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Trigonometry urgent help please :)

Hi guys,
Please could someone kindly help me with the question in the attachment. The first part was to prove the first part in red and I managed to do that. But I'm stuck on finding the values of X.

Any help will be appreciated.
Thanks in advance :smile:
image.jpeg134.6KB
Original post by Aty100
Hi guys,
Please could someone kindly help me with the question in the attachment. The first part was to prove the first part in red and I managed to do that. But I'm stuck on finding the values of X.

Any help will be appreciated.
Thanks in advance :smile:


What have you tried so far for the second part? :h:
Reply 2
Avatar for Aty100
Aty100
OP
15
Original post by SeanFM
What have you tried so far for the second part? :h:


This is what I thought about doing but I'm not sure.

Also that should say square root 😂
(edited 7 years ago)
image.jpeg116.7KB
Original post by Aty100
This is what I thought about doing but I'm not sure.

Also that should say square root 😂


tan⁡2(x)=(tan⁡(x))2\tan^2(x)=(\tan (x))^2

tan⁡2(x)=1\tan^2(x)=1
tan⁡(x)=±1=±1\tan (x) = \pm \sqrt{1}=\pm 1

You know from here yeah? :smile:
(edited 7 years ago)
Original post by The-Spartan
tan⁡2(x)=(tan⁡(x))2\tan^2(x)=(\tan (x))^2

tan⁡2(x)=1\tan^2(x)=1
tan⁡(x)=1=1\tan (x) = \sqrt{1}=1

You know from here yeah? :smile:


there is another possibility ....when you do the square root ?
Original post by Aty100
This is what I thought about doing but I'm not sure.

Also that should say square root 😂


A substitution like y = tanx may help you, so you have y^2 = 1. Then you could say that y^2 - 1 = 0 and then use something to help you progress.
Original post by the bear
there is another possibility ....when you do the square root ?


Haha just noticed that as you replied, silly me XD
Reply 7
Avatar for Aty100
Aty100
OP
15
Original post by The-Spartan
tan⁡2(x)=(tan⁡(x))2\tan^2(x)=(\tan (x))^2

tan⁡2(x)=1\tan^2(x)=1
tan⁡(x)=1=±1\tan (x) = \sqrt{1}=\pm 1

You know from here yeah? :smile:


Omg I totally forgot 😂 Thank You!

So would the answers be

1/4 pi
5/4 pi
3/4 pi
7/4 pi
Original post by Aty100
Omg I totally forgot 😂 Thank You!

So would the answers be

1/4 pi
5/4 pi
3/4 pi
7/4 pi


Right on :smile:
Now you have the answers:
@SeanFM stated another useful method that you could use,
y=tan⁡(x)y=\tan (x) and so y2=1  ⟺  y2−1=0y^2=1 \iff y^2-1=0
Difference of two squares was what the trick here was,
(y−1)(y+1)=0(y-1)(y+1)=0
And so you get
y=±1y=\pm 1.
Original post by The-Spartan
tan⁡2(x)=(tan⁡(x))2\tan^2(x)=(\tan (x))^2
tan⁡2(x)=1\tan^2(x)=1
tan⁡(x)=1=±1\tan (x) = \sqrt{1}=\pm 1

√1 = 1
±√1 = ±1
Original post by 04MR17
√1 = 1
±√1 = ±1


Erm... No.
1=±1\sqrt{1} = \pm 1
To show, lets take the inverse:
12=11^2 = 1,
(−1)2=1(-1)^2=1.
∴1=±1\therefore \sqrt{1}=\pm 1
:smile:
Original post by The-Spartan
Erm... No.
1=±1\sqrt{1} = \pm 1
To show, lets take the inverse:
12=11^2 = 1,
(−1)2=1(-1)^2=1.
∴1=±1\therefore \sqrt{1}=\pm 1
Irrelevant. If you are going to put + Ans. Then you have to have a + before the √.
Original post by 04MR17
Irrelevant. If you are going to put + Ans. Then you have to have a + before the √.


Oh, thought you were saying that the square root does not equal plus or minus :redface:
Hmm, ive always written it as 1=±1\sqrt{1}=\pm 1 :frown:
Original post by The-Spartan
Oh, thought you were saying that the square root does not equal plus or minus :redface:
Hmm, ive always written it as 1=±1\sqrt{1}=\pm 1 :frown:


This thread should clear things up :h:
Original post by SeanFM
This thread should clear things up :h:


Oh gawd, i've omitted the plus/minus on all of my square roots my whole life and only put them in the answer :rofl:
Think i need a break from maths for a couple days :colonhash:
Original post by The-Spartan
Oh gawd, i've omitted the plus/minus on all of my square roots my whole life and only put them in the answer :rofl:
Think i need a break from maths for a couple days :colonhash:


It's okay :redface: at A-level you can kind of get away with not knowing, apart from integration by substitution where you need to know.

Urgh, latex is being a nightmare.

If you had a substitution with u = x^(1/2) and the limits of x are 0 and 9, then the limits of u are 0 and 3 rather than 0 and -3.
(edited 7 years ago)
Original post by SeanFM
It's okay :redface: at A-level you can kind of get away with not knowing, apart from integration by substitution where you need to know.

Urgh, latex is being a nightmare.

If you had a substitution with u = x^(1/2) and the limits of x are 0 and 9, then the limits of u are 0 and 3 rather than 0 and -3.


Yeah no i get the whole roots thing, my notation was just off :frown: instead of writing
±x=±y\pm \sqrt{x} = \pm y where i needed the ±\pm i would just write x=±y\sqrt{x}=\pm y. If i didn't need the negative root id omit the ±\pm entirely, which is wrong really :smile:
Another one of my notation errors, along with the implies arrows (i use   ⟺  \iff nearly all the time, when i can't xD)
Original post by The-Spartan
Oh, thought you were saying that the square root does not equal plus or minus :redface:
Hmm, ive always written it as 1=±1\sqrt{1}=\pm 1 :frown:
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