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Solving trig function...
given, [tex]csc(6b+pi/8) = sec(2b-pi/8)[/tex]
solve for b
I managed to simplify it to:
[tex]cos(2b-pi/8) = sin(6b+pi/8)[/tex]
How would i solve for b
well i know that [tex]sinx = cos(pi/2-x)[/tex] so...
[tex]cos(2b-pi/8) = cos[pi/2-(6b+pi/8)][/tex]
[tex]cos(2b-pi/8) - cos[pi/2-(6b+pi/8)] = 0[/tex]
[tex]cos[(2b-pi/8)-(pi/2-6b-pi/8)] = 0[/tex]
[tex]8b-pi/2 = cos-1(0)[/tex]
[tex]8b = pi[/tex]
[tex]b=pi/8[/tex]
but it doesn't work :sad:
Homework Statement
given, [tex]csc(6b+pi/8) = sec(2b-pi/8)[/tex]
solve for b
Homework Equations
The Attempt at a Solution
I managed to simplify it to:
[tex]cos(2b-pi/8) = sin(6b+pi/8)[/tex]
How would i solve for b
well i know that [tex]sinx = cos(pi/2-x)[/tex] so...
[tex]cos(2b-pi/8) = cos[pi/2-(6b+pi/8)][/tex]
[tex]cos(2b-pi/8) - cos[pi/2-(6b+pi/8)] = 0[/tex]
[tex]cos[(2b-pi/8)-(pi/2-6b-pi/8)] = 0[/tex]
[tex]8b-pi/2 = cos-1(0)[/tex]
[tex]8b = pi[/tex]
[tex]b=pi/8[/tex]
but it doesn't work :sad:
Last edited: