Write The Expression In Terms Of Sin & Cos and Simplify

[tristatefabricatorsinc]

Write The Expression In Terms Of Sin & Cos and Simplify It

A) cos^ 2 x - sin^2 x
---------------------- I am unsure how to solve this (struggling)
sin x cos x

B) sec x - cosx

On this one I know sec x = 1 / cos x

but I do not know how to get past

1
------ - cos x
cos x

How can you simplify this more? The answer the book came up with I beleive had a cot in it and something else, but I cannot fully remember (Dont have my book with me)

Any assistance would be greatly appreciated!

Thanks

 

[soroban]

Re: Write The Expression In Terms Of Sin & Cos and Simpl

Hello, tristatefabricatorsinc!

Write the expression in terms of sin & cos and simplify.

$\displaystyle A)\;\;\frac{\cos^2x\,-\,\sin^2x}{\sin x\cdot\cos x}$

There are a number of solutions . . .

If you recognize the Double-angle identities:
$\displaystyle \;\;\cos^2\theta\,-\,\sin^2\theta\:=\:\cos2\theta\;\;\;\;\;2\cdot\sin\theta\cdot\cos\theta\:=\:\sin2\theta$

the problem becomes: \(\displaystyle \L\,\frac{\cos2x}{\frac{1}{2}\cdot\sin2x}\)$\displaystyle \:=\;2\cdot\cot2x$


Or we can make two fractions:
\(\displaystyle \L\;\;\;\frac{\cos^2x}{\sin x\cdot\cos x}\,-\,\frac{\sin^2}{\sin x\cdot\cos }\:=\:\frac{\cos x}{\sin x}\,-\,\frac{\sin x}{\cos x}\)$\displaystyle \:=\:\cot x\,-\,\tan x$

$\displaystyle B)\;\;\sec x\,-\,\cos x$

On this one I know $\displaystyle \sec x\,=\,\frac{1}{\cos x}$

but I do not know how to get past: $\displaystyle \,\frac{1}{\cos x}\,-\,\cos x$

We have: \(\displaystyle \L\:\frac{1}{\cos x}\,-\,\frac{\cos x}{1}\)

Get a common denominator: \(\displaystyle \L\:\frac{1}{\cos x}\,-\,\frac{\cos x}{1}\cdot\frac{\cos x}{\cos x} \;=\;\frac{1}{\cos x} \,-\,\frac{\cos^2x}{\cos x} \;=\;\frac{1\,-\,\cos^2x}{\cos x}\;=\;\frac{\sin^2x}{\cos x}\)

\(\displaystyle \L\;\;\;=\:\frac{\sin x}{1}\cdot\frac{\sin x}{\cos x}\)$\displaystyle \;=\;\sin x\cdot\tan x$