Math / Trigonometry

Question1. Simplify the expressions to a single trigonometric function: a. $\frac{\csc ^{2}(t)-1}{\csc ^{2}(t)}$ b. $\cos (x) \tan (x)+\sin (x)$
2. Consider $\cot (x)=\frac{1}{\sqrt{3}}$ a. What quadrant is the angle located? b. Will the cosine of this angle be positive or negative? c. Will the sine of this angle be positive or negative?
3. Consider the function $\frac{x}{\sqrt{9-x^{2}}}$ . Rewrite this expression by substituting $x=3 \sin \theta$ , and simplifying completely.

Studdy Solution

STEP 1

What is this asking? We're playing with trig identities to simplify expressions, and then doing some detective work with a specific cotangent value to figure out the quadrant and signs of cosine and sine.
Finally, we'll do a trig substitution and simplify a funky expression! Watch out! Remember those Pythagorean identities, and don't forget there are two possible quadrants for any given cotangent value.
Also, be super careful with your algebra when simplifying – it's easy to drop a negative sign or mix up your fractions.

STEP 2

1. Simplify Expression a
2. Simplify Expression b
3. Determine Quadrant and Signs
4. Trig Substitution and Simplification

STEP 3

Alright, let's rewrite that $\csc^2(t)$ in the numerator using the Pythagorean identity: $\csc^2(t) = 1 + \cot^2(t)$ .
So, our expression becomes $\frac{(1 + \cot^2(t)) - 1}{\csc^2(t)}$ .

STEP 4

Now, look at that numerator!
The 1s add to zero, leaving us with $\frac{\cot^2(t)}{\csc^2(t)}$ .
Remember, $\cot(t) = \frac{\cos(t)}{\sin(t)}$ and $\csc(t) = \frac{1}{\sin(t)}$ .

STEP 5

Let's substitute those definitions in: $\frac{\frac{\cos^2(t)}{\sin^2(t)}}{\frac{1}{\sin^2(t)}}$ .
This looks like a hot mess, but it's about to get way cleaner.

STEP 6

We're dividing by $\frac{1}{\sin^2(t)}$ , which is the same as multiplying by its reciprocal, $\sin^2(t)$ .
So, we have $\frac{\cos^2(t)}{\sin^2(t)} \cdot \sin^2(t)$ .

STEP 7

Boom! The $\sin^2(t)$ terms divide to one, and we're left with $\cos^2(t)$ .
Nice and tidy!

STEP 8

Let's rewrite $\tan(x)$ as $\frac{\sin(x)}{\cos(x)}$ .
Our expression becomes $\cos(x) \cdot \frac{\sin(x)}{\cos(x)} + \sin(x)$ .

STEP 9

The $\cos(x)$ terms divide to one, giving us $\sin(x) + \sin(x)$ .

STEP 10

Adding those together, we get $2\sin(x)$ .
Super simple!

STEP 11

We're given $\cot(x) = \frac{1}{\sqrt{3}}$ .
Since cotangent is positive, $x$ must be in either Quadrant I or Quadrant III.

STEP 12

In Quadrant I, both cosine and sine are positive.

STEP 13

In Quadrant III, both cosine and sine are negative.

STEP 14

We're given the expression $\frac{x}{\sqrt{9 - x^2}}$ and we're told to substitute $x = 3\sin(\theta)$ .
Let's do it!
We get $\frac{3\sin(\theta)}{\sqrt{9 - (3\sin(\theta))^2}}$ .

STEP 15

Squaring the term inside the square root gives us $\frac{3\sin(\theta)}{\sqrt{9 - 9\sin^2(\theta)}}$ .

STEP 16

We can factor out a 9 from the terms under the square root: $\frac{3\sin(\theta)}{\sqrt{9(1 - \sin^2(\theta))}}$ .

STEP 17

Using the Pythagorean identity, $1 - \sin^2(\theta) = \cos^2(\theta)$ , we get $\frac{3\sin(\theta)}{\sqrt{9\cos^2(\theta)}}$ .

STEP 18

Taking the square root of the denominator gives us $\frac{3\sin(\theta)}{3\cos(\theta)}$ .

STEP 19

The 3s divide to one, leaving us with $\frac{\sin(\theta)}{\cos(\theta)}$ , which is just $\tan(\theta)$ !

STEP 20

1. a. $\cos^2(t)$ b. $2\sin(x)$
2. a. Quadrant I or Quadrant III b. Positive in Quadrant I, negative in Quadrant III c. Positive in Quadrant I, negative in Quadrant III
3. $\tan(\theta)$

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Studdy Solution

STEP 1

STEP 2

1. Simplify Expression a2. Simplify Expression b3. Determine Quadrant and Signs4. Trig Substitution and Simplification

STEP 3

Alright, let's **rewrite** that csc⁡2(t)\csc^2(t)csc2(t) in the numerator using the **Pythagorean identity**: csc⁡2(t)=1+cot⁡2(t)\csc^2(t) = 1 + \cot^2(t)csc2(t)=1+cot2(t).So, our expression becomes (1+cot⁡2(t))−1csc⁡2(t)\frac{(1 + \cot^2(t)) - 1}{\csc^2(t)}csc2(t)(1+cot2(t))−1​.

STEP 4

Now, look at that numerator!The 1s **add to zero**, leaving us with cot⁡2(t)csc⁡2(t)\frac{\cot^2(t)}{\csc^2(t)}csc2(t)cot2(t)​.Remember, cot⁡(t)=cos⁡(t)sin⁡(t)\cot(t) = \frac{\cos(t)}{\sin(t)}cot(t)=sin(t)cos(t)​ and csc⁡(t)=1sin⁡(t)\csc(t) = \frac{1}{\sin(t)}csc(t)=sin(t)1​.

STEP 5

Let's **substitute** those definitions in: cos⁡2(t)sin⁡2(t)1sin⁡2(t)\frac{\frac{\cos^2(t)}{\sin^2(t)}}{\frac{1}{\sin^2(t)}}sin2(t)1​sin2(t)cos2(t)​​.This looks like a hot mess, but it's about to get *way* cleaner.

STEP 6

We're **dividing** by 1sin⁡2(t)\frac{1}{\sin^2(t)}sin2(t)1​, which is the same as **multiplying** by its reciprocal, sin⁡2(t)\sin^2(t)sin2(t).So, we have cos⁡2(t)sin⁡2(t)⋅sin⁡2(t)\frac{\cos^2(t)}{\sin^2(t)} \cdot \sin^2(t)sin2(t)cos2(t)​⋅sin2(t).

STEP 7

Boom! The sin⁡2(t)\sin^2(t)sin2(t) terms **divide to one**, and we're left with cos⁡2(t)\cos^2(t)cos2(t).Nice and tidy!

STEP 8

Let's **rewrite** tan⁡(x)\tan(x)tan(x) as sin⁡(x)cos⁡(x)\frac{\sin(x)}{\cos(x)}cos(x)sin(x)​.Our expression becomes cos⁡(x)⋅sin⁡(x)cos⁡(x)+sin⁡(x)\cos(x) \cdot \frac{\sin(x)}{\cos(x)} + \sin(x)cos(x)⋅cos(x)sin(x)​+sin(x).

STEP 9

The cos⁡(x)\cos(x)cos(x) terms **divide to one**, giving us sin⁡(x)+sin⁡(x)\sin(x) + \sin(x)sin(x)+sin(x).

STEP 10

Adding those together, we get 2sin⁡(x)2\sin(x)2sin(x).Super simple!

STEP 11

We're given cot⁡(x)=13\cot(x) = \frac{1}{\sqrt{3}}cot(x)=3​1​.Since cotangent is positive, xxx must be in either **Quadrant I or Quadrant III**.

STEP 12

In **Quadrant I**, *both* cosine and sine are positive.

STEP 13

In **Quadrant III**, *both* cosine and sine are negative.

STEP 14

We're given the expression x9−x2\frac{x}{\sqrt{9 - x^2}}9−x2​x​ and we're told to **substitute** x=3sin⁡(θ)x = 3\sin(\theta)x=3sin(θ).Let's do it!We get 3sin⁡(θ)9−(3sin⁡(θ))2\frac{3\sin(\theta)}{\sqrt{9 - (3\sin(\theta))^2}}9−(3sin(θ))2​3sin(θ)​.

STEP 15

**Squaring** the term inside the square root gives us 3sin⁡(θ)9−9sin⁡2(θ)\frac{3\sin(\theta)}{\sqrt{9 - 9\sin^2(\theta)}}9−9sin2(θ)​3sin(θ)​.

STEP 16

We can **factor out** a 9 from the terms under the square root: 3sin⁡(θ)9(1−sin⁡2(θ))\frac{3\sin(\theta)}{\sqrt{9(1 - \sin^2(\theta))}}9(1−sin2(θ))​3sin(θ)​.

STEP 17

Using the **Pythagorean identity**, 1−sin⁡2(θ)=cos⁡2(θ)1 - \sin^2(\theta) = \cos^2(\theta)1−sin2(θ)=cos2(θ), we get 3sin⁡(θ)9cos⁡2(θ)\frac{3\sin(\theta)}{\sqrt{9\cos^2(\theta)}}9cos2(θ)​3sin(θ)​.

STEP 18

Taking the **square root** of the denominator gives us 3sin⁡(θ)3cos⁡(θ)\frac{3\sin(\theta)}{3\cos(\theta)}3cos(θ)3sin(θ)​.

STEP 19

The 3s **divide to one**, leaving us with sin⁡(θ)cos⁡(θ)\frac{\sin(\theta)}{\cos(\theta)}cos(θ)sin(θ)​, which is just tan⁡(θ)\tan(\theta)tan(θ)!

STEP 20

1. a. cos⁡2(t)\cos^2(t)cos2(t) b. 2sin⁡(x)2\sin(x)2sin(x)2. a. Quadrant I or Quadrant III b. Positive in Quadrant I, negative in Quadrant III c. Positive in Quadrant I, negative in Quadrant III3. tan⁡(θ)\tan(\theta)tan(θ)

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1. Simplify Expression a
2. Simplify Expression b
3. Determine Quadrant and Signs
4. Trig Substitution and Simplification

Alright, let's rewrite that $\csc^2(t)$ in the numerator using the Pythagorean identity: $\csc^2(t) = 1 + \cot^2(t)$ .
So, our expression becomes $\frac{(1 + \cot^2(t)) - 1}{\csc^2(t)}$ .

Now, look at that numerator!
The 1s add to zero, leaving us with $\frac{\cot^2(t)}{\csc^2(t)}$ .
Remember, $\cot(t) = \frac{\cos(t)}{\sin(t)}$ and $\csc(t) = \frac{1}{\sin(t)}$ .

Let's substitute those definitions in: $\frac{\frac{\cos^2(t)}{\sin^2(t)}}{\frac{1}{\sin^2(t)}}$ .
This looks like a hot mess, but it's about to get way cleaner.

We're dividing by $\frac{1}{\sin^2(t)}$ , which is the same as multiplying by its reciprocal, $\sin^2(t)$ .
So, we have $\frac{\cos^2(t)}{\sin^2(t)} \cdot \sin^2(t)$ .

Boom! The $\sin^2(t)$ terms divide to one, and we're left with $\cos^2(t)$ .
Nice and tidy!

Let's rewrite $\tan(x)$ as $\frac{\sin(x)}{\cos(x)}$ .
Our expression becomes $\cos(x) \cdot \frac{\sin(x)}{\cos(x)} + \sin(x)$ .

The $\cos(x)$ terms divide to one, giving us $\sin(x) + \sin(x)$ .

Adding those together, we get $2\sin(x)$ .
Super simple!

We're given $\cot(x) = \frac{1}{\sqrt{3}}$ .
Since cotangent is positive, $x$ must be in either Quadrant I or Quadrant III.

In Quadrant I, both cosine and sine are positive.

In Quadrant III, both cosine and sine are negative.

We're given the expression $\frac{x}{\sqrt{9 - x^2}}$ and we're told to substitute $x = 3\sin(\theta)$ .
Let's do it!
We get $\frac{3\sin(\theta)}{\sqrt{9 - (3\sin(\theta))^2}}$ .

Squaring the term inside the square root gives us $\frac{3\sin(\theta)}{\sqrt{9 - 9\sin^2(\theta)}}$ .

We can factor out a 9 from the terms under the square root: $\frac{3\sin(\theta)}{\sqrt{9(1 - \sin^2(\theta))}}$ .

Using the Pythagorean identity, $1 - \sin^2(\theta) = \cos^2(\theta)$ , we get $\frac{3\sin(\theta)}{\sqrt{9\cos^2(\theta)}}$ .

Taking the square root of the denominator gives us $\frac{3\sin(\theta)}{3\cos(\theta)}$ .

The 3s divide to one, leaving us with $\frac{\sin(\theta)}{\cos(\theta)}$ , which is just $\tan(\theta)$ !

1. a. $\cos^2(t)$ b. $2\sin(x)$
2. a. Quadrant I or Quadrant III b. Positive in Quadrant I, negative in Quadrant III c. Positive in Quadrant I, negative in Quadrant III
3. $\tan(\theta)$