Trigonometry Question (1 Viewer)

[bored of sc]

If sin -3/ 5 and cos x < 0, find the exact value of cot x and cosec x.

For the answer I drew a right angled triangle and labelled the opposite side to x 3, the hypotenuse -5 (since cos x < 0 and cos = adjacent / hypotenuse (so hypotenuse is negative to make cos angle negative)), then used pythagoras to get the adjacent side as 4 ( adjacent^2 = (-5)^2 - 3^2 )

so then to find cot, you do 1/tan x = 1/ 3/4 = 4/3

and cosec x = 1/ sin x = 1/ 3/-5 = -5/3

is this correct? please help, my exam is tomorrow and i haven't had a chance to ask my teacher this week

 

[lyounamu]

If sin -3/ 5 and cos x < 0, find the exact value of cot x and cosec x.

For the answer I drew a right angled triangle and labelled the opposite side to x 3, the hypotenuse -5 (since cos x < 0 and cos = adjacent / hypotenuse (so hypotenuse is negative to make cos angle negative)), then used pythagoras to get the adjacent side as 4 ( adjacent^2 = (-5)^2 - 3^2 )

so then to find cot, you do 1/tan x = 1/ 3/4 = 4/3

and cosec x = 1/ sin x = 1/ 3/-5 = -5/3

is this correct? please help, my exam is tomorrow and i haven't had a chance to ask my teacher this week

Your answer is correct but your method is wrong. You will potentially lose marks.

Well. Hypotenuse can never be negative. It is always positive as it is a length.

You are actually given conditions here. So you need to find the specific domain. Sine is negative in 3rd & 4th quadrants & cosine is negative in 2nd & 3rd quadrants. Therefore, the specific quadrant that meets the requirement of the conditions is 3rd quadrant. In the 3rd quadrant, both opposite & adjacent sides are negative (as the adjacent moves left with the opposite side moving downwards). As I said earlier, hypotenuse is positive (always) as it is a length.

So you have -3 as your opposite side and -4 as your adjacent side.

From the diagram, you can see that tan x = -3/-4 = 3/4 and sin x = -3/5

Cot x = 1/tanx
Therefore, cot x = 1/(3/4)
= 4/3

Cosec x = 1/(sinx)
= 1/(-3/5)
= -5/3

 

[x.Exhaust.x]

If sin -3/ 5 and cos x < 0, find the exact value of cot x and cosec x.

For the answer I drew a right angled triangle and labelled the opposite side to x 3, the hypotenuse -5 (since cos x < 0 and cos = adjacent / hypotenuse (so hypotenuse is negative to make cos angle negative)), then used pythagoras to get the adjacent side as 4 ( adjacent^2 = (-5)^2 - 3^2 )

so then to find cot, you do 1/tan x = 1/ 3/4 = 4/3

and cosec x = 1/ sin x = 1/ 3/-5 = -5/3

is this correct? please help, my exam is tomorrow and i haven't had a chance to ask my teacher this week

Edit: Cosec X = -5/3
Cot X = 4/3

Got it . Lyounamu got the same answer as well so we can't be wrong, ay .

 

[lyounamu]

Edit: Cosec X = -5/3
Cot X = 4/3

Got it . Lyounamu got the same answer as well so we can't be wrong, ay .

Hahahahahaha... I am always wrong. I make so many mistakes while posting answer up (due to my poor typing). I always forget something. I am much better when it comes to writing on paper. I rarely make mistake.

However, posting answers up on this website repeatedly increases my aptitude.

By the way, when I do on computer I tend to do questions really quickly which even makes my answer more unreliable.

 

[x.Exhaust.x]

Hahahahahaha... I am always wrong. I make so many mistakes while posting answer up (due to my poor typing). I always forget something. I am much better when it comes to writing on paper. I rarely make mistake.

However, posting answers up on this website repeatedly increases my aptitude.

By the way, when I do on computer I tend to do questions really quickly which even makes my answer more unreliable.

I agree on how it increases your aptitude, but I disagree on how you're always wrong .

 

[bored of sc]

Oh! Right, the quadrants thing.... damn, there goes my confidence :mad1:

 

[tommykins]

You should always work with absolute values when you draw the triangles, then look back at the initial answer to determine it's domain and whether it is positive or negative.

 

[bored of sc]

Thanks guys, there was a question identical to this on the test I had today.