Math / Algebra

Question[-/2 Points] DETAILS MY NOTES MCKTRIG7 5.1.
Multiply the numerator and denominator of the fraction by the conjugat (a) $\frac{1-\sqrt{2}}{1+\sqrt{2}}$ $\square$ (b) $\frac{1-\sin x}{1+\sin x}$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to multiply the top and bottom of each fraction by its conjugate, which is just like its twin but with a different sign in the middle! Watch out! Don't forget to distribute carefully when multiplying by the conjugate, and remember those special product formulas – they're lifesavers!

STEP 2

1. Solve (a)
2. Solve (b)

STEP 3

Alright, let's multiply the top and bottom of $\frac{1-\sqrt{2}}{1+\sqrt{2}}$ by the conjugate of the bottom, which is $1-\sqrt{2}$ .
This is a slick trick to get rid of those pesky square roots in the denominator!

STEP 4

The numerator becomes $(1-\sqrt{2})\cdot(1-\sqrt{2})$ .
Let's FOIL this out: $1 - \sqrt{2} - \sqrt{2} + (\sqrt{2})^2 = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}$ .
So our new numerator is $3 - 2\sqrt{2}$ .

STEP 5

Now for the denominator: $(1+\sqrt{2})\cdot(1-\sqrt{2})$ .
This is a difference of squares, so it simplifies to $1^2 - (\sqrt{2})^2 = 1 - 2 = -1$ .
Boom! No more square roots on the bottom!

STEP 6

Our fraction is now $\frac{3 - 2\sqrt{2}}{-1}$ .
We can simplify this to $-3 + 2\sqrt{2}$ .
Look at that, nice and clean!

STEP 7

Now, let's tackle $\frac{1-\sin x}{1+\sin x}$ .
We'll multiply the top and bottom by the conjugate of the denominator, which is $1-\sin x$ .

STEP 8

The numerator becomes $(1-\sin x)\cdot(1-\sin x) = (1-\sin x)^2$ .
We can expand this to $1 - 2\sin x + \sin^2 x$ .

STEP 9

The denominator is $(1+\sin x)\cdot(1-\sin x)$ .
This is another difference of squares, which simplifies to $1^2 - (\sin x)^2 = 1 - \sin^2 x$ .

STEP 10

Remember the Pythagorean identity: $\sin^2 x + \cos^2 x = 1$ .
This means $1 - \sin^2 x = \cos^2 x$ .
So our denominator becomes $\cos^2 x$ .

STEP 11

Our fraction now looks like this: $\frac{1 - 2\sin x + \sin^2 x}{\cos^2 x}$ .
We can leave it like this, or we can split it into separate fractions: $\frac{1}{\cos^2 x} - \frac{2\sin x}{\cos^2 x} + \frac{\sin^2 x}{\cos^2 x}$ .

STEP 12

(a) $-3 + 2\sqrt{2}$ (b) $\frac{1 - 2\sin x + \sin^2 x}{\cos^2 x}$

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Question[-/2 Points] DETAILS MY NOTES MCKTRIG7 5.1.
Multiply the numerator and denominator of the fraction by the conjugat (a) $\frac{1-\sqrt{2}}{1+\sqrt{2}}$ $\square$ (b) $\frac{1-\sin x}{1+\sin x}$ $\square$

Studdy Solution

STEP 1

STEP 2

1. Solve (a)
2. Solve (b)

STEP 3

Alright, let's multiply the top and bottom of $\frac{1-\sqrt{2}}{1+\sqrt{2}}$ by the conjugate of the bottom, which is $1-\sqrt{2}$ .
This is a slick trick to get rid of those pesky square roots in the denominator!

STEP 4

The numerator becomes $(1-\sqrt{2})\cdot(1-\sqrt{2})$ .
Let's FOIL this out: $1 - \sqrt{2} - \sqrt{2} + (\sqrt{2})^2 = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}$ .
So our new numerator is $3 - 2\sqrt{2}$ .

STEP 5

Now for the denominator: $(1+\sqrt{2})\cdot(1-\sqrt{2})$ .
This is a difference of squares, so it simplifies to $1^2 - (\sqrt{2})^2 = 1 - 2 = -1$ .
Boom! No more square roots on the bottom!

STEP 6

Our fraction is now $\frac{3 - 2\sqrt{2}}{-1}$ .
We can simplify this to $-3 + 2\sqrt{2}$ .
Look at that, nice and clean!

STEP 7

Now, let's tackle $\frac{1-\sin x}{1+\sin x}$ .
We'll multiply the top and bottom by the conjugate of the denominator, which is $1-\sin x$ .

STEP 8

The numerator becomes $(1-\sin x)\cdot(1-\sin x) = (1-\sin x)^2$ .
We can expand this to $1 - 2\sin x + \sin^2 x$ .

STEP 9

The denominator is $(1+\sin x)\cdot(1-\sin x)$ .
This is another difference of squares, which simplifies to $1^2 - (\sin x)^2 = 1 - \sin^2 x$ .

STEP 10

Remember the Pythagorean identity: $\sin^2 x + \cos^2 x = 1$ .
This means $1 - \sin^2 x = \cos^2 x$ .
So our denominator becomes $\cos^2 x$ .

STEP 11

Our fraction now looks like this: $\frac{1 - 2\sin x + \sin^2 x}{\cos^2 x}$ .
We can leave it like this, or we can split it into separate fractions: $\frac{1}{\cos^2 x} - \frac{2\sin x}{\cos^2 x} + \frac{\sin^2 x}{\cos^2 x}$ .

STEP 12

(a) $-3 + 2\sqrt{2}$ (b) $\frac{1 - 2\sin x + \sin^2 x}{\cos^2 x}$

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Question[-/2 Points] DETAILS MY NOTES MCKTRIG7 5.1.Multiply the numerator and denominator of the fraction by the conjugat (a) 1−21+2\frac{1-\sqrt{2}}{1+\sqrt{2}}1+2​1−2​​ □\square□ (b) 1−sin⁡x1+sin⁡x\frac{1-\sin x}{1+\sin x}1+sinx1−sinx​ □\square□

Studdy Solution

STEP 1

STEP 2

1. Solve (a)2. Solve (b)

STEP 3

Alright, let's **multiply** the top and bottom of 1−21+2\frac{1-\sqrt{2}}{1+\sqrt{2}}1+2​1−2​​ by the conjugate of the bottom, which is 1−21-\sqrt{2}1−2​.This is a slick trick to get rid of those pesky square roots in the denominator!

STEP 4

STEP 5

Now for the denominator: (1+2)⋅(1−2)(1+\sqrt{2})\cdot(1-\sqrt{2})(1+2​)⋅(1−2​).This is a **difference of squares**, so it simplifies to 12−(2)2=1−2=−11^2 - (\sqrt{2})^2 = 1 - 2 = -112−(2​)2=1−2=−1.Boom! No more square roots on the bottom!

STEP 6

Our fraction is now 3−22−1\frac{3 - 2\sqrt{2}}{-1}−13−22​​.We can **simplify** this to −3+22-3 + 2\sqrt{2}−3+22​.Look at that, nice and clean!

STEP 7

Now, let's tackle 1−sin⁡x1+sin⁡x\frac{1-\sin x}{1+\sin x}1+sinx1−sinx​.We'll **multiply** the top and bottom by the conjugate of the denominator, which is 1−sin⁡x1-\sin x1−sinx.

STEP 8

The numerator becomes (1−sin⁡x)⋅(1−sin⁡x)=(1−sin⁡x)2(1-\sin x)\cdot(1-\sin x) = (1-\sin x)^2(1−sinx)⋅(1−sinx)=(1−sinx)2.We can **expand** this to 1−2sin⁡x+sin⁡2x1 - 2\sin x + \sin^2 x1−2sinx+sin2x.

STEP 9

The denominator is (1+sin⁡x)⋅(1−sin⁡x)(1+\sin x)\cdot(1-\sin x)(1+sinx)⋅(1−sinx).This is another **difference of squares**, which simplifies to 12−(sin⁡x)2=1−sin⁡2x1^2 - (\sin x)^2 = 1 - \sin^2 x12−(sinx)2=1−sin2x.

STEP 10

Remember the **Pythagorean identity**: sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1sin2x+cos2x=1.This means 1−sin⁡2x=cos⁡2x1 - \sin^2 x = \cos^2 x1−sin2x=cos2x.So our denominator becomes cos⁡2x\cos^2 xcos2x.

STEP 11

STEP 12

(a) −3+22-3 + 2\sqrt{2}−3+22​ (b) 1−2sin⁡x+sin⁡2xcos⁡2x\frac{1 - 2\sin x + \sin^2 x}{\cos^2 x}cos2x1−2sinx+sin2x​

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Question[-/2 Points] DETAILS MY NOTES MCKTRIG7 5.1.
Multiply the numerator and denominator of the fraction by the conjugat (a) $\frac{1-\sqrt{2}}{1+\sqrt{2}}$ $\square$ (b) $\frac{1-\sin x}{1+\sin x}$ $\square$

1. Solve (a)
2. Solve (b)

Alright, let's multiply the top and bottom of $\frac{1-\sqrt{2}}{1+\sqrt{2}}$ by the conjugate of the bottom, which is $1-\sqrt{2}$ .
This is a slick trick to get rid of those pesky square roots in the denominator!

Now for the denominator: $(1+\sqrt{2})\cdot(1-\sqrt{2})$ .
This is a difference of squares, so it simplifies to $1^2 - (\sqrt{2})^2 = 1 - 2 = -1$ .
Boom! No more square roots on the bottom!

Our fraction is now $\frac{3 - 2\sqrt{2}}{-1}$ .
We can simplify this to $-3 + 2\sqrt{2}$ .
Look at that, nice and clean!

Now, let's tackle $\frac{1-\sin x}{1+\sin x}$ .
We'll multiply the top and bottom by the conjugate of the denominator, which is $1-\sin x$ .

The numerator becomes $(1-\sin x)\cdot(1-\sin x) = (1-\sin x)^2$ .
We can expand this to $1 - 2\sin x + \sin^2 x$ .

The denominator is $(1+\sin x)\cdot(1-\sin x)$ .
This is another difference of squares, which simplifies to $1^2 - (\sin x)^2 = 1 - \sin^2 x$ .

Remember the Pythagorean identity: $\sin^2 x + \cos^2 x = 1$ .
This means $1 - \sin^2 x = \cos^2 x$ .
So our denominator becomes $\cos^2 x$ .

(a) $-3 + 2\sqrt{2}$ (b) $\frac{1 - 2\sin x + \sin^2 x}{\cos^2 x}$