# Solving Trigonometric Equations

• Mar 9th 2010, 09:27 AM
bcavanaugh34
Solving Trigonometric Equations
I am having some serious problems with this section. My book is no use to me. Thanks for all the help! It says: use the basic identities to change the expression to one involving only sines and cosines. Then simplify to a basic trigometric function.

1). (sinx)(tanx+cotx)

2). sinx-tan(x)cos(x)+ cos(pi/2-x)

3). sin(x)cos(x)tan(x)sec(x)csc(x)
• Mar 9th 2010, 10:18 AM
Quote:

Originally Posted by bcavanaugh34
I am having some serious problems with this section. My book is no use to me. Thanks for all the help! It says: use the basic identities to change the expression to one involving only sines and cosines. Then simplify to a basic trigometric function.

1). (sinx)(tanx+cotx)

2). sinx-tan(x)cos(x)+ cos(pi/2-x)

3). sin(x)cos(x)tan(x)sec(x)csc(x)

The basic identities for 1. are...

$\displaystyle tanx=\frac{sinx}{cosx}$

$\displaystyle cotx=\frac{cosx}{sinx}$

You now need to simplify

$\displaystyle sinx\left(\frac{sinx}{cosx}+\frac{cosx}{sinx}\righ t)$

For 3. you need two more

$\displaystyle secx=\frac{1}{cosx}$

$\displaystyle cscx=\frac{1}{sinx}$

I recommend you try those first as you should find 2. more challenging
• Mar 9th 2010, 10:27 AM
masters
Quote:

Originally Posted by bcavanaugh34
I am having some serious problems with this section. My book is no use to me. Thanks for all the help! It says: use the basic identities to change the expression to one involving only sines and cosines. Then simplify to a basic trigometric function.

1). (sinx)(tanx+cotx)

2). sinx-tan(x)cos(x)+ cos(pi/2-x)

3). sin(x)cos(x)tan(x)sec(x)csc(x)

Hi bcavanaugh34,

[1] $\displaystyle \sin x(\tan x + \cos x)$

$\displaystyle =\sin x \left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)$

$\displaystyle =\sin x \left(\frac{\sin^2 x+\cos^2 x}{\sin x \cos x}\right)$

$\displaystyle {\color{red}\sin^2 x+\cos^2 x = 1}$

$\displaystyle =\frac{1}{\cos x}$

[2] $\displaystyle \sin x - \tan x \cos x + \cos \left(\frac{\pi}{2}-x\right)$

$\displaystyle {\color{red}\cos \left(\frac{\pi}{2}-x\right)=\sin x}$ <===== Co-Function Identity

$\displaystyle =\sin x - \frac{\sin x}{\cos x}\cos x + \sin x$

$\displaystyle =\sin x - \sin x + \sin x$

$\displaystyle =\sin x$

You can try [3] now.

Sorry, Archie, I'm working too slow. But, anyway.....