proving that an equation is an identity

[Guest]

Hi, i need help with these questions

4/cos^2x -5 = 4 tan ^2 x-1

(sin x + cos x )2 = 1 + 2sin x cos x

thanks

 

[galactus]

4/cos^2x -5 = 4 tan ^2 x-1

Do you mean the following?

$\displaystyle \frac{4}{\cos^{2}{(x)}\,-\,5}\,=\,4\tan^{2}{(x)}\,-\,1$

Use appropriate parentheses when posting lest your problem may seem ambiguous.

 

[Guest]

yes

 

[Mrspi]

4/cos^2x -5 = 4 tan ^2 x-1

(sin x + cos x )2 = 1 + 2sin x cos x

I think I can help you with the second one, IF you mean this:

(sin x + cos x)[sup:m0nvekmv]2[/sup:m0nvekmv] = 1 + 2 sin x cos x

Square the binomial on the left side. (sin x + cos x)[sup:m0nvekmv]2[/sup:m0nvekmv] is
sin[sup:m0nvekmv]2[/sup:m0nvekmv] x + 2 sin x cos x + cos[sup:m0nvekmv]2[/sup:m0nvekmv] x

Rearrange:

(sin[sup:m0nvekmv]2[/sup:m0nvekmv] x + cos[sup:m0nvekmv]2[/sup:m0nvekmv] x) + 2 sin x cos x

Now, what is sin[sup:m0nvekmv]2[/sup:m0nvekmv] x + cos[sup:m0nvekmv]2[/sup:m0nvekmv] x?

Substitute, and you're done!

 

[Guest]

^ thanks

 

[soroban]

Hello, Aka!

Please check the first equation. As written, it is not true

$\displaystyle \frac{4}{\cos^2x\,-\,5}$$\displaystyle \:=\:4\tan^2x\,-\,1$

$\displaystyle \text{Let }x\,=\,\frac{\pi}{4}$

$\displaystyle \text{Left side: }\,\frac{4}{\cos^2\left(\frac{\pi}{4}\right)\,-\,5}\; = \;\frac{4}{\left(\frac{1}{\sqrt{2}}\right)^2\,-\,5}\; = \;\frac{4}{\frac{1}{2}\,-\,5}\; = \;\frac{4}{-\frac{9}{2}} \; = \;-\frac{8}{9}$

$\displaystyle \text{Right side: }\,4\tan^2\left(\frac{\pi}{4}\right)\,-\,1\;=\;4\cdot1^2\,-\,1\;=\;3$

 

[galactus]

$\displaystyle \frac{1}{cos^{2}(x)}=sec^{2}(x)$

Now, use the identity: $\displaystyle sec^{2}(x)=tan^{2}(x)+1$

 

[pbrandon]

Thank you.

Paul