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# Trig mk5 watch

1. Again where 0°≤x≤360°

cos²(x/2)=1+sin(x/2)

2. Use this identity to form a quadratic in
3. (Original post by B_9710)

Use this identity to form a quadratic in
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?
4. (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 .
5. (Original post by B_9710)
Should be .
but i took sin(x/2) out as a factor? or does that not work?
6. (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?
7. (Original post by Zacken)

damn i forgot, i'm being an idiot again, thanks a bunch guys
8. (Original post by Zacken)
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?
9. (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 , 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.
10. (Original post by Zacken)
Remember it's , 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 cheers
you can also guarantee i'll make a trig mk6 thread xD
11. (Original post by thefatone)
i guess logic and subbing values is the way to go on this one then 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...
12. (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

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Updated: February 18, 2016
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