Math 30-1 trigonometry (grade 12, Canada)

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kris.yarmak123

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Can someone please provide me a solution? We’ve never done something like that with tan. I’d be very grateful.
The question is:
If the general solutions to the equation 2tan^22x=2 are written as (pi/k) + (pi/4)n where n€I, then k= ___
 
Is it:

[MATH]\tan^{22}(x)[/MATH]
or:

[MATH]\tan^{2}(2x)[/MATH] ?
 
I'm pretty sure you mean 2tan^2(2x) = 2. Parentheses are very useful ...

Please show your work as far as you can take it, and then we can discuss the details. (For example, can you isolate the tangent?)

I suspect you might find the answer before you get to the part that is new to you, so we'll want to talk about things even after you know what k is.
 
Now let's see your work. (Don't let the fact that you've seen part of the answer distract you from the fact that you need to know how to solve it yourself.)
 
[MATH][SUB][/SUB][/MATH]
Dr.Peterson said:
I'm pretty sure you mean 2tan^2(2x) = 2. Parentheses are very useful ...

Please show your work as far as you can take it, and then we can discuss the details. (For example, can you isolate the tangent?)

I suspect you might find the answer before you get to the part that is new to you, so we'll want to talk about things even after you know what k is.
Click to expand...
I did these steps:
tan²(2x)=1
(4tan²x)/(1-tan²x)²=1
(4tan²x)=(1-tan²x)²
4tan²x=(1-tan⁴x)
4tan²x=(1-tan²x)(1+tan²x)
4tan²x=(1-tan²x)(sec²x)
(4sin²x)/(cos²x)=(1-(sin²x)/(cos²x))(1/cos²x)
(4sin²x)/(cos²x)=(cos²x-sin²x)/(cos⁴x)
4sin²x=(cos²x-sin²x)/(cos²x)
(4sin²x)(cos²x)=cos²x-sin²x
(4sin²x)(cos²x)=cos2x
....and from this step I get lost...
 
kris.yarmak123 said:
[MATH][SUB][/SUB][/MATH]I did these steps:
tan²(2x)=1
(4tan²x)/(1-tan²x)²=1
(4tan²x)=(1-tan²x)²
4tan²x=(1-tan⁴x)
4tan²x=(1-tan²x)(1+tan²x)
4tan²x=(1-tan²x)(sec²x)
(4sin²x)/(cos²x)=(1-(sin²x)/(cos²x))(1/cos²x)
(4sin²x)/(cos²x)=(cos²x-sin²x)/(cos⁴x)
4sin²x=(cos²x-sin²x)/(cos²x)
(4sin²x)(cos²x)=cos²x-sin²x
(4sin²x)(cos²x)=cos2x
....and from this step I get lost...
Click to expand...
It's not necessary to switch to tan(x). Solve for 2x. Then for x.
 
Thanks for showing your work; that allows us to really help you.

There are several points in your work where you took the hard way rather than the quick way.

My hint was to isolate the tangent, meaning tan(2x). If tan²(2x)=1, what is tan(2x)?

In general, unless you have both x and 2x as arguments in an equation, you shouldn't use the double-angle identities.

Here are some other errors:
kris.yarmak123 said:
[MATH][SUB][/SUB][/MATH]I did these steps:
tan²(2x)=1
(4tan²x)/(1-tan²x)²=1
(4tan²x)=(1-tan²x)²
4tan²x=(1-tan⁴x) <-- wrong; (a-b)^2 is not a^2 - b^2
4tan²x=(1-tan²x)(1+tan²x)
4tan²x=(1-tan²x)(sec²x) <-- I would have stayed with only one trig function, expanding to a quadratic form
(4sin²x)/(cos²x)=(1-(sin²x)/(cos²x))(1/cos²x)
(4sin²x)/(cos²x)=(cos²x-sin²x)/(cos⁴x) <-- wrong; how is 1 - sin^2(x) = cos^2(x) - sin^2(x)?
4sin²x=(cos²x-sin²x)/(cos²x) <-- wrong; you can't cancel only one term in the numerator
(4sin²x)(cos²x)=cos²x-sin²x
(4sin²x)(cos²x)=cos2x
....and from this step I get lost...
Click to expand...
You got lost long before that. Try the way we suggested.
 
Oh, I got the idea
(tan2x-1)(tan2x+1)
tan2x=1 or tan2x=-1
2x=tan^-1(1) or 2x=tan^-1(-1)
45°,225° 135°,315°
22.5°,112.5° 67.5°,157.5°
π/4, 5π/4 3π/4, 7π/4
π/8, 5π/8 3π/8, 7π/8
Therefore, k=8 :)
 
Yes, given:

[MATH]\tan^{2}(2x)=1[/MATH]
Then:

[MATH]\tan(2x)=\pm1[/MATH]
Hence:

[MATH]2x=\frac{\pi}{4}+\frac{\pi}{2}k[/MATH] where \(k\in\mathbb{Z}\)

Or:

[MATH]x=\frac{\pi}{8}+\frac{\pi}{4}k[/MATH]
 
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