Question

the solution to \[b*\sin(ax)=\sqrt{2}*\cos(ax)\] is \[\frac{ \pi }{ 4 }\].

What are the possible values of a,b.

Help anyone. I don't know how should i solve this.

Thanks.

Answer 1

Wow you have some weird tricky problems lol

Answer 2

\[b*\tan(ax)=\sqrt{2}\]
i think it will simplify to this.

Answer 3

yes. it's my homework.

Answer 4

\[\large\rm \tan(ax)=\frac{\sqrt2}{b}\]I'm still trying to figure this one out.
But here is what I'm thinking so far...

By plugging in key values for a,
you can come up with corresponding b values that work.

For example:
a=1,
well tangent of pi/4 is 1, ya?
So b=sqrt2.

Answer 5

a cannot equal 0, that would cause a problem for b.
a cannot equal 2 as tangent of pi/2 is not defined.

But then we get all these weird numbers in between...
When a=4/3, tangent of pi/3 gives us sqrt3,
so b= ummm.. like sqrt2/sqrt3, ya?

Answer 6

yes. you are right a=1 and b=sqrt{2}

Answer 7

if i sub these values in a calculator, it gives me somewhat pi/4

Answer 8

You are a genius. Thanks for the help again.

Answer 9

I'm not sure how we get the entire collection of a and b's, hmm :\

Answer 10

don't worry, i just need one possible value

Answer 11

Just one? Oh interesting :D

Answer 12

yeah !!