Math / Trigonometry

QuestionWrite the expression as a single trigonometric function. $\sin 9 x \cos 8 x+\cos 9 x \sin 8 x$ $\square$ Submit Answer 12. [-/1 Points] DETAILS MY NOTES MCKTRIC
Write the expression as a single trigonometric function. $\cos 3 x \cos x-\sin 3 x \sin x$ $\square$ Need Help? Watch It Submit Answer 13. [-/1 Points] DETAILS MY NOTES MCKTRIG
Write the expression as a single trigonometric function. $\sin 3 x \cos x-\cos 3 x \sin x$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to rewrite some expressions with multiple trigonometric functions into a single trigonometric function. Watch out! Remember those sum and difference formulas!
It's easy to mix them up, so take your time and double-check!

STEP 2

1. Solve the first expression
2. Solve the second expression
3. Solve the third expression

STEP 3

The sum formula for sine is $\sin(a + b) = \sin a \cos b + \cos a \sin b$ .
This looks just like our expression!

STEP 4

Let's match the parts of our expression $\sin 9x \cos 8x + \cos 9x \sin 8x$ to the formula.
We see that $a = 9x$ and $b = 8x$ .
So, we can rewrite our expression as $\sin(9x + 8x)$ .

STEP 5

Adding those together gives us $\sin(17x)$ !
Boom!

STEP 6

This time we have $\cos 3x \cos x - \sin 3x \sin x$ .
The relevant formula is $\cos(a + b) = \cos a \cos b - \sin a \sin b$ .

STEP 7

Matching the parts, we have $a = 3x$ and $b = x$ .
So, our expression becomes $\cos(3x + x)$ .

STEP 8

Adding gives us $\cos(4x)$ !
Fantastic!

STEP 9

Our expression is $\sin 3x \cos x - \cos 3x \sin x$ .
This time, it's the difference formula for sine: $\sin(a - b) = \sin a \cos b - \cos a \sin b$ .

STEP 10

We have $a = 3x$ and $b = x$ .
Plugging those in, we get $\sin(3x - x)$ .

STEP 11

Subtracting gives us $\sin(2x)$ !
Wonderful!

STEP 12

The solutions are: $\sin(17x)$ $\cos(4x)$ $\sin(2x)$

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Studdy Solution

STEP 1

What is this asking? We need to rewrite some expressions with multiple trigonometric functions into a single trigonometric function. Watch out! Remember those sum and difference formulas!
It's easy to mix them up, so take your time and double-check!

STEP 2

1. Solve the first expression
2. Solve the second expression
3. Solve the third expression

STEP 3

The sum formula for sine is $\sin(a + b) = \sin a \cos b + \cos a \sin b$ .
This looks just like our expression!

STEP 4

Let's match the parts of our expression $\sin 9x \cos 8x + \cos 9x \sin 8x$ to the formula.
We see that $a = 9x$ and $b = 8x$ .
So, we can rewrite our expression as $\sin(9x + 8x)$ .

STEP 5

Adding those together gives us $\sin(17x)$ !
Boom!

STEP 6

This time we have $\cos 3x \cos x - \sin 3x \sin x$ .
The relevant formula is $\cos(a + b) = \cos a \cos b - \sin a \sin b$ .

STEP 7

Matching the parts, we have $a = 3x$ and $b = x$ .
So, our expression becomes $\cos(3x + x)$ .

STEP 8

Adding gives us $\cos(4x)$ !
Fantastic!

STEP 9

Our expression is $\sin 3x \cos x - \cos 3x \sin x$ .
This time, it's the difference formula for sine: $\sin(a - b) = \sin a \cos b - \cos a \sin b$ .

STEP 10

We have $a = 3x$ and $b = x$ .
Plugging those in, we get $\sin(3x - x)$ .

STEP 11

Subtracting gives us $\sin(2x)$ !
Wonderful!

STEP 12

The solutions are: $\sin(17x)$ $\cos(4x)$ $\sin(2x)$

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Studdy Solution

STEP 1

What is this asking? We need to rewrite some expressions with multiple trigonometric functions into a single trigonometric function. Watch out! Remember those *sum and difference formulas*!It's easy to mix them up, so take your time and double-check!

STEP 2

1. Solve the first expression2. Solve the second expression3. Solve the third expression

STEP 3

The sum formula for sine is sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b\sin(a + b) = \sin a \cos b + \cos a \sin bsin(a+b)=sinacosb+cosasinb.This looks *just* like our expression!

STEP 4

Let's **match** the parts of our expression sin⁡9xcos⁡8x+cos⁡9xsin⁡8x\sin 9x \cos 8x + \cos 9x \sin 8xsin9xcos8x+cos9xsin8x to the formula.We see that a=9xa = 9xa=9x and b=8xb = 8xb=8x.So, we can rewrite our expression as sin⁡(9x+8x)\sin(9x + 8x)sin(9x+8x).

STEP 5

Adding those together gives us sin⁡(17x)\sin(17x)sin(17x)!Boom!

STEP 6

This time we have cos⁡3xcos⁡x−sin⁡3xsin⁡x\cos 3x \cos x - \sin 3x \sin xcos3xcosx−sin3xsinx.The relevant formula is cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b\cos(a + b) = \cos a \cos b - \sin a \sin bcos(a+b)=cosacosb−sinasinb.

STEP 7

Matching the parts, we have a=3xa = 3xa=3x and b=xb = xb=x.So, our expression becomes cos⁡(3x+x)\cos(3x + x)cos(3x+x).

STEP 8

Adding gives us cos⁡(4x)\cos(4x)cos(4x)!Fantastic!

STEP 9

Our expression is sin⁡3xcos⁡x−cos⁡3xsin⁡x\sin 3x \cos x - \cos 3x \sin xsin3xcosx−cos3xsinx.This time, it's the *difference* formula for sine: sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b\sin(a - b) = \sin a \cos b - \cos a \sin bsin(a−b)=sinacosb−cosasinb.

STEP 10

We have a=3xa = 3xa=3x and b=xb = xb=x.Plugging those in, we get sin⁡(3x−x)\sin(3x - x)sin(3x−x).

STEP 11

Subtracting gives us sin⁡(2x)\sin(2x)sin(2x)!Wonderful!

STEP 12

The solutions are: sin⁡(17x)\sin(17x)sin(17x) cos⁡(4x)\cos(4x)cos(4x)sin⁡(2x)\sin(2x)sin(2x)

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We need to rewrite some expressions with multiple trigonometric functions into a single trigonometric function. Watch out! Remember those sum and difference formulas!
It's easy to mix them up, so take your time and double-check!

1. Solve the first expression
2. Solve the second expression
3. Solve the third expression

The sum formula for sine is $\sin(a + b) = \sin a \cos b + \cos a \sin b$ .
This looks just like our expression!

Let's match the parts of our expression $\sin 9x \cos 8x + \cos 9x \sin 8x$ to the formula.
We see that $a = 9x$ and $b = 8x$ .
So, we can rewrite our expression as $\sin(9x + 8x)$ .

Adding those together gives us $\sin(17x)$ !
Boom!

This time we have $\cos 3x \cos x - \sin 3x \sin x$ .
The relevant formula is $\cos(a + b) = \cos a \cos b - \sin a \sin b$ .

Matching the parts, we have $a = 3x$ and $b = x$ .
So, our expression becomes $\cos(3x + x)$ .

Adding gives us $\cos(4x)$ !
Fantastic!

Our expression is $\sin 3x \cos x - \cos 3x \sin x$ .
This time, it's the difference formula for sine: $\sin(a - b) = \sin a \cos b - \cos a \sin b$ .

We have $a = 3x$ and $b = x$ .
Plugging those in, we get $\sin(3x - x)$ .

Subtracting gives us $\sin(2x)$ !
Wonderful!

The solutions are: $\sin(17x)$ $\cos(4x)$ $\sin(2x)$