Confused!?!

[cosmic]

Hi guys,

Given sin(x)=19\51, I'm asked to "find the exact value of cos(x)".

I know that I can find (x) by taking the inverse of sin(x). Then use the angle (x) to find cos(x) and come out with 0.928012515. But is this the exact value of cos(x) like the question requires? I'm a little confused any help in this regard would be greatly appreciated.

Thanks in advance

 

[Quaid]

Given sin(x)=19\51, I'm asked to "find the exact value of cos(x)".

Hi there.

Draw the typical right triangle, with angle x, and opposite side 19 and hypotenuse 51, as sin(x) is opposite/hypotenuse.

After you calculate the length of the adjacent side, you may write cos(x) as adjacent/hypotenuse.

Cheers

 

[mmm4444bot]

find cos(x) and come out with 0.928012515. But is this the exact value of cos(x)?

No, it's a decimal approximation.

Whenever a calculator gives such a value (for trig, radicals, logs, etc), it is an estimate, rounded to the number of places displayed.

 

[cosmic]

Hi there.

Draw the typical right triangle, with angle x, and opposite side 19 and hypotenuse 51, as sin(x) is opposite/hypotenuse.

After you calculate the length of the adjacent side, you may write cos(x) as adjacent/hypotenuse.

Cheers

Hi,

Thank you for your reply. I used the cosine rule to find the adj because the question doesn't say that the triangle is a right-angled triangle. Anyway I worked the adj to be 34.11.... which again would not give me an exact value. Any ideas?

Thank you

No, it's a decimal approximation.

Whenever a calculator gives such a value (for trig, radicals, logs, etc), it is an estimate, rounded to the number of places displayed.

Hi,

Thanks for your reply. Yeah I kind of guessed as much. Would be grateful if you could point me in the right direction.

Thanks.

 

[Quaid]

the question doesn't say that the triangle is a right-angled triangle.

Please post the complete exercise statement.

What did they tell you about a triangle? Did you calculate the 19/51 ratio, or was it given?

Ciao

 

[cosmic]

Please post the complete exercise statement.

What did they tell you about a triangle? Did you calculate the 19/51 ratio, or was it given?

Ciao

Hi,

Thank you for your response.

The question just says...Given that (x) is an acute angle with sin(x)=19/51, find the exact value of cos(x)

Thanks.

 

[cosmic]

Please post the complete exercise statement.

What did they tell you about a triangle? Did you calculate the 19/51 ratio, or was it given?

Ciao

Based on a right-angled triangle the adj side which I worked out is 47.328.....which again doesn't give me an exact value. I have no idea what I'm supposed to be doing to getting an exact value. :lol:

 

[pka]

The question just says...Given that (x) is an acute angle with sin(x)=19/51, find the exact value of cos(x)

Assuming that $\displaystyle 0<x<\frac{\pi}{2}$ the exact value is $\displaystyle \cos(x)=\dfrac{8\sqrt{35}}{51}~.$

 

[cosmic]

Assuming that $\displaystyle 0<x<\frac{\pi}{2}$ the exact value is $\displaystyle \cos(x)=\dfrac{8\sqrt{35}}{51}~.$

Hi,

Thanks you for your reply. Yes (x) is around 22 degrees. But how did you manage to work that out?

 

[Deleted member 4993]

Hi,

Thanks you for your reply. Yes (x) is around 22 degrees. But how did you manage to work that out?

You can use

sin²(x) + cos²(x) = 1

 

[cosmic]

You can use

sin²(x) + cos²(x) = 1

Hi,

Thank you for your reply. I've tried that but I didn't get exactly 1. It was more like 0.99977...

Thanks.

 

[pka]

Hi,

Thank you for your reply. I've tried that but I didn't get exactly 1. It was more like 0.99977...

Put your calculator away, they are absolutely useless if the question uses the word exact
NO CALCULATOR CAN BE EXACT.

 

[cosmic]

Put your calculator away, they are absolutely useless if the question uses the word exact
NO CALCULATOR CAN BE EXACT.

:lol: I just sent you a pm. I'm putting my calculator away but the answer that you obtained came out exact on the calculator. So do you think I should just use the identity and the exact value as 1?

How did you manage to get your answer anyhow, I'm intrigued.

Thanks :grin:

 

[pka]

I just sent you a pm. I'm putting my calculator away but the answer that you obtained came out exact on the calculator. So do you think I should just use the identity and the exact value as 1?
How did you manage to get your answer anyhow, I'm intrigued.

Of course you know that $\displaystyle \sin^2(x)+\cos^2(x)=1$.
So look at this

 

[cosmic]

Of course you know that $\displaystyle \sin^2(x)+\cos^2(x)=1$.
So look at this

Thanks, you've been a great help.

I initially tried the identity but like you pointed out I should just put my calculator away because all it did was confuse me more. :grin:

All the best.

 

[Quaid]

Yes (x) is around 22 degrees. But how did you manage to work that out?

I'm confused. You asked for the exact value of cos(x).

Is 22 degrees what you desired?

Otherwise, if the exact value of cos(x) is still the question, why can't you use right-triangle trig? :???:

 

[cosmic]

I'm confused. You asked for the exact value of cos(x).

Is 22 degrees what you desired?

Otherwise, if the exact value of cos(x) is still the question, why can't you use right-triangle trig? :???:

Sorry, just to clarify sin(x) is 19/51 which is around 22 degrees not exactly 22 degrees.

 

[Deleted member 4993]

Sorry, just to clarify sin(x) is 19/51 which is around 22 degrees not exactly 22 degrees.

But that information is irrelevant for your problem!

sin(x) = 19/51

sin²(x) = 19²/51²

cos²(x) = 1- 19²/51² = (2601 - 361)/2601 = 2240/2601

cos(x) = ±(√2240)/51 = ±8*√(35)/51

assuming 0≤ x ≤ π/2, the positive value should be chosen.

 

[cosmic]

But that information is irrelevant for your problem!

sin(x) = 19/51

sin²(x) = 19²/51²

cos²(x) = 1- 19²/51² = (2601 - 361)/2601 = 2240/2601

cos(x) = ±(√2240)/51 = ±8*√(35)/51

assuming 0≤ x ≤ π/2, the positive value should be chosen.

Hi,

Yes I know I was just clarifying it for Quaid.

But like you just stated I rearranged the identity for cos(x) and obtained an exact value.

Thanks.