How can I solve for b in the equation csc(6b+pi/8) = sec(2b-pi/8)?

[I Like Pi]

Solving trig function...

Homework Statement

given, csc(6b+pi/8) = sec(2b-pi/8)

solve for b

Homework Equations

The Attempt at a Solution

I managed to simplify it to:

cos(2b-pi/8) = sin(6b+pi/8)

How would i solve for b

well i know that sinx = cos(pi/2-x) so...
cos(2b-pi/8) = cos[pi/2-(6b+pi/8)]
cos(2b-pi/8) - cos[pi/2-(6b+pi/8)] = 0
cos[(2b-pi/8)-(pi/2-6b-pi/8)] = 0
8b-pi/2 = cos-1(0)
8b = pi
b=pi/8

but it doesn't work :sad:

 

[Mentallic]

cos^{-1}(cosA+cosB)\neq A+B

which is what you have done.

And just as a word of advice, if you do happen to do something to the equation such as take the inverse cosine of it, you need to do it all in one go - i.e. don't take the inverse of one side, then the inverse of the other side. Do it altogether.

Using sin(x)=cos(\pi/2-x)

sin(6b+\pi/8)=cos(3\pi/8-6b)=cos(-3(2b-\pi/8))

Now use the fact that cos(x)=cos(-x) and you should be all good from there.