verify trigonometric functions

[zachwiggles]

sin^2(x) + cos^2(x) + tan^2(x) = sec^2(x)

 

[galactus]

Prove:

\(\displaystyle sin^{2}(x) + cos^{2}(x) + tan^{2}(x) = sec^{2}(x)\)

When dealing with these problems, it is a good idea to convert everything to sin and cos.

Remember that \(\displaystyle cos^{2}(x)=1-sin^{2}(x)\) and \(\displaystyle sec^{2}(x)=\frac{1}{cos^{2}(x)}=\frac{1}{1-sin^{2}(x)}\)

\(\displaystyle sin^{2}(x)+(1-sin^{2}(x))+\frac{sin^{2}(x)}{1-sin^{2}(x)}=\frac{1}{1-sin^{2}(x)}\)

Hey, we got it all in terms of sin. That's even better. Can you finish?.

 

[soroban]

Hello, zachwiggles!

\(\displaystyle \text{Prive: }\:\sin^2\!x + \cos^2\!x + \tan^2\!x \:=\: \sec^2\!x\)

\(\displaystyle \underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} + \tan^2\!x\ \;=\;\underbrace{1 + \tan^2\!x}_{\text{This is }sec^2\!x} \;=\;\sec^2\!x\)

 

[galactus]

There ya' go. That's better than my myopic method.