Trig Nightmare: Solve Cosx/1 - Sinx | Free Math Help Forum

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You are using an out of date browser. It may not display this or other websites correctly.You should upgrade or use an alternative browser. trig nightmare: solve cosx/1 - sinx

• Thread starter seekingh
• Start date Jun 22, 2006

S

seekingh

New member

Joined Jun 22, 2006 Messages 3 can you help me solve cos x/1- sin x

pka

Elite Member

Joined Jan 29, 2005 Messages 11,978 "To solve" means that there is an equation! Unless I have missed something, there is no equal sign in your question. Thus there is nothing to solve. S

seekingh

New member

Joined Jun 22, 2006 Messages 3 restate problem misquote on my part, the instructions are to use fundamental identies to simplify expression

stapel

Super Moderator

Staff member Joined Feb 4, 2004 Messages 16,550 Re: restate problem

seekingh said: misquote on my part, the instructions are to use fundamental identies to simplify expression Click to expand...

What you have posted means one of the following: . . . . .\(\displaystyle \L \frac{\cos{(x)}}{1}\,-\,\sin{(x)}\) . . . . .\(\displaystyle \L \cos{\left(\frac{x}{1}\right)}\,-\,\sin{(x)}\) Is either of these what you meant? When you reply with confirmation or correction, please include everything you have tried thus far. Thank you. Eliz. S

seekingh

New member

Joined Jun 22, 2006 Messages 3 no, neith er one is correct. the expression is wriiten cos x as the numerator and 1-sin x as the denominator. I have tried using the basic formulas of y/r for sin and x/r for cos. However I getting sec x+ tan x, which is not a correct answer according to my book

stapel

Super Moderator

Staff member Joined Feb 4, 2004 Messages 16,550

seekingh said: the expression is wriiten cos x as the numerator and 1-sin x as the denominator. Click to expand...

So you mean your expression to be as follows...? . . . . .\(\displaystyle \L \frac{\cos{(x)}}{1\,-\,\sin{(x)}}\)

seekingh said: I have tried using the basic formulas of y/r for sin[e] and x/r for cos[ine]. Click to expand...

What have you done? What steps did you take?

seekingh said: However I getting sec x+ tan x, which is not a correct answer according to my book Click to expand...

This expression is already fairly simple. It is difficult to say what the book might have in mind. If you have "the" answer, it might help if you provided that, as this could guide our advice to you. Thank you. Eliz.

skeeter

Elite Member

Joined Dec 15, 2005 Messages 3,204 \(\displaystyle (\frac{cosx}{1-sinx})(\frac{1+sinx}{1+sinx}) =\) \(\displaystyle \frac{cosx(1+sinx)}{1-sin^2x} =\) \(\displaystyle \frac{cosx(1+sinx)}{cos^2x} =\) \(\displaystyle \frac{1+sinx}{cosx}\) note that the last expression also equals \(\displaystyle secx + tanx\), which is what you stated earlier ... so, what did the "book" have as an answer? You must log in or register to reply here. Share: Facebook X (Twitter) Reddit Pinterest Tumblr WhatsApp Email Share Link

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