# HOW CAN I SOLVE THIS? sin(350°+θ)cos(110°+θ)-cos(350°+θ)sin(110°+θ)

## Use the sin (A - B) identity:

Sin(A - B) = sinAcosB - cosAsinB

here you A = 350+θ , and B = 110+θ

sin(350°+θ)cos(110°+θ)-cos(350°+θ)sin(110°+θ)
= sin(350+θ - (110+θ) )
= sin(240°)
= - sin 60°
= - √3 / 2

## Well, it looks like you've stumbled upon a trigonometric expression. Don't worry, solving it is easier than catching a clown's nose!

To simplify this expression, let's use some handy trigonometric identities. First, we have the sum of angles identity for sine: sin(A+B) = sin(A)cos(B) + cos(A)sin(B).

Let's apply that identity twice:

sin(350°+θ)cos(110°+θ) - cos(350°+θ)sin(110°+θ)
= (sin(350°)cos(θ)cos(110°)sin(θ)) - (cos(350°)sin(θ)sin(110°)cos(θ))

Now, since sin(350°) = -sin(10°) and cos(350°) = cos(10°), we can rewrite the expression further:

= (-sin(10°)cos(θ)cos(110°)sin(θ)) - (cos(10°)sin(θ)sin(110°)cos(θ))

No clowning around here! Now, let's simplify it even more:

= -cos(θ)sin(θ)(sin(10°)cos(110°) + cos(10°)sin(110°))

But wait, there's more! We can simplify further by using the sum of angles identity for cosine: cos(A+B) = cos(A)cos(B) - sin(A)sin(B).

Now, let's apply that identity:

= -cos(θ)sin(θ)(cos(10° + 110°))
= -cos(θ)sin(θ)(cos(120°))
= -cos(θ)sin(θ)(-0.5)

Ta-da! We have simplified the expression to -0.5cos(θ)sin(θ).

So, to solve this expression, you can plug in the value of θ that you have, and then calculate -0.5cos(θ)sin(θ). Just remember to wear a giant clown nose while doing it for maximum accuracy!

## To simplify the expression sin(350°+θ)cos(110°+θ)-cos(350°+θ)sin(110°+θ), you can use the following trigonometric identity:

sin(A + B) = sin(A)cos(B) + cos(A)sin(B)

Applying this identity, you can rewrite the expression as:

sin(350°+θ)cos(110°+θ) - cos(350°+θ)sin(110°+θ)
= sin(350°+θ + 110°+θ)
= sin(460°+2θ)

Since the sine function is periodic with a period of 360°, you can subtract 360° from the angle to bring it within one period:

sin(460°+2θ) = sin((460°+2θ) - 360°)
= sin(100°+2θ)

So, the simplified expression is sin(100°+2θ).

## To solve the given expression, you can use the trigonometric identities to simplify it. Let's break it down step by step:

Step 1: Expand the terms using the sum and difference formulas for sine and cosine:
sin(350° + θ)cos(110° + θ) - cos(350° + θ)sin(110° + θ)

Using the sum formula for sine:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B)

Using the difference formula for cosine:
cos(A - B) = cos(A)cos(B) + sin(A)sin(B)

We can rewrite the expression as:

[sin(350°)cos(110°) + cos(350°)sin(110°)]·cos(θ) - [cos(350°)cos(110°) - sin(350°)sin(110°)]·sin(θ)

Step 2: Simplify the trigonometric terms:
sin(350°) = sin(360° - 10°) = -sin(10°)
cos(350°) = cos(360° - 10°) = cos(10°)
sin(110°) = sin(90° + 20°) = sin(90°)cos(20°) + cos(90°)sin(20°) = cos(20°)
cos(110°) = cos(90° + 20°) = cos(90°)cos(20°) - sin(90°)sin(20°) = -sin(20°)

Substituting these values into the expression:

[-sin(10°)cos(20°) + cos(10°)sin(20°)]·cos(θ) - [cos(10°)cos(20°) + sin(10°)sin(20°)]·sin(θ)

Step 3: Simplify further:
Using the commutative property of multiplication, we can rearrange the terms:
[-cos(20°)sin(10°) + sin(20°)cos(10°)]·cos(θ) - [cos(10°)cos(20°) + sin(10°)sin(20°)]·sin(θ)

Notice that the terms in the brackets represent sin(α + β) and cos(α + β) respectively, where α = 10° and β = 20°. Since sin(α + β) = sin(β + α) and cos(α + β) = cos(β + α), we can rearrange the terms:

sin(20° + 10°)·cos(θ) - cos(20° + 10°)·sin(θ)

Step 4: Use the angle addition formula:
sin(30°)·cos(θ) - cos(30°)·sin(θ)

We know that sin(30°) = 1/2 and cos(30°) = √3/2:

(1/2)·cos(θ) - (√3/2)·sin(θ)

Finally, we have simplified the expression. The final result is:

(1/2)cos(θ) - (√3/2)sin(θ)

So, the solution to the given expression is (1/2)cos(θ) - (√3/2)sin(θ).