Help Proving Trig Identities

**vinny2318** · June 4th 2012, 01:59 PM

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!!

**Reckoner** · June 4th 2012, 02:18 PM

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.

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 $\sin^2x$ ? Because even then I can't figure out how to interpret that first identity. $\cos^2x\neq\sin^2x+\cos^2x - 1$ , for example.

**Soroban** · June 4th 2012, 06:17 PM

Hello, vinny2318!

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

This one is not an identity.
Please check for typos.

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

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

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

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

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

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

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

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

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

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