(Less than) 21 Trig Questions from Amrita

Agnishom · 2014-05-17 21:23:06

I cannot understand what I did either

Please do me a favor and tell me where the sum is borrowed from. If RD Sharma, then where in the book is it given.

Bob · 2014-05-17 21:24:57

hi Amrita,

The solution requires the use of inverse trig functions which are multi-valued. So you have to be careful not to lose any solutions. Gourish's solution is much neater than mine but I cannot see how to deduce both solutions from it, whereas mine does lead to both. Here it is:

so

Apply inverse cos

cancel a pi

Trial of different values of 'n' shows that only n = 0 leads to valid trig values (ie. -1 ≤ sin/cos theta ≤ +1

So

At this point you can square to get to Gourish's solution. But I didn't spot that so here's what I did:

and so

so

and

multiplying together:

But have I lost any solutions due to multi-values?

Cosine can be positive in the 1st and 4th quadrant. So the sine could come out negative as well and the only way to check is to check the values numerically.

So plus or minus 3/4.

Bob

Amrita · 2014-05-17 21:26:06

Thank u Bob.

Amrita · 2014-05-17 21:28:21

Agnishom, Bob had done it finally. See it.

Agnishom · 2014-05-17 21:29:11

Bob, What did I do wrong?

Bob · 2014-05-17 21:35:43

When you squared both sides, you didn't; ie. 0.5 squared = 0.25

Bob

Agnishom · 2014-05-17 21:37:02

Oh Snap!

:mad

Bob · 2014-05-17 21:37:13

Latex is working btw. But you have to type the square brackets bit yourself.

Bob

Agnishom · 2014-05-17 21:44:38

Yes, Latex is working.

anonimnystefy · 2014-05-17 22:55:23

Hi Bob

I think a solution is lost in the third step, where you get rid of the outer cosine. The stuff on the left can also be -sin(theta)+2n.

Bob · 2014-05-18 00:57:53

hi Stefy,

Hhmmm. If you meant in my answer, thanks for the suggestion but it still seems to lead to +3/4.

If you meant Gourish's, you'll have to explain why that is a valid step for inv sin.

If you can write out some steps it would help my old brain.

Bob

anonimnystefy · 2014-05-18 09:14:00

I meant yours.

I went through your solution step by step and it seems to me that you lose the solution in the very last step.

Bob · 2014-05-18 09:35:23

But I said that.

But have I lost any solutions due to multi-values?
Cosine can be positive in the 1st and 4th quadrant. So the sine could come out negative as well and the only way to check is to check the values numerically.
So plus or minus 3/4.

Bob

anonimnystefy · 2014-05-18 09:37:58

Sorry, missed that one. I don't see why it must be done numerically. You can just say cos(2 theta)=sqrt(7)/4 => sin(2 theta)=+/- 3/4.

Agnishom · 2014-05-18 15:21:56

2.If a,b are two values of x (0 <= a,b <= 2 π ) which satisfy the equation 6 cos x + 8 sin x = 9, Find sin (a+b)

Bob · 2014-05-18 18:45:58

I don't see why it must be done numerically. You can just say cos(2 theta)=sqrt(7)/4 => sin(2 theta)=+/- 3/4.

Earlier, squaring was involved, so there's no guarantee that either answer is correct. You have to check them by substitution into the original problem.

Bob

Just looking at the new question now.

Bob · 2014-05-18 18:58:20

6 cos x + 8 sin x = 9

Divide through by root (6^2 + 8^2)= 10

cosy.cosx + siny.sinx = 0.9 where y = arctan 8/6

So in that range the two solutions will be related (sketch a graph) so a + b is a simple value.

Bob

Bob · 2014-05-18 19:58:38

Now I've tried it I think you'll need a bit more.

I've got a busy morning but I'll try to check this out properly soon.

Bob

Agnishom · 2014-05-18 20:06:18

Good Morning, Bob.

I do not understand what you did there.

Bob · 2014-05-18 22:29:30

Hi Agnishom,

Don't worry because I now realise it was 90% kabbably doo. Here's a sensible version:

This allows you to use the compound angle formula (but for sine(A+B) not cosine as I suggested earlier)

where tan(alpha) = 6/8

So we can sketch this sine curve and see that 0.9 crosses it in two places in the range, call them a and b.

Look at the point exactly half way between a and b, (a+b)/2

So what is sin((a+b)/2) ?

Hence use the double angle formula to get sin(a+b).

I hope this works. (fingers crossed smiley)

Bob

anonimnystefy · 2014-05-19 02:31:37

bob bundy wrote:

I don't see why it must be done numerically. You can just say cos(2 theta)=sqrt(7)/4 => sin(2 theta)=+/- 3/4.
Earlier, squaring was involved, so there's no guarantee that either answer is correct. You have to check them by substitution into the original problem.
Bob
Just looking at the new question now.

Well, in the squaring method, the other solution, with -sin(x)+2n is needed.

Also, the answer for this one seems to be 0.96, but that was M's doing.

Last edited by anonimnystefy (2014-05-19 02:32:10)

Bob · 2014-05-19 03:28:07

Thank you M. That's what I got ... without a CAS again! You risk having your brain atrophy altogether!!!

Bob

Agnishom · 2014-05-19 03:31:19

I'm sorry bob, but I do not see why you are dividing by 10. And how that helps you in using the sin(a+b) formula?

anonimnystefy · 2014-05-19 04:27:48

Well, I'm using it just to get an answer. I was searching for an elegant way to solve it.

Bob · 2014-05-19 04:30:06

Any expression of the form

a.cos(x) + b.sin(x) can be written as a single sine or cosine by this method. We want a/something and b/something to be the sine and cosine of a single angle. So imagine a right angled triangle with opposite = 6 and adjacent = 8. The hypotenuse is 10 and this is what you divide by.

Then 6/10 can be written as sin(alpha) and 8/10 as cosine(alpha). Then use the compound angle formula to write as sine(x + alpha). We know the shape of this graph; it's a standard sine curve but displaced by alpha to the left.

You can sketch it and mark the line y = 0.9 to show the positions of a and b. They are symmetrically placed either side of the point with x coordinate (a+b)/2.

You will see it is when the sine curve has its maximum. ie. sine(x+alpha) = 1

But we want sine((a+b)/2). You can use the function grapher like I did to find this x value and hence its sine, or (challenge) do it by considering the displacement of this graph from the standard sine curve. Once you have that you can get cosine((a+b)/2) and hence sine(a+b)

Bob

graph below.

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#51 2014-05-17 21:23:06

Re: (Less than) 21 Trig Questions from Amrita

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