practice exam help? (2 Viewers)

timinator1993

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also does anyone know if there are any solutions or answers to above paper???
 

4025808

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hia, im doing this practice exam for maths 2u,
Bored of Studies - Student online community, resources, notes, past exams, UAI Calculator, ENTER Calculator, HSC, VCE
and for question 1b, do they mean find what cos is in the first quadrant, then use exact values ti find it? or on my calculator do i just type in cos210, or is it SHIFT cos210?? thanks for you helps
just type in cos210 and you'll get the right answer

if you do shift cos then you wont get an answer because cos has a range between -1 and 1.
 

timinator1993

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also how do i do 1d? how do you know when to shift sin it or just sin it? just sin gives a very small number, but shift sin gives a larger, more realistic number??? how my logic works is that if i find sin of 0.725 then do 180-(sin of 725) that should be my answer, obvuiusly this is wrong... any help?
ill type the questio out
if sinx = 0.725, find the values of:
sin(180-x)
sin(-x)
sin(180+x)
help??
thanks guys
 

random-1005

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also how do i do 1d? how do you know when to shift sin it or just sin it? just sin gives a very small number, but shift sin gives a larger, more realistic number??? how my logic works is that if i find sin of 0.725 then do 180-(sin of 725) that should be my answer, obvuiusly this is wrong... any help?
ill type the questio out
if sinx = 0.725, find the values of:
sin(180-x)
sin(-x)
sin(180+x)
help??
thanks guys

sin(180-x)=sinx

sin(-x)= -sin(x)

sin(180+x) =-sinx

theres some of the properties, the second one is easy to remember from the graph (it has rotational symetry ( lol i know i cant spell, my interests dont lie in english) about origin, hence odd), all the others will be obvious if you draw ASTC circle.

ie if you drew an angle (180 +x) where x is an acute angle, you would see that you would end up in the third quadrant (where sin is negative) with a relative angle (angle with the x axis) of x ( thus -sinx), all the other trig functions cos and tan can be done exact same way
 
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hscishard

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also how do i do 1d? how do you know when to shift sin it or just sin it? just sin gives a very small number, but shift sin gives a larger, more realistic number??? how my logic works is that if i find sin of 0.725 then do 180-(sin of 725) that should be my answer, obvuiusly this is wrong... any help?
ill type the questio out
if sinx = 0.725, find the values of:
sin(180-x)
sin(-x)
sin(180+x)
help??
thanks guys
It's a bit obvious that you never really payed attention in year 9 and 10.
Shift sin finds the angles whereas sin finds the sides of triangles.

At least you can think outside the box.
 

timinator1993

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okkk, thanks sort of, but what i mean is im given sinx=0.725m how do i find sin(180-x), could you please run me threw the steps for just one of them? thanks
 

random-1005

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okkk, thanks sort of, but what i mean is im given sinx=0.725m how do i find sin(180-x), could you please run me threw the steps for just one of them? thanks

i just did lol

another example, say tan(180 +x)

this lies in third quad, with related angle x, tan is positive in thrid quad, hence tan(180+x)= tan(x)

Its very important you know which quad you are in so you know what sign to make it


A question for you, what would cos(180+x) be ??
 
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timinator1993

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ok well i do shift sin of 0.725 giving 46.5, (degrees), 180-ANS = 133, so if we know that x=133, than sin(180-x) must = sin(133), which = gives 0.73, is tha right?
 

random-1005

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ok well i do shift sin of 0.725 giving 46.5, (degrees), 180-ANS = 133, so if we know that x=133, than sin(180-x) must = sin(133), which = gives 0.73, is tha right?

no, put the calculator away, its not gona help you lol, got nothing to do with shift
 

hscishard

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okkk, thanks sort of, but what i mean is im given sinx=0.725m how do i find sin(180-x), could you please run me threw the steps for just one of them? thanks
It's still just 0.725m

sin(180-x) = sinx
sin(180+x)= -sinx
sin(360-x) = - sinx

All of this comes from ASTC.
180-x = 2nd quad
180+x = 3rd quad
360-x = 4th quad
Try and think of that.
 

timinator1993

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sorry, ill run through what im doing.ok so it would be
sinx=0.725, find what sin(180-x) equals
cause 180-x is in 2nd quad where sin is positive, it would just be sin0.725, which = 48d28m? degrre/minutes
 

random-1005

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sorry, ill run through what im doing.ok so it would be
sinx=0.725, find what sin(180-x) equals
cause 180-x is in 2nd quad where sin is positive, it would just be sin0.725, which = 48d28m? degrre/minutes

no, we know sin(180-x) = sinx { see previous posts}

hence sin(180-x) = 0.725 {thats it, simple}
 

hscishard

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sorry, ill run through what im doing.ok so it would be
sinx=0.725, find what sin(180-x) equals
cause 180-x is in 2nd quad where sin is positive, it would just be sin0.725, which = 48d28m? degrre/minutes
That's correct. Though the bolded step is not needed. It's not asking to find the angle. It wants the answer of sin(whatever)

Oh wait wtf.
No. You don't go sin(0.725)

0.725 is the answer ok.
x is an angle.
sin(angle) = 0.725
sin(180-angle) = sin(angle)
Therefore
sin(180-angle) = 0.725
 
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timinator1993

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ohh ok, no working out needed, they gave me the answer, so i was trying to find tha answer from the answer if u get what i mean, i had sinx, but i was trying to do sinx of sinx....
 

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