Andiio's SM Questions Thread

[Andiio]

What would be the best method to do this question?

Given the domain of sin x and cos x are restricted to [-pi/2 , pi/2] and [0 , pi] respectively, define the implied domain and range for: y = sin^-1 (cos x).

Thanks!

[brightsky]

Let y = sin^(-1)(cos x)
sin(y) = cos(x)
We know that -pi/2 =< y =< pi/2, giving our range.
This means -1 =< sin(y) =< 1.
So -1 =< cos(x) =< 1, which fits the domain of cos(x) perfectly.
Hence range = [-pi/2, pi/2], domain = [0, pi]

[pi]

You could also treat it as a composite function and work like methods from there, but brightsky's method is probably better/easier...

[Andiio]

Thanks brightsky!

How about this one?

Solve 'sin x + cos x = 1' for x E [0,2pi]

[brightsky]

There is a trick for this one:

sinx + cosx = 1
sqrt(2)/2 * sinx + sqrt(2)/2 * cosx = sqrt(2)/2
sinx sin(pi/4) + cosx cos(pi/4) = sqrt(2)/2
cos(x - pi/4) = sqrt(2)/2
Solve from there.

[Andiio]

Thanks!

The worked solutions are quite obscure in this one:

Let sec ß = b where ß E [pi/2 , pi]. Find, in terms of ß, two values of x in the range [-pi , pi] which satisfy the following equation:

sec x = -b

I'm not quite sure how you would go about this? Could you please explain the process for me?

Thanks!

[TrueTears]

brightsky's trick is nice, i learnt that trick from ahmad in 2008, however just for interest although definitely a less elegant method, square both sides and you can solve it from there, but rmb to know when to exclude solutions

[Andiio]

brightsky's trick is nice, i learnt that trick from ahmad in 2008, however just for interest although definitely a less elegant method, square both sides and you can solve it from there, but rmb to know when to exclude solutions

The trick with the (sin x + cos x = 1) or the earlier one?

[TrueTears]

former

[Andiio]

former

Could you please explain the 'trick' in greater depth? Like where I can apply it and such?

[brightsky]

former

Could you please explain the 'trick' in greater depth? Like where I can apply it and such?

Basically, you are trying to exploit the compound angle formula for cosine in order to change the equation into something with only one trig function (in this case cosine). Can't think of another question where you might apply this specific trick on the top of my head but it's really just the logic behind it that you need to know.

[evaever]

former

Could you please explain the 'trick' in greater depth? Like where I can apply it and such?

(i) sinx -cosx = 1
(ii) tanx + 1 = secx
(iii) tanx - 1 = secx
(iv) 1 + cotx = cosecx
(v) 1 - cotx = cosecx

[TrueTears]

former

Could you please explain the 'trick' in greater depth? Like where I can apply it and such?

well, i guess it just comes with practise, you shouldn't treat it as a method and knowing when to apply it.

it's more of wishful thinking, a spark of ingenuity at times

[Andiio]

Thanks!

The worked solutions are quite obscure in this one:

Let sec ß = b where ß E [pi/2 , pi]. Find, in terms of ß, two values of x in the range [-pi , pi] which satisfy the following equation:

sec x = -b

I'm not quite sure how you would go about this? Could you please explain the process for me?

Thanks!

Could anyone please help me with this question?

[brightsky]

Same rules as with cosine:

cos(b) = 1/b
We want cos(x) = -1/b
cos(pi - b) = -1/b
cos(b - pi) = -1/b

So x = pi - b or b - pi.