Verifying Six Trigonometric Identities

[math-a-phobic]

Hi! I was trying to verify some identities and got stuck. Please help! I'm really confused. The only problem I actually finished was problem 6, but I'm not positive its right. The work I did can be seen below.


Problem 1:

Problem 2:

Problem 3:

Problem 4:

Problem 5:

Problem 6:

Thanks for the help! I really appreciate it. Sorry this is such a long topic.

 

[galactus]

Try subbing some known identities instead of x and y.

For example, #1, $\displaystyle tan^{2}(x)=sec^{2}(x)-1$

 

[soroban]

Re: Verifying Identities

Hello, math-a-phobic!

These are pretty straight-forward.
They use the standard identities:
$\displaystyle \;\;\sin^2\theta\,+\,\cos^2\theta\:=\:1$
$\displaystyle \;\;\sec^2\theta\:=\:\tan^2\theta\,+\,1$
$\displaystyle \;\;\csc^2\theta\:=\:\cot^2\theta\,+\,1$
Only #6 requires a certain "trick".

$\displaystyle 1)\;4\tan^4x\,+\,\tan^2x\,-\,3\;=\;\sec^2x(4\tan^2x\,-\,3)$

Factor the left side: $\displaystyle \,(\underbrace{\tan^2x\,+\,1})(4\tan^2x\,-\,3)$

. . . . . And we have: $\displaystyle \;\;\;\sec^2x(4\tan^2x\,-\,3)$

$\displaystyle 2)\;\csc^4x\,-\,2\csc^2x\,+\,1\;=\;\cot^4x$

Factor the left side: $\displaystyle \,(\csc^2x\,-\,1)^2$

. . . . And we have: $\displaystyle \;\;\;(\cot^2x)^2\:=\:\cot^4x$

$\displaystyle 3)\;\sin x(1\,-\,2\cos^2x\,+\,\cos^4x)\;=\;\sin^5x$

Factor the left side: $\displaystyle \,\sin x(\underbrace{1\,-\,\cos^2x})^2$

. . . . And we have: $\displaystyle \;\;\;\sin x(\sin^2x)^2 \;= \;\sin x(\sin^4x) \;= \; \sin^5x$

$\displaystyle 4)\;\sec^4\theta\,-\,\tan^4\theta \;= \;1\,+\,2\tan^2\theta$

Factor the left side: $\displaystyle \,(\underbrace{\sec^2\theta\,-\,\tan^2\theta})(\sec^2\,+\,\tan^2\theta)$

. . . . And we have: $\displaystyle \;\;\;\;\;\;\;\;1\,\cdot\,[(\overbrace{1\,+\,\tan^2\theta})\,+\,\tan^2\theta) \;= \;1\,+\,2\tan^2\theta$

$\displaystyle 5)\;\csc^4\theta\,-\,\cot^4\theta\;=\;2\csc^2\theta\,-\,1$

Factor the left side: $\displaystyle \,(\underbrace{\csc^2\theta - \cot^2\theta})(\csc^2\,+\,\cot^2\theta)$

. . . . And we have: $\displaystyle \;\;\;\;\;\;\;1\,\cdot\,[\csc^2\theta\,+\,(\overbrace{\csc^2\theta\,-\,1})] \;= \;2\csc^2\theta\,-\,1$

\(\displaystyle 6)\L\;\frac{\sin\beta}{1\,-\,\cos\beta} \;= \; \frac{1\,+\,\cos\beta}{\sin\beta}\)

On the left side, multiply top and bottom by $\displaystyle (1\,+\,\cos\beta):$

\(\displaystyle \L\;\frac{\sin\beta}{1\,-\,\cos\beta}\,\cdot\,\frac{1\,+\,\cos\beta}{1\,+\,\cos\beta}\;=\;\frac{\sin\beta(1\,+\,\cos\beta)}{1\,-\,\cos^2\beta} \;= \;\frac{\sin\beta(1\,+\,\cos\beta)}{\sin^2\beta} \;= \;\frac{1\,+\,\cos\beta}{\sin\beta}\)