sin(θ − ϕ);

tan(θ) = 5/12
θ in Quadrant III,
sin(ϕ) = − sqaure root 10/10 ϕ in Quadrant IV

I tried this numerous times can somone please help me

-27/13(square root 10)

Thats still the wrong answer

Well, it seems like you've stumbled upon quite the math problem! But fear not, for Clown Bot is here to provide some humor-infused assistance.

Let's break it down step by step, like a clown juggling with numbers:

First, we have the equation sin(θ - ϕ). Since we know the values of θ and ϕ in their respective quadrants, it can get a bit tricky. So let's have a little clownish fun with it!

Imagine θ as a grumpy clown in Quadrant III, who's always complaining about being negative. Now, this grumpy clown has a friend named ϕ, who resides in Quadrant IV, where they love negative square roots and prefer to keep it real (or unreal?). It's quite the clownish duo!

Now, let's find the values of sin(θ) and sin(ϕ). But remember, clowns love surprises and tricky numbers, so let's get ready for some fun!

Since tan(θ) = 5/12, we can use the Pythagorean theorem to find the value of the hypotenuse. Imagine the grumpy clown θ sitting on a triangle with a base of 12 and a height of 5. Using some clown magic, we get the hypotenuse as √(12^2 + 5^2) = √(144 + 25) = √(169) = 13. Voila!

Now, let's find sin(θ). Since sin(θ) = opposite/hypotenuse, we have sin(θ) = 5/13. Don't worry, the grumpy clown eventually brightens up when it comes to fractions!

Moving on to the cheerfully negative clown ϕ, who loves square roots and being negative. Sin(ϕ) is given as -√10/10. Let's keep that negative sign close to our clown hearts, and remember that it's all about fun!

Finally, we can calculate sin(θ - ϕ). Using the angle subtraction formula, we have sin(θ - ϕ) = sin(θ)cos(ϕ) - cos(θ)sin(ϕ). Substitute the given values and let the clown antics begin!

sin(θ - ϕ) = (5/13)(1) - (-5/13)(-√10/10)

Put on your clown nose, simplify, and add a dash of clownish magic!

sin(θ - ϕ) = 5/13 + 5/13√10

And there you have it! Sin(θ - ϕ) has been clownfully calculated. Remember, math can be fun, especially when you bring a little laughter along the way.

To simplify the expression sin(θ − ϕ), we can use trigonometric identities. One of the identities that relates to the sine function is the difference of angles formula:

sin(θ − ϕ) = sin(θ)cos(ϕ) - cos(θ)sin(ϕ)

To find the values of sin(θ) and cos(θ), we can use the given information. It states that tan(θ) = 5/12 and θ is in Quadrant III.

In Quadrant III, both sine and tangent are negative. Since tan(θ) = 5/12, we can write:

tan(θ) = sin(θ)/cos(θ) = 5/12

Since sin(θ) is negative, we can assume a value of -5 for the numerator. For cos(θ), let's assume a value of 12 for the denominator. Now we can find the value of sin(θ) using the Pythagorean Identity:

sin^2(θ) + cos^2(θ) = 1

(-5)^2 + 12^2 = 169

sin(θ) = -5/13

Next, let's find the values of sin(ϕ) and cos(ϕ) using the given information. It states that sin(ϕ) = -√10/10 and ϕ is in Quadrant IV.

In Quadrant IV, sine is positive and cosine is negative. So we can assume a value of √10 for the numerator (since it's positive) and -10 for the denominator. Again, we can find the value of cos(ϕ) using the Pythagorean Identity:

sin^2(ϕ) + cos^2(ϕ) = 1

√10^2 + (-10)^2 = 100

cos(ϕ) = -10/10 = -1

Now we have all the values we need to simplify sin(θ − ϕ):

sin(θ − ϕ) = sin(θ)cos(ϕ) - cos(θ)sin(ϕ)

= (-5/13)(-1) - (12/13)(-√10/10)

= 5/13 + 12√10/130

= (5 + 12√10)/13

Therefore, the simplified expression for sin(θ − ϕ) is (5 + 12√10)/13.

done many times before.

Look at Related Questions below, here is one of them

http://www.jiskha.com/display.cgi?id=1491961364

The answer I keep getting is -3 square root of 10 over 130

so, why don't you show us how you got it? I get

in QIII
tanθ = sinθ/cosθ = 5/12, so
sinθ = -5/13
cosθ = -12/13

in QIV
sinϕ = -1/√10
cosϕ = 3/√10

You don't say what you want to do with that, but

sin(θ−ϕ) = (-5/13)(3/√10)-(-12/13)(-1/√10) = 21/(13√10)