Trigonometry

While you are technically correct you can get to the mentioned answer by using [imath]u=4x[/imath], then [imath]2\sin u = \sqrt{2-\sqrt{3}}[/imath], square both parts and use some trigonometric identities. Can you work out the rest?
 
blamocur said:
While you are technically correct you can get to the mentioned answer by using [imath]u=4x[/imath], then [imath]2\sin u = \sqrt{2-\sqrt{3}}[/imath], square both parts and use some trigonometric identities. Can you work out the rest?
Click to expand...
I seem to be going in circles. IMG_20220605_134937.jpg
 
blamocur said:
While you are technically correct you can get to the mentioned answer by using [imath]u=4x[/imath], then [imath]2\sin u = \sqrt{2-\sqrt{3}}[/imath], square both parts and use some trigonometric identities. Can you work out the rest?
Click to expand...
I got it, kinda. The only difference is I got kPi/4 instead of kPi/2 and 5Pi/48 instead of 11Pi/48IMG_20220605_145229.jpg
IMG_20220605_145236.jpg
 
Loki123 said:
I made a mistake, I know why it's supposed to be 11Pi, but what about kPi/4 instead of kPi/2
Click to expand...
I've made the same mistake of getting the period of [imath]\frac{k\pi}{4}[/imath] instead of [imath]\frac{k\pi}{2}[/imath]. The error arose when we squared the equation :
[imath]4\sin^2 u = 2-\sqrt{3}[/imath] is not equivalent to [imath]2\sin u = \sqrt{2-\sqrt{3}}[/imath], but it is equivalent to [imath]\pm \left( \sqrt {2-\sqrt{3}}\right)[/imath]. I.e., by squaring we've introduce additional "solutions" (there is a term for them, but I don't remember what it is).
 
You must log in or register to reply here.
Top