alpha)$ is negative ($-\frac{4}{5}$). Cosine is negative in the second and third quadrants. Now, let's think about tangent. Tangent is positive in the first and third quadrants and negative in the second and fourth quadrants. So, if cosine is negative and we want to find the tangent, we need to consider both possibilities.

Let's calculate the tangent for both cases:

Case 1: $\sin(\alpha) = \frac{3}{5}$

$tg(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}$

Case 2: $\sin(\alpha) = -\frac{3}{5}$

$tg(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{-\frac{3}{5}}{-\frac{4}{5}} = \frac{3}{4}$

So, we have two possible values for $tg(\alpha)$: $-\frac{3}{4}$ and $\frac{3}{4}$. Since cosine is negative, $\alpha$ is in either the second or third quadrant. Tangent is negative in the second quadrant and positive in the third quadrant. Without further information to constrain the possibilities, we would accept both values, but carefully consider the restrictions imposed. This is what makes trigonometry so interesting, guys! The signs really matter.

3. Constructing Angle $A$ when $\sin A = \frac{3}{4}$

Alright, this problem is a little different. Instead of calculating a value, we need to construct an angle $A$ such that $\sin A = \frac{3}{4}$. This involves some geometry, which is pretty cool. Here's how we can do it:

1. Draw a ray: Start by drawing a horizontal ray (a line that extends infinitely in one direction). Let's call the endpoint of this ray $O$. This will be one side of our angle.
2. Draw a perpendicular line: At point $O$, draw a line perpendicular to the ray we just drew. This creates a right angle.
3. Determine the sides: Remember that $\sin A = \frac{\text{opposite}}{\text{hypotenuse}}$. So, in our case, the side opposite angle $A$ should have a length of 3 units, and the hypotenuse should have a length of 4 units. These are ratios, guys, so we can choose any convenient unit of measurement (centimeters, inches, whatever!).
4. Mark the opposite side: On the perpendicular line, mark a point (let's call it $B$) that is 3 units away from $O$. So, the length of the segment $OB$ is 3 units.
5. Draw an arc: Now, we need to find the point on the horizontal ray that will form the hypotenuse of our right triangle. Take a compass and set its radius to 4 units (our hypotenuse length). Place the compass point at $B$ and draw an arc that intersects the horizontal ray. This intersection point is crucial!
6. Mark the intersection: Let's call the point where the arc intersects the horizontal ray $C$. Now, we have a triangle $OBC$ where $OB = 3$ units, and $BC = 4$ units (our chosen units from before).
7. Draw the hypotenuse: Draw a line segment connecting points $O$ and $C$. This is the hypotenuse of our right triangle.

Now, we have a right triangle $OBC$ with a right angle at $O$. The side opposite angle $C$ is $OB$, which has a length of 3 units, and the hypotenuse $BC$ has a length of 4 units. Therefore, $\sin(\angle C) = \frac{OB}{BC} = \frac{3}{4}$. So, angle $C$ is our desired angle $A$! Cool, huh? We just constructed an angle with a specific sine value. This really showcases the geometric representation of trigonometric functions, guys. It's not just numbers, it's shapes!

4. Finding the Angle Between Ray OA and the Positive x-axis

The last problem is a bit vague, but we can interpret it and solve it. It says we have a ray $OA$ that intersects a unit semicircle. We need to find the angle between this ray and the positive x-axis. This problem is all about visualizing the unit circle and how angles are defined within it. This concept is super important for higher-level trigonometry and calculus, guys!

A unit circle is a circle with a radius of 1 centered at the origin (0, 0) in the coordinate plane. A semicircle is simply half of a circle. In this case, we're dealing with the upper half of the unit circle (the part where the y-coordinates are positive).

When we talk about angles in the unit circle, we measure them counterclockwise from the positive x-axis. The positive x-axis is considered 0 degrees (or 0 radians). As we move counterclockwise around the circle, the angle increases. The top of the semicircle is 90 degrees ($\frac{\pi}{2}$ radians), the negative x-axis is 180 degrees ($\pi$ radians), and so on.

So, to solve this problem, we need more information! We need to know the coordinates of point $A$ where the ray $OA$ intersects the unit semicircle. Let's say the coordinates of point $A$ are $(x, y)$. Because it's a unit semicircle, we know that the equation of the circle is $x^2 + y^2 = 1$, and $y$ will be greater or equal to 0 (since we are on the upper semicircle).

The angle $\theta$ between the ray $OA$ and the positive x-axis can be found using trigonometric functions. Specifically, we can use the sine and cosine functions:

• $\cos(\theta) = x$ (the x-coordinate of point A)
• $\sin(\theta) = y$ (the y-coordinate of point A)

Once we have the values of $\sin(\theta)$ and $\cos(\theta)$, we can use the inverse trigonometric functions (arcsin, arccos, or arctan) to find the angle $\theta$. For example:

• $\theta = \arccos(x)$
• $\theta = \arcsin(y)$
• $\theta = arctan(\frac{y}{x})$

However, we need to be careful about the range of the inverse trigonometric functions. For example, $\arccos$ returns angles between 0 and 180 degrees. We should then ascertain if the sine is positive or negative in order to know in which quadrant we are, and adjust the angle if necessary.

Without the coordinates of point A, we can't find a specific numerical value for the angle. But, we've outlined the process. We can determine the angle if we know the coordinates. This highlights how trigonometric functions connect angles and coordinates on the unit circle, guys. It's a beautiful relationship!

Wrapping Up

So, there you have it! We've worked through four different trigonometry problems, covering everything from finding cosine and tangent to constructing angles and understanding the unit circle. Remember the key identities and relationships, think about the quadrants, and you'll be well on your way to mastering trigonometry. Keep practicing, guys, and you'll become trig wizards in no time!