The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Trig mk5

Again where 0°≤x≤360°

cos²(x/2)=1+sin(x/2)
Reply 1
Avatar for B_9710
B_9710
18
cos⁡2(x/2)=1−sin⁡2(x/2) \displaystyle \cos^2(x/2)=1-\sin^2(x/2)

Use this identity to form a quadratic in sin⁡(x/2) \sin(x/2)
Reply 2
Avatar for thefatone
thefatone
OP
6
Original post by B_9710
cos⁡2(x/2)=1−sin⁡2(x/2) \displaystyle \cos^2(x/2)=1-\sin^2(x/2)

Use this identity to form a quadratic in sin⁡(x/2) \sin(x/2)


i have come to

1-sin²(x/2)=1+sin(x/2)
sin²(x/2)+sin(x/2)=0
sin(x/2)(sinx+1)=0

where did i go wrong?
Reply 3
Avatar for B_9710
B_9710
18
Original post by thefatone
i have come to

1-sin²(x/2)=1+sin(x/2)
sin²(x/2)+sin(x/2)=0
sin(x/2)(sinx+1)=0

where did i go wrong?

Should be sin⁡(x/2)+1 \sin(x/2)+1 .
Reply 4
Avatar for thefatone
thefatone
OP
6
Original post by B_9710
Should be sin⁡(x/2)+1 \sin(x/2)+1 .


but i took sin(x/2) out as a factor? or does that not work?
Reply 5
Avatar for Zacken
Zacken
22
Original post by thefatone
i have come to

1-sin²(x/2)=1+sin(x/2)
sin²(x/2)+sin(x/2)=0
sin(x/2)(sinx+1)=0

where did i go wrong?


sin⁡2x2+sin⁡x2=sin⁡x2(sin⁡x2+1)\displaystyle \sin^2 \frac{x}{2} + \sin \frac{x}{2} = \sin \frac{x}{2} \left(\sin \frac{x}{2} + 1\right)
Reply 6
Avatar for thefatone
thefatone
OP
6
Original post by Zacken
sin⁡2x2+sin⁡x2=sin⁡x2(sin⁡x2+1)\displaystyle \sin^2 \frac{x}{2} + \sin \frac{x}{2} = \sin \frac{x}{2} \left(\sin \frac{x}{2} + 1\right)




damn i forgot, i'm being an idiot again, thanks a bunch guys :biggrin:
Reply 7
Avatar for thefatone
thefatone
OP
6
Original post by Zacken
sin⁡2x2+sin⁡x2=sin⁡x2(sin⁡x2+1)\displaystyle \sin^2 \frac{x}{2} + \sin \frac{x}{2} = \sin \frac{x}{2} \left(\sin \frac{x}{2} + 1\right)


why doesn't the value for x = 180° work?
when i draw the sin graph i get 3 values for where sin=0
logically if i sub in 180 they don't work

what about the other one aswell?
where sin x= -1 there's only one value which is 270°
of course if i sub it in 270° won't work but why?
Reply 8
Avatar for Zacken
Zacken
22
Original post by thefatone
why doesn't the value for x = 180° work?
when i draw the sin graph i get 3 values for where sin=0
logically if i sub in 180 they don't work

what about the other one aswell?
where sin x= -1 there's only one value which is 270°
of course if i sub it in 270° won't work but why?


Remember it's x2\frac{x}{2}, so so if you put 180 degrees in, you'll get sin(180/2) = sin 90 = 1. You need x/2 = 0, 180, 360, etc... not x.
Reply 9
Avatar for thefatone
thefatone
OP
6
Original post by Zacken
Remember it's x2\frac{x}{2}, so so if you put 180 degrees in, you'll get sin(180/2) = sin 90 = 1. You need x/2 = 0, 180, 360, etc... not x.


i guess logic and subbing values is the way to go on this one then :biggrin: cheers
you can also guarantee i'll make a trig mk6 thread xD
Reply 10
Avatar for Zacken
Zacken
22
Original post by thefatone
i guess logic and subbing values is the way to go on this one then :biggrin: cheers
you can also guarantee i'll make a trig mk6 thread xD


No, you have sin a = 0 when a = 0, 180, 360. Now let a = x/2. Then x/2 = 0, 180, 360. So x = 0, 360, 720.

sin a = -1, when a = 270. Now let a = x/2. Then x/2 = 270 so x = 2 * 270, etc...
Reply 11
Avatar for thefatone
thefatone
OP
6
Original post by Zacken
No, you have sin a = 0 when a = 0, 180, 360. Now let a = x/2. Then x/2 = 0, 180, 360. So x = 0, 360, 720.

sin a = -1, when a = 270. Now let a = x/2. Then x/2 = 270 so x = 2 * 270, etc...


ah that's what i was looking for thanks so much, i think over the holiday i would've peeled my brain from my skull if i didn't have any help from all u guys at tsr :biggrin:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you