unstoppable C3 trig questions - help

1) Given that arcsin k = a, where 0 < k < 1 and a is in radians write down in terms of a the first two positive values of x satisfying the equation sin x = k ????

2) Given that x satisfies arcsin x = k, where 0 < k < 90

a) state the range of the possible values of x

I got 0 < x < 1 which is correct, but part b is killing me

b) express in terms of x;
i ) cosk ii) tan k

I want some epic help on both Questions please

Reply 1

Farhan.Hanif93

Original post by iAre Teh Lejend

1) Given that arcsin k = a, where 0 < k < 1 and a is in radians write down in terms of a the first two positive values of x satisfying the equation sin x = k ????

Draw the graph of y=sinx and y=k for some value k. It should be obvious what the two values are.

2) Given that x satisfies arcsin x = k, where 0 < k < 90

a) state the range of the possible values of x

I got 0 < x < 1 which is correct, but part b is killing me

b) express in terms of x;
i ) cosk ii) tan k

I want some epic help on both Questions please

\arcsin x = k \implies \sin (k)=x

Then draw yourself a right angled triangle with one of the other angles as k, label the side opposite it of length x and label the hypotenuse at 1. Using SOHCAHTOA, you can clearly see that this triangle satisfies sin k = x.

Can you use this to find cosk and tank?

(edited 12 years ago)

Reply 2

Chicken Bacon

Original post by Farhan.Hanif93

Draw the graph of sinx=k for some value k. It should be obvious what the two values are.

\arcsin x = k \implies \sin (k)=x

I'm confused for both questions still...

Reply 3

Farhan.Hanif93

Original post by iAre Teh Lejend

I'm confused for both questions still...

OK, have you managed to work out what the smallest positive solution of sinx=k is? (this should be simple C2 stuff).

So you've sketched y=sinx and y=k, and seen the two smallest positive intersections right? Those are your solutions. Can you see any way of getting an expression for the second solution in terms of the first one?

For the second question, have you drawn the triangle? Can you see why we're using that particular triangle for help? Since it's right angled, you can use pythagorus to work out the other side. Then using SOHCAHTOA (GCSE stuff) on it, you can find what cos k and tan k are.

I'm not supposed to do the question for you so it'll help if you posted some working following my hints.

Reply 4

Chicken Bacon

But why do I label the opposite as 'x' ? :s-smilie:

Reply 5

Farhan.Hanif93

Original post by iAre Teh Lejend

But why do I label the opposite as 'x' ? :s-smilie:

\sin k = \dfrac{x}{1}

and by SOHCAHTOA,

\sin k = \dfrac{opposite}{adjacent}

. That's one way to think of it.

Reply 6

ps1265A

Original post by Farhan.Hanif93

\sin k = \dfrac{x}{1}

and by SOHCAHTOA,

\sin k = \dfrac{opposite}{adjacent}

. That's one way to think of it.

Why does tan(k) remain the same for part c)? Shouldn't it be negative?

Reply 7

TenOfThem

Original post by ps1265A

Why does tan(k) remain the same for part c)? Shouldn't it be negative?

There is no part (c)

If you are looking at 2(b)(ii) then the question stated that the angle k was between 0 and 90
Why would tan(k) be negative

Reply 8

ps1265A

Original post by TenOfThem

There is no part (c)

If you are looking at 2(b)(ii) then the question stated that the angle k was between 0 and 90
Why would tan(k) be negative

In my book there's a part c) to this question. It says if the k was -pi/2<k<0, how would the value of cos and tan change. I understand that cos doesn't change as the angle moves to the 4th quadrant. But tan is negative in this quadrant, yet the answer in my book states that the value of tan doesn't change.

Reply 9

TenOfThem

Original post by ps1265A

In my book there's a part c) to this question.

It says if the k was -pi/2<k<0, how would the value of cos and tan change. I understand that cos doesn't change as the angle moves to the 4th quadrant. But tan is negative in this quadrant, yet the answer in my book states that the value of tan doesn't change.

I have no idea why your book is wrong - it seems odd that he question would shift from degrees to radians in that way - are you sure that you are looking at the correct question/answer

Anyway - yes between -90 and 0 tan is negative

Reply 10

ps1265A

Original post by TenOfThem

Wouldn't it have been useful to mention that rather than just making a rather random post that makes no sense in context

In fact - why not just make a thread that asks your question rather than finding one that is 3 years old

I have no idea why your book is wrong - it seems odd that he question would shift from degrees to radians in that way - are you sure that you are looking at the correct question/answer

Anyway - yes between -90 and 0 tan is negative