Trigonometric Equation Homework Help: Solving 3sinx + 4cosx = 1

[Bogrune]

Homework Statement

I took a test in my Precalculus class about 3 weeks ago, and the tests were handed back this week. I saw that I didn't do so well because I was stumped on this equation:

3sinx + 4cosx = 1

Homework Equations

We were to use the following formulas to solve this equation:

sin(A+B)= sinAcosB + cosAsinB
cos(A+B)= cosAcosB - sinAsinB
sin² + cos² = 1

The Attempt at a Solution

I tried to apply one of the formulas, but it just didn't make sense:

3sin²x + 4cos²x = 1

I then tried to use the addition identities, but it still didn't make sense since both expressions(3sinx + 4cosx) only have one term. Anyone know how to solve this?

 

[Mark44]

Homework Statement

I took a test in my Precalculus class about 3 weeks ago, and the tests were handed back this week. I saw that I didn't do so well because I was stumped on this equation:

3sinx + 4cosx = 1

Homework Equations

We were to use the following formulas to solve this equation:

sin(A+B)= sinAcosB + cosAsinB
cos(A+B)= cosAcosB - sinAsinB
sin² + cos² = 1

The Attempt at a Solution

I tried to apply one of the formulas, but it just didn't make sense:

3sin²x + 4cos²x = 1

You can't do this. You are apparently squaring both sides, but to do that, you have to square the quantity 3sin(x) + 4cos(x).

If you had an equation a + b = 1, and you squared both sides, you would get
(a + b)² = 1²,
or a² + 2ab + b² = 1

You would NOT get
a² + b² = 1²

I then tried to use the addition identities, but it still didn't make sense since both expressions(3sinx + 4cosx) only have one term. Anyone know how to solve this?

Divide both sides of the equation by 5. There's a reason I picked this number, which I'll get to later.

This gives you
3/5 sin(x) + 4/5 cos(x) = 1/5

Since 3/5 and 4/5 are between -1 and 1, we can consider them to be the cosine and sine, respectively of some angle A, so let cos(A) = 3/5 and sin(A) = 4/5.

Now rewrite the equation above as
cos(A) sin(x) + sin(A)cos(x) = 1/5

You can use one of your relevant identities to rewrite the left side in terms of either sin or cos of something.

The reason I picked 5 in an earlier step is that I wanted to turn 3 and 4 into numbers that would be suitable values for a sine and cosine. What I did was take the square root of 3² + 4² = sqrt(25) = 5.

 

[Bogrune]

Divide both sides of the equation by 5. There's a reason I picked this number, which I'll get to later.

This gives you
3/5 sin(x) + 4/5 cos(x) = 1/5

Since 3/5 and 4/5 are between -1 and 1, we can consider them to be the cosine and sine, respectively of some angle A, so let cos(A) = 3/5 and sin(A) = 4/5.

Now rewrite the equation above as
cos(A) sin(x) + sin(A)cos(x) = 1/5

You can use one of your relevant identities to rewrite the left side in terms of either sin or cos of something.

The reason I picked 5 in an earlier step is that I wanted to turn 3 and 4 into numbers that would be suitable values for a sine and cosine. What I did was take the square root of 32 + 42 = sqrt(25) = 5.

I'm sorry, I'm known to be pretty slow. So am I to turn the equation:

cosAsinx + sinAcosx = 1/5
into
sin(A+x) = 1/5?

 

[Mark44]

I'm sorry, I'm known to be pretty slow. So am I to turn the equation:

cosAsinx + sinAcosx = 1/5
into
sin(A+x) = 1/5?

Yep, but you'll know what A is.

 

[Bogrune]

Okay, thanks. So will my solution (with the work shown, of course) be:

sin(A+x) = 1/5

A+x = sin^-1(1/5) +2πn

x= 4/5 + sin^-1(1/5) +2πn

 

[Mark44]

Okay, thanks. So will my solution (with the work shown, of course) be:

sin(A+x) = 1/5

A+x = sin^-1(1/5) +2πn

x= 4/5 + sin^-1(1/5) +2πn

A is not 4/5. sin(A) = 4/5, so what is A? You also know that cos(A) = 3/5.

 

[Bogrune]

Oh, I think I get it now. So it's really

sin^-1(4/5) + sin^-1(1/5) + 2πn

since I believe that sin(A) = 4/5, A = sin^-1(4/5)?

 

[tiny-tim]

this has got rather confusing

you're rewriting the original equation as

(3/5)sinx + (4/5)cosx = 1/5 …

carry on from there ​

 

[Mark44]

Okay, thanks. So will my solution (with the work shown, of course) be:

sin(A+x) = 1/5

A+x = sin^-1(1/5) +2πn

x= 4/5 + sin^-1(1/5) +2πn

As I noted earlier, the equation above is wrong because you replaced A by 4/5.

The equation above it is correct, although I don't believe you're getting all of the solutions. If sin(x) = y, then sin(x + 2π) = y, as well. What you're forgetting is that it's also true that sin(x) = sin(π/2 - x), or angles that are symmetric around π/2 have the same sines.

It might be best to find a single solution for x, and then generalize to all possible values of x later on.

You have sin(A + x) = 1/5, so A + x = sin^-1 (1/5), so x = -A + sin^-1 (1/5).

Substitute in for A (it is NOT 3/5 or 4/5) to find x.

Oh, I think I get it now. So it's really

sin^-1(4/5) + sin^-1(1/5) + 2πn

since I believe that sin(A) = 4/5, A = sin^-1(4/5)?

 

[Bogrune]

The equation above it is correct, although I don't believe you're getting all of the solutions. If sin(x) = y, then sin(x + 2π) = y, as well. What you're forgetting is that it's also true that sin(x) = sin(π/2 - x), or angles that are symmetric around π/2 have the same sines.

Please pardon me for saying this: I understand that part, but isn't sin(x) really equal to sin(π-x), since the function of sin(x) is an odd function?

 

[tiny-tim]

yup! he meant "sin(x) = sin(π - x), or angles that are symmetric around π/2 have the same sines."

 

[Mark44]

Yep, that's what I meant. What I was trying to say was that sin(x + π/2) = sin(x - π/2).