Help Proving Trig Identities

I have missedeEvery trig unit ever in school by accident and I just always winged the tests but now I've worked so hard to keep a 90% average (normally i get like a 68% in math) I just don't want it to drop. I have an assignment where I have to complete over 30 questions and get them perfect. I am not asking you to complete the assignment for me but if I post an example of a few could you show me what your doing to get the answer so I can look off that and use it to help me do the rest that would be great. By the way it's proving trig identities.

4.cos^2=sin^2+cos^2-1

15. sin+tan=tan(1+cos)

22.sec^4-tan^4=1+2tan^2

28. sec^2-sin^2=cos^2+tan^2

If you could help i would help me soo much!!

Re: Help Proving Trig Identities

Quote:

Originally Posted by

**vinny2318** I have missedeEvery trig unit ever in school by accident and I just always winged the tests but now I've worked so hard to keep a 90% average (normally i get like a 68% in math)

Don't you have a textbook that you could learn from? If not, there are plenty of websites that you could reference.

Quote:

Originally Posted by

**vinny2318** 4.cos^2=sin^2+cos^2-1

15. sin+tan=tan(1+cos)

22.sec^4-tan^4=1+2tan^2

28. sec^2-sin^2=cos^2+tan^2

None of these make sense. Sine, cosine, and tangent are *functions*. They each require an input. One cannot evaluate "sin" by itself, you need to specify what you are taking the sine of. When you wrote sin^2 Did you mean to write $\displaystyle \sin^2x$? Because even then I can't figure out how to interpret that first identity. $\displaystyle \cos^2x\neq\sin^2x+\cos^2x - 1$, for example.

Re: Help Proving Trig Identities

Hello, vinny2318!

Quote:

$\displaystyle 4.\;\cos^2\!x\:=\:\sin^2\!x+\cos^2\!x-1$

This one is not an identity.

Please check for typos.

Quote:

$\displaystyle 15.\;\sin x+\tan x \:=\:\tan x(1+\cos x)$

The left side is: .$\displaystyle \sin x + \frac{\sin x}{\cos x} \;=\;\frac{\sin x \cos x + \sin x}{\cos x}$

. . . . . . . . . . $\displaystyle =\;\frac{\sin x}{\cos x}(\cos x + 1) \;=\;\tan x(1 + \cos x)$

Quote:

$\displaystyle 22.\;\sec^4\!x-\tan^4\!x\:=\:1+2\tan^2\!x$

The left side is: .$\displaystyle \underbrace{(\sec^2\!x-\tan^2\!x)}_{\text{This is 1}}(\sec^2\!x + \tan^2\!x) \;=\; \underbrace{\sec^2\!x}_{1+\tan^2\!x} + \tan^2\!x $

. . . . . . . . . . $\displaystyle =\;(1 +\tan^2\!x) + \tan^2\!x \;=\;1 + 2\tan^2\!x$

Quote:

$\displaystyle 28.\;\sec^2x-\sin^2x \:=\:\cos^2x+\tan^2x$

The left side is: .$\displaystyle \frac{1}{\cos^2\!x} - (1 - \cos^2\!x) \;=\; \frac{\overbrace{1 - \cos^2\!x}^{\sin^2\!x} + \cos^4\!x}{\cos^2\!x} \;=\; \frac{\sin^2\!x + \cos^4\!x}{\cos^2\!x}$

. . . . . . . . . . $\displaystyle =\;\frac{\sin^2\!x}{\cos^2\!x} + \frac{\cos^4\!x}{\cos^2\!x} \;=\; \tan^2\!x + \cos^2\!x$