A Few Simple Calculus Problems

[DoctorReynaldo]

Good morning folks, I have been recently stumped by a few of my son's homework calculus problems and was hoping for a little bit of help. I'm sure they are simple, the thing is it has been 20 years since I took this.

2cosx=sinx
sin(pi)/2x=1
sin^2x+sin=1 (find zeros)
coscot=2cos
tan^2/cos(90-x)
(sinx^x-cosx^4)/(sinx^2-cosx^2)=1
cosx(secx-cosx)

I appreciate all your help

 

[cookiemonster]

1. 2cosx = sinx
Divide through by cosx
2 = tanx
x = arctan2

2. sin(pi)/(2x) = 1
2x = sin(pi)
sin(pi) is equal to 0, so
x = 0

3. (sinx)^2 + sinx - 1 = 0
Factor using the quadratic formula
$$\sin x = \frac{-1 \pm \sqrt{1 - 4*1*(-1)}}{2}$$
Then just take the arcsin.

4. I don't understand this notation.

5. We're short an equals sign on a right-hand sign here. But this identity might help:
cos[Pi/2 - x] = sin[x]

6. & 7. You should probably clean up the notation in these ones before we attempt them. It's a little ambiguous and different interpretations really change the problems. And there's no equals sign in 7.

cookiemonster

 

[DeadWolfe]

For question 4:

"coscot=2cos"

I assume that would be (cosx)(cotx) = 2cosx"?

In which case

cot x = 2
(1/tan x) = 2
1 = 2 tanx
tan x = 1/2
x = arctan 1/2

Helpful?

Also, if 7 was supposed to be "cosx = (secx-cosx)"

Then:

cos x = sec x - cosx
sec x = 0
(cos x)^-1 = 0

Undefined ( divide by zero)

But I'm guessing that's not what you meant.

Probably kinda late now anyway, but oh well.

 

[Wooh]

Originally posted by DoctorReynaldo
Good morning folks, I have been recently stumped by a few of my son's homework calculus problems and was hoping for a little bit of help. I'm sure they are simple, the thing is it has been 20 years since I took this.

2cosx=sinx
sin(pi)/2x=1
sin^2x+sin=1 (find zeros)
coscot=2cos
tan^2/cos(90-x)
(sinx^x-cosx^4)/(sinx^2-cosx^2)=1
cosx(secx-cosx)

I appreciate all your help

For number 6. I am assuming you mean sinx^4-cosx^4? If so, then (sinx^2+cosx^2)(sinx^2-cosx^2)/(sinx^2-cosx^2)
sinx^2+cosx^2
1=1

 

[Muzza]

Originally posted by cookiemonster
1. 2cosx = sinx
Divide through by cosx
2 = tanx
x = arctan2

Just remember that the tangent has a period of $$\pi$$, so $$x = \arctan{2} + n\pi$$. It's also good to check that $$\cos{x} \neq 0$$ before dividing with it...

 

[Maria]

Need some help..

Can someone please help me with this one?

cos2x = 2 cos x sin x :uhh:

 

[Muzza]

It's equivalent to cos(2x) = sin(2x), or 1 = sin(2x)/cos(2x)...

 

[Maria]

help..

I need to find 4 angles..

 

[Zurtex]

I need to find 4 angles..

Generally best to make your own thread.

$$\sin 2x \equiv 2 \cos x \sin x$$

So as stated above your problem is the same as:

$$\sin 2x = \cos 2x$$

Which is the same as:

$$\frac{\sin 2x}{\cos 2x} = 1$$

Now you should remember a simple identity about sin over cos which makes this really simple.

 

[Maria]

I`ll that.. Thanks

 

[Maria]

Thanks

I got it right.. thanks a lot :rofl: