Question

sin θ cos 3θ + cos θ sin 3θ = 0

Use an Addition or Subtraction Formula to simplify the equation.

Answer 1

Using the Sine addition formula I get Sin4θ, but when I enter that as an answer my homework site marks it as incorrect.

Answer 2

No, it's as it's shown.

Answer 3

There are no powers in this problem.

Answer 4

Consider the left hand side.
sin θ cos 3θ + cos θ sin 3θ
= sin (θ+3θ)
= sin 4θ
Do you need to solve it? If not, then I'm pretty sure it is correct. Perhaps you should type s in Sin without capitalising it

Answer 5

It just asks for me to simplify the equation using the Addition or substraction formula.

Answer 6

Is there a way to derive Sin4θ into something else?

Answer 7

Yes. Use double angle formula
sin4θ = 2sin2θcos2θ = 2(2sinθcosθ)cos2θ = 4sinθcosθcos2θ
I'm pretty sure that would be something undesired

Answer 8

sin(alpha+beta) = sinalphacosbeta+sinbetacosalpha
(1)
sin(alpha-beta) = sinalphacosbeta- sinbetacosalpha
(2)
cos(alpha+beta) = cosalphacosbeta-sinalphasinbeta
(3)
cos(alpha-beta) = cosalphacosbeta+sinalphasinbeta

ok so we have

sin θ cos 3θ + cos θ sin 3θ = 0

sinθ(cos(θ)cos(2θ)-sin(θ)sin(2θ)) + cos(θ)(sin(θ)cos(2θ)+sin(2θ)cos(θ)) = 0

Answer 9

sin(3θ) = sin(2θ + θ)

Answer 10

sinθ(cos(θ)cos(2θ)-sin(θ)sin(2θ)) + cos(θ)(sin(θ)cos(2θ)+sin(2θ)cos(θ)) = 0
sinθcos(θ)cos(2θ) - sin^(2)(θ)sin(2θ) + cos(θ)sin(θ)cos(2θ) + sin(2θ)cos^(2)(θ) = 0
- sin^(2)(θ)sin(2θ) + sin(2θ)cos^(2)(θ) = 0

Answer 11

- sin^(2)(θ)(sin(θ)cos(θ)+sin(θ)cos(θ))+ (sin(θ)cos(θ)+sin(θ)cos(θ))cos^(2)(θ) = 0
-sin^(2)θsin(θ)cos(θ) - sin^(2)(θ)cos(θ) + sin(θ)cos(θ)cos^(2)(θ) + cos^(2)(θ)sin(θ)cos(θ) = 0

Answer 12

-sin^(2)θsin(θ)cos(θ) - sin^(2)(θ)cos(θ) + sin(θ)cos(θ)cos^(2)(θ) + cos^(2)(θ)sin(θ)cos(θ) = 0
sin(θ)cos(θ)(-sin^(2)(θ) - sin(θ) + cos^(2)(θ) + cos^(2)(θ)) = 0

Answer 13

-sin^(2)(θ) - sin(θ) + cos^(2)(θ) + cos^(2)(θ) = 0
simplify using identites
-1+cos^(2)(θ) - sin(θ) +cos^(2)(θ) + cos^(2)(θ) = 0
3cos^(2)(θ) - sin(θ) = 0

Answer 14

actually probably best to convert things interms of sin(θ) so I can factor

Answer 15

-sin^(2)(θ) - sin(θ) + 1 - sin^(2)(θ) + 1 - sin^(2)(θ) = 0
-3sin^(2)(θ) - sin(θ) + 2 = 0

Answer 16

You don't need to factor at all. You only need to use the addition or subtraction formula.

Answer 17

@Australopithecus Actually, you can directly apply the formula
\[sin(\alpha+\beta)=sin\alpha cos\beta +sin\beta cos\alpha\]
Put α = θ and β =3θ , you can see the the question is actually the same as the right hand side. So, you can get sin (θ+3θ) = sin4θ immediately.

Answer 18

I know that the answer Sin4Theta is correct, but for some reason Web Assign (my homework site) marks it incorrectly.

Answer 19

Email your instructor

Answer 20

https://www.wolframalpha.com/input/?i=sin+%CE%B8+cos+3%CE%B8+%2B+cos+%CE%B8+sin+3%CE%B8+%3D+0+

Answer 21

That would be a tech problem, not a math problem anymore!

Answer 22

(n /π)/4

I've also tried that, but it's marked as incorrect as well.

Answer 23

I hope it is a tech problem, I've already emailed my instructor. Thanks anyways.

Answer 24

the answer is pin/4 where n is an element of Integers
you are correct wolfram alpha confirms email instructor