in exercises 39–46, write each expression in factored form as an algebraic expression of a single trigonometric function (e.g., ((2sin x + 3)(sin x - 1))). 39. (cos^2 x + 2cos x + 1) 40. (1 - 2sin x + sin^2 x) 41. (1 - 2sin x + (1 - cos^2 x)) 42. (sin x - cos^2 x - 1) 43. (cos x - 2sin^2 x + 1) 44. (sin^2 x + \frac{2}{csc x} + 1) 45. (4 an^2 x - \frac{4}{cot x} + sin x csc x) 46. (sec^2 x - sec x + an^2 x)

Question

in exercises 39–46, write each expression in factored form as an algebraic expression of a single trigonometric function (e.g., ((2sin x + 3)(sin x - 1))). 39. (cos^2 x + 2cos x + 1) 40. (1 - 2sin x + sin^2 x) 41. (1 - 2sin x + (1 - cos^2 x)) 42. (sin x - cos^2 x - 1) 43. (cos x - 2sin^2 x + 1) 44. (sin^2 x + \frac{2}{csc x} + 1) 45. (4	an^2 x - \frac{4}{cot x} + sin x csc x) 46. (sec^2 x - sec x + 	an^2 x)

Accepted Answer

Let's solve each problem one by one:

### Problem 39: $\boldsymbol{\cos^2 x + 2\cos x + 1}$
# Explanation:
## Step 1: Recognize the perfect square trinomial
The expression $\cos^2 x + 2\cos x + 1$ is a perfect square trinomial of the form $a^2 + 2ab + b^2$, where $a = \cos x$ and $b = 1$.
The formula for a perfect square trinomial is $a^2 + 2ab + b^2=(a + b)^2$.
## Step 2: Apply the perfect square formula
Substituting $a = \cos x$ and $b = 1$ into the formula, we get:
$(\cos x + 1)^2$
# Answer:
$(\cos x + 1)^2$

### Problem 40: $\boldsymbol{1 - 2\sin x + \sin^2 x}$
# Explanation:
## Step 1: Rearrange the terms
Rearrange the expression as $\sin^2 x - 2\sin x + 1$, which is a perfect square trinomial of the form $a^2 - 2ab + b^2$, where $a=\sin x$ and $b = 1$.
The formula for a perfect square trinomial is $a^2 - 2ab + b^2=(a - b)^2$.
## Step 2: Apply the perfect square formula
Substituting $a=\sin x$ and $b = 1$ into the formula, we get:
$(\sin x - 1)^2$
# Answer:
$(\sin x - 1)^2$

### Problem 41: $\boldsymbol{1 - 2\sin x + (1 - \cos^2 x)}$
# Explanation:
## Step 1: Use the Pythagorean identity
Recall the Pythagorean identity $\sin^2 x+\cos^2 x = 1$, so $1-\cos^2 x=\sin^2 x$.
Substitute $1 - \cos^2 x$ with $\sin^2 x$ in the expression:
$1-2\sin x+\sin^2 x$
## Step 2: Recognize the perfect square trinomial
The expression $1-2\sin x+\sin^2 x$ is a perfect square trinomial of the form $a^2 - 2ab + b^2$, where $a = \sin x$ and $b=1$.
Using the formula $a^2 - 2ab + b^2=(a - b)^2$, we have:
$(\sin x - 1)^2$
# Answer:
$(\sin x - 1)^2$

### Problem 42: $\boldsymbol{\sin x - \cos^2 x - 1}$
# Explanation:
## Step 1: Rearrange the terms
Rearrange the expression as $-\cos^2 x+\sin x - 1= - (\cos^2 x + 1-\sin x)$
## Step 2: Use the Pythagorean identity
Recall $\cos^2 x=1 - \sin^2 x$. Substitute $\cos^2 x$ in the expression:
$-((1 - \sin^2 x)+1-\sin x)=- (2 - \sin x-\sin^2 x)$
## Step 3: Factor the quadratic in terms of $\sin x$
Factor $2 - \sin x-\sin^2 x=-( \sin^2 x+\sin x - 2)$
Factor $\sin^2 x+\sin x - 2$: we need two numbers that multiply to $- 2$ and add to $1$. The numbers are $2$ and $- 1$.
$\sin^2 x+\sin x - 2=\sin^2 x + 2\sin x-\sin x - 2=\sin x(\sin x + 2)-1(\sin x + 2)=(\sin x + 2)(\sin x - 1)$
So $-( \sin^2 x+\sin x - 2)=-(\sin x + 2)(\sin x - 1)=( - \sin x - 2)(\sin x - 1)=(1 - \sin x)(\sin x + 2)$
Wait, let's go back to the original expression:
$\sin x - \cos^2 x - 1=\sin x-( \cos^2 x + 1)=\sin x-(1 - \sin^2 x + 1)=\sin x-(2 - \sin^2 x)=\sin^2 x+\sin x - 2$
Factor $\sin^2 x+\sin x - 2$ as $(\sin x + 2)(\sin x - 1)$
# Answer:
$(\sin x + 2)(\sin x - 1)$

### Problem 43: $\boldsymbol{\cos x - 2\sin^2 x + 1}$
# Explanation:
## Step 1: Rearrange the terms
Rearrange the expression as $- 2\sin^2 x+\cos x + 1$
## Step 2: Use the Pythagorean identity
Recall $\sin^2 x=1 - \cos^2 x$. Substitute $\sin^2 x$:
$-2(1 - \cos^2 x)+\cos x + 1=-2 + 2\cos^2 x+\cos x + 1=2\cos^2 x+\cos x - 1$
## Step 3: Factor the quadratic in terms of $\cos x$
We need to factor $2\cos^2 x+\cos x - 1$. Let $y = \cos x$, then the quadratic is $2y^2+y - 1$.
We need two numbers that multiply to $2	imes(- 1)=-2$ and add to $1$. The numbers are $2$ and $- 1$.
$2y^2+y - 1=2y^2+2y - y - 1=2y(y + 1)-1(y + 1)=(2y - 1)(y + 1)$
Substitute back $y=\cos x$:
$(2\cos x - 1)(\cos x + 1)$
# Answer:
$(2\cos x - 1)(\cos x + 1)$

### Problem 44: $\boldsymbol{\sin^2 x+\frac{2}{\csc x}+1}$
# Explanation:
## Step 1: Use the reciprocal identity
Recall that $\csc x=\frac{1}{\sin x}$, so $\frac{1}{\csc x}=\sin x$. Then $\frac{2}{\csc x}=2\sin x$
## Step 2: Substitute and factor
The expression becomes $\sin^2 x+2\sin x + 1$
This is a perfect square trinomial of the form $a^2 + 2ab + b^2$, where $a=\sin x$ and $b = 1$.
Using the formula $a^2 + 2ab + b^2=(a + b)^2$, we get:
$(\sin x + 1)^2$
# Answer:
$(\sin x + 1)^2$

### Problem 45: $\boldsymbol{4	an^2 x-\frac{4}{\cot x}+\sin x\csc x}$
# Explanation:
## Step 1: Use reciprocal identities
Recall that $\cot x=\frac{\cos x}{\sin x}$, so $\frac{1}{\cot x}=	an x$ and $\csc x=\frac{1}{\sin x}$, so $\sin x\csc x = 1$
Substitute these into the expression:
$4	an^2 x-4	an x + 1$
## Step 2: Recognize the perfect square trinomial
The expression $4	an^2 x-4	an x + 1$ is a perfect square trinomial of the form $a^2 - 2ab + b^2$, where $a = 2	an x$ and $b = 1$.
Using the formula $a^2 - 2ab + b^2=(a - b)^2$, we have:
$(2	an x - 1)^2$
# Answer:
$(2	an x - 1)^2$

### Problem 46: $\boldsymbol{\sec^2 x - \sec x + 	an^2 x}$
# Explanation:
## Step 1: Use the Pythagorean identity
Recall the Pythagorean identity $\sec^2 x=1+	an^2 x$, so $	an^2 x=\sec^2 x - 1$
Substitute $	an^2 x$ in the expression:
$\sec^2 x-\sec x+\sec^2 x - 1=2\sec^2 x-\sec x - 1$
## Step 2: Factor the quadratic in terms of $\sec x$
Let $y = \sec x$, then the quadratic is $2y^2-y - 1$
Factor $2y^2-y - 1$: we need two numbers that multiply to $2	imes(-1)=-2$ and add to $- 1$. The numbers are $-2$ and $1$.
$2y^2-y - 1=2y^2-2y + y - 1=2y(y - 1)+1(y - 1)=(2y + 1)(y - 1)$
Substitute back $y = \sec x$:
$(2\sec x + 1)(\sec x - 1)$
# Answer:
$(2\sec x + 1)(\sec x - 1)$

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QUESTION IMAGE

Question

Explanation:

Problem 39: $\boldsymbol{\cos^2 x + 2\cos x + 1}$

Step 1: Recognize the perfect square trinomial

Step 2: Apply the perfect square formula

Step 1: Rearrange the terms

Step 2: Apply the perfect square formula

Step 1: Use the Pythagorean identity

Step 2: Recognize the perfect square trinomial

Answer:

Problem 40: $\boldsymbol{1 - 2\sin x + \sin^2 x}$