# Proving trig identities

1. Jun 22, 2008

### projection

exam coming up...need some help with these identities for practise.

prove the following:

A)

$$\frac{tan^3x}{1+tan^2x}+\frac{cot^3x}{1+cot^2x}=\frac{1-2sin^2xcos^2x}{sinxcos}$$

B)

$$sec^6x-tan^6x=1+3tan^2xsec^x$$

C)

$$cos^4x=\frac{3}{8}+\frac{1}{2}cos2x+\frac{1}{8}cos4x$$

D)
$$cos\frac{x}{2}=\pm\sqrt{\frac{1+cosx}{2}}$$

i tried really hard but can't seem to get any progress. i tried to get common denominator, use other identies to transform but nothing, can i get some advice on what method to use or where to start?

2. Jun 22, 2008

### gunch

I have only had a go at A), but it seems to simply be repeated application of common identities. For instance creating a common denominator in the first we get,
$$\frac{\tan(x)(\tan^2(x) + 1) + \cot(x)(\cot^2(x) + 1)}{2+\tan^2(x) + \cot^2(x)} = \frac{1-2\sin^2(x)\cos^2(x)}{\sin(x)\cos(x)}$$
We can replace some of the tangents and cotangents by sines and cosine by multiplying through with $$\sin^2(x)\cos^2(x)$$ which yields:
$$\frac{\frac{\sin^3(x)}{\cos(x)}(\sin^2(x) + \cos^2(x)) + \frac{\cos^3(x)}{\sin(x)}(\sin^2(x) + \cos^2(x))}{2\sin^2(x)\cos^2(x)+\sin^4(x) + \cos^4(x)} = \frac{1-2\sin^2(x)\cos^2(x)}{\sin(x)\cos(x)}$$
The $$\sin^4(x) + \cos^4(x)$$ can easily be simplified to $$(\sin^2(x) + \cos^2(x))^2 - 2\sin^2(x)\cos^2(x)$$. At the same time we see that multiplying both sides by $$\sin(x)\cos(x)$$ will simplify the expression extremely:
$$\frac{\sin^4(x) + \cos^4(x)}{2\sin^2(x)\cos^2(x) + 1 - 2\sin^2(x)\cos^2(x) } = 1-2\sin^2(x)\cos^2(x)$$

$$1-2\sin^2(x)\cos^2(x) = 1-2\sin^2(x)\cos^2(x)$$

I hope this helps. Simply try experimenting with the expressions. Reduce the number of different trigonometric functions you're working with, see if you can get $$cos^2 x + \sin^2 x$$ introduced somehow, and simplify. Also for C and D remember the double-angle formulas.

3. Jun 22, 2008

### projection

I think that were are supposed to prove them in the LHS and RHS formate. so i don;t think multilpying something out is allowed...

4. Jun 22, 2008

### the1ceman

$$\frac{\tan^3x}{1+\tan^2x}+\frac{\cot^3x}{1+\cot^2x}=\frac{1-2\sin^2x\cos^2x}{\sinx\cosx}$$
A good place to start is to convert everything to sin and cos on the LHS(i will use s for sin(x) and c for cos(x)):

$$\frac{\tan^3x}{1+\tan^2x}+\frac{\cot^3x}{1+\cot^2x}=\frac{\frac{s^3}{c^3}}{1+\frac{s^2}{c^2}}+\frac{\frac{c^3}{s^3}}{1+\frac{c^2}{s^2}}=\frac{\frac{s^3}{c}}{c^2+s^2}+\frac{\frac{c^3}{s}}{s^2+c^2}=\frac{s^3}{c}+\frac{c^3}{s}$$

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

$$=\frac{s^4+c^4}{sc}$$

So the denominator is done. Just need to show that s^4+c^4=1-2(s^2)(c^2).Well

$$s^4+c^4=(1-c^2)^2+c^4=1-2(c^2-c^4)=1-2c^2(1-c^2)=1-2c^2s^2$$

And we are done! Of course there are many ways to do trig identities, a good place to usually start is convert everything to sines ans cosines and then use the easy trig identities.

5. Jun 22, 2008

### Defennder

Firstly try to express the LHS as fractions with $$\cos x \ \mbox{and} \ \sin x$$ as denominators. Use the trigo identities for tan,sec and csc, in order to express everything in sin and cos. From there on it's just algebra once you get the fraction with the correct denominator.

You have $$\tan^2 x + 1 = \sec^2 x$$ Cube them on both sides and apply the result. This is the only trigo identity you'll need for this. And I believe you meant [itex]3\tan^2x(\sec^2x)[/tex].

For this you'll probably want to show the reverse; work on RHS and try to reduce it to LHS. Note that the LHS is simply cos^4 x. That means you should try to convert all the trigo functions on the RHS to functions of cos x and it's powers only.

For this it's easier to start on the right and reduce it to the left. Just use the identity $$\cos (2x) = \cos^2 x -\sin^2 x$$.

6. Jun 22, 2008

### tiny-tim

Hi projection!
Oh come on guys …

you're all supposed to know trigonometric identities like 1 + tan²x = sec²x, 1 + cot²x = cosec²x.
Hint: use the algebraic formula for factoring a³ - b³.

And for C) and D) you'll need the same elementary trig identities.