Simple Trig Questions, arg! (1 Viewer)

Finx

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1. If sin x = 0.3, find:

i) cos(90-x)
ii) cos² x
iii) cosec x
iv) sin(180+x)
v) sin(360-x)

(You probably don't have to all of these, just explain what I'm supposed to do here =P)

2. What is the exact value of cos(-150°)?


5. If cos(theta)=0.96, find the exact value of cot(theta).

Thanks in advance! =]
 

fishy89sg

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Finx said:
1. If sin x = 0.3, find:

i) cos(90-x)
ii) cos² x
iii) cosec x
iv) sin(180+x)
v) sin(360-x)

(You probably don't have to all of these, just explain what I'm supposed to do here =P)

2. What is the exact value of cos(-150°)?


5. If cos(theta)=0.96, find the exact value of cot(theta).

Thanks in advance! =]

1 i) cos(90-x) = sinx = 0.3
ii) cos² x = 1 - sin ² x = 1 - (0.3)(0.3) = 0.91
iii) cosecx = 1/sinx = 1/0.3 = 3.3333
iv) sin(180+x) = -sinx = -0.3
v) sin(360-x) = -sinx = -0.3

2. -sqrt3 /2
5. cos theta = 96/100
therefore constructing a right angled triangle, your third side is sqrt(100² - 96²) = 28
so tan theta = O/A = 28/96
cot theta = 96/28
 
Last edited:

x.Exhaust.x

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With the 1st question, just manipulate them with your 3 trigonometric rules until you get sine. Then substitute the value in and if necessary, use ASTC to determine if it's positive or negative

Edit: Thumbs up for above post.
 

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Finx said:
1. If sin x = 0.3, find:

i) cos(90-x)
ii) cos² x
iii) cosec x
iv) sin(180+x)
v) sin(360-x)

(You probably don't have to all of these, just explain what I'm supposed to do here =P)

2. What is the exact value of cos(-150°)?


5. If cos(theta)=0.96, find the exact value of cot(theta).

Thanks in advance! =]
For 2. find the value cos(150) since the cos wave is an even function.

For 5 draw a triangle with 24 as the opposite side and 25 as the hypotenuse. Then find the third angle using Pythagoras and the rest is simple.
 

kaz1

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Finx said:
^ What? o_O

How'd you get that.
In a right angled triangle you have and angle theta and (90-theta).
Notice how cos (90-theta)=sin theta.
 

Finx

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kaz1 said:
Notice how cos (90-theta)=sin theta.
No, I didn't notice this. We've probably done it in class and I wasn't paying attention. I'll look up a text book for it.
 

fishy89sg

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Finx said:
No, I didn't notice this. We've probably done it in class and I wasn't paying attention. I'll look up a text book for it.
if you wanna know how its derived... you use the following 3unit formula:
cos(a-b) = cosacosb + sinasinb

so with a=90, b=theta:
cos(90 - theta) = cos90costheta + sin90sintheta = 0 + sintheta = sintheta

try it in your calculator... it sersly works
 

Finx

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fishy89sg said:
iv) sin(180+x) = -sinx = -0.3
v) sin(360-x) = -sinx = -0.3
I got v), but I don't get iv).

Is it possible to expand sin(180+x)?
Therefore giving sin180+sinx
= 0 + 0.3
= 0.3?
 

lolman12567

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sin(180+x)

does the 180 + x show that the ray must lie in the 3rd quadrants where only tan is positive therefore making sin negative?

dunno, i'm quite confused about this topic myself
 

Finx

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lolman12567 said:
sin(180+x)

does the 180 + x show that the ray must lie in the 3rd quadrants where only tan is positive therefore making sin negative?

dunno, i'm quite confused about this topic myself
I think you're right.
 

tommykins

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lolman12567 said:
sin(180+x)

does the 180 + x show that the ray must lie in the 3rd quadrants where only tan is positive therefore making sin negative?

dunno, i'm quite confused about this topic myself
Yes
 

lolokay

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Finx said:
I got v), but I don't get iv).

Is it possible to expand sin(180+x)?
Therefore giving sin180+sinx
= 0 + 0.3
= 0.3?
you can't expand a function that way
for sine, the expansion is:
sin(A+B) = sinAcosB + sinBcosA
so sin(180+x) = sin180cosx + cos180sinx
= -sin[x]

it's the negative of sin[x] because, by adding 180 you're flipping it - so the opposite side will have the same magnitude but be pointing downwards.
Similarly, cos(180+x) = -cos[x] because the adjacent has been flipped.
but tan(180+x) = tan[x] because both the opposite and adjacent have been flipped

sin[90-x] and similar identities can be found by observing the right-angled triangle, since the cosine at 1 angle is the same as the sine of the other.
 

Finx

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lolokay said:
you can't expand a function that way
for sine, the expansion is:
sin(A+B) = sinAcosB + sinBcosA
so sin(180+x) = sin180cosx + cos180sinx
= -sin[x]
You could've just said cos180 = -1.. lol =]

I understand that expansion, but isn't that 3 unit stuff? These questions were taken from a 2u paper.
 

lolokay

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well, you asked about the sine expansion, and it says your doing 3u
you don't need that formula to figure out these questions, just think about it like I was saying - eg. the angle gets flipped either 180' or about the x or y axis, or they're complimentary angles, or whatever else, to figure out what it will be in terms of another angle

eg. sin(180-x),
180-x means that the angle will have been flipped about the y axis, but since the opposite side is still positive it will = sin[x]
 

fishy89sg

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seriously guys, its soo much easier to sketch the graphs

when you sketch y=sin(180+x), the curve shifts 180 degrees to the left, so in effect, you get the curve y=-sinx

and y=sin(-x) = -sinx so sin(360-x) = -sin(360+x) and when thats sketched, its the same as -sinx since the sinx curve repeats itself every 360 degrees
 

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