1. ## Trigonometric Identities

Verify these identities are true.

$cos(4\theta) = cos^4\theta - 6 sin^2\theta cos^2\theta+sin^4\theta$

$cos^4x=\frac{1}{8}(3 + 4cos(2x) + cos(4x))$

$sec^2\frac{U}{2} = \frac{2sec U}{secU+1}$

------------------

For the first one I am thinking, turn all the sin's into cos'. And try to get it look like $cos4\theta = 2cos^2\theta - 1$

So I converted the all the sin^4 to sin^2 * sin^2, replacing it with 1-cos^2 each, along with the one in the original to get an equation that looks like this:

$cos4\theta = cos^4\theta-6(1-cos^2\theta)cos^2\theta+(1-cos^2\theta)(1-cos^2\theta)$

and tried to simplify...

$cos4\theta = cos^4\theta-(6cos^2\theta-6cos^4\theta)+(1-cos^2\theta)(1-cos^2\theta)$

however I keep getting lost and having difficulty.

The second one I'm completely lost on.

The third one is clearly a half angle identity but how to simplify it is beyond me at this moment.

My professor gave us 40 identities to study for the upcoming test and I've been working on them all night. I'm just a tiny bit frustrated at this point.

2. ## Re: Trigonometric Identities

$cos(4\theta) = cos(2\theta + 2\theta)$

3. ## Re: Trigonometric Identities

Originally Posted by trighelp
Verify these identities are true.

$cos(4\theta) = cos^4\theta - 6 sin^2\theta cos^2\theta+sin^4\theta$

$cos^4x=\frac{1}{8}(3 + 4cos(2x) + cos(4x))$

$sec^2\frac{U}{2} = \frac{2sec U}{secU+1}$

------------------

For the first one I am thinking, turn all the sin's into cos'. And try to get it look like $cos4\theta = 2cos^2\theta - 1$

So I converted the all the sin^4 to sin^2 * sin^2, replacing it with 1-cos^2 each, along with the one in the original to get an equation that looks like this:

$cos4\theta = cos^4\theta-6(1-cos^2\theta)cos^2\theta+(1-cos^2\theta)(1-cos^2\theta)$

and tried to simplify...

$cos4\theta = cos^4\theta-(6cos^2\theta-6cos^4\theta)+(1-cos^2\theta)(1-cos^2\theta)$

however I keep getting lost and having difficulty.

The second one I'm completely lost on.

The third one is clearly a half angle identity but how to simplify it is beyond me at this moment.

My professor gave us 40 identities to study for the upcoming test and I've been working on them all night. I'm just a tiny bit frustrated at this point.
\displaystyle \begin{align*}\cos{4\theta} &= \cos{2\cdot 2\theta} \\ &= \cos^2{2\theta} - \sin^2{2\theta} \\ &= \left(\cos^2{\theta} - \sin^2{\theta}\right)^2 - \left(2\sin{\theta}\cos{\theta}\right)^2 \\ &= \cos^4{\theta} - 2\sin^2{\theta}\cos^2{\theta} + \sin^4{\theta} - 4\sin^2{\theta}\cos^2{\theta} \\ &= \cos^4{\theta} - 6\sin^2{\theta}\cos^2{\theta} + \sin^4{\theta} \end{align*}

4. ## Re: Trigonometric Identities

Originally Posted by trighelp
Verify these identities are true.

$cos(4\theta) = cos^4\theta - 6 sin^2\theta cos^2\theta+sin^4\theta$

$cos^4x=\frac{1}{8}(3 + 4cos(2x) + cos(4x))$

$sec^2\frac{U}{2} = \frac{2sec U}{secU+1}$

------------------

For the first one I am thinking, turn all the sin's into cos'. And try to get it look like $cos4\theta = 2cos^2\theta - 1$

So I converted the all the sin^4 to sin^2 * sin^2, replacing it with 1-cos^2 each, along with the one in the original to get an equation that looks like this:

$cos4\theta = cos^4\theta-6(1-cos^2\theta)cos^2\theta+(1-cos^2\theta)(1-cos^2\theta)$

and tried to simplify...

$cos4\theta = cos^4\theta-(6cos^2\theta-6cos^4\theta)+(1-cos^2\theta)(1-cos^2\theta)$

however I keep getting lost and having difficulty.

The second one I'm completely lost on.

The third one is clearly a half angle identity but how to simplify it is beyond me at this moment.

My professor gave us 40 identities to study for the upcoming test and I've been working on them all night. I'm just a tiny bit frustrated at this point.
\displaystyle \begin{align*} \frac{1}{8}(3 + 4\cos{2x} + \cos{4x}) &= \frac{1}{8}[3 + 4(2\cos^2{x} - 1) + 2\cos^2{2x} - 1] \\ &= \frac{1}{8}[3 + 8\cos^2{x} - 4 + 2\cos^2{2x} - 1] \\ &= \frac{1}{8}(8\cos^2{x} + 2\cos^2{2x} - 2) \\ &= \frac{1}{8}[8\cos^2{x} + 2(2\cos^2{x} - 1)^2 - 2] \\ &= \frac{1}{8}[8\cos^2{x} + 2(4\cos^4{x} - 4\cos^2{x} + 1) - 2] \\ &= \frac{1}{8}[8\cos^2{x} + 8\cos^4{x} - 8\cos^2{x} + 2 - 2] \\ &= \frac{1}{8}(8\cos^4{x}) \\ &= \cos^4{x} \end{align*}

5. ## Re: Trigonometric Identities

Originally Posted by trighelp
Verify these identities are true.

$cos(4\theta) = cos^4\theta - 6 sin^2\theta cos^2\theta+sin^4\theta$

$cos^4x=\frac{1}{8}(3 + 4cos(2x) + cos(4x))$

$sec^2\frac{U}{2} = \frac{2sec U}{secU+1}$

------------------

For the first one I am thinking, turn all the sin's into cos'. And try to get it look like $cos4\theta = 2cos^2\theta - 1$

So I converted the all the sin^4 to sin^2 * sin^2, replacing it with 1-cos^2 each, along with the one in the original to get an equation that looks like this:

$cos4\theta = cos^4\theta-6(1-cos^2\theta)cos^2\theta+(1-cos^2\theta)(1-cos^2\theta)$

and tried to simplify...

$cos4\theta = cos^4\theta-(6cos^2\theta-6cos^4\theta)+(1-cos^2\theta)(1-cos^2\theta)$

however I keep getting lost and having difficulty.

The second one I'm completely lost on.

The third one is clearly a half angle identity but how to simplify it is beyond me at this moment.

My professor gave us 40 identities to study for the upcoming test and I've been working on them all night. I'm just a tiny bit frustrated at this point.
\displaystyle \begin{align*} \sec{2\theta} &= \frac{1}{\cos{2\theta}} \\ &= \frac{1}{\cos^2{\theta} - \sin^2{\theta}} \end{align*}

\displaystyle \begin{align*} \sec{2\theta} + 1 &= \frac{1}{\cos^2{\theta} - \sin^2{\theta}} + 1 \\ &= \frac{1 + \cos^2{\theta} - \sin^2{\theta}}{\cos^2{\theta} - \sin^2{\theta}} \\ &= \frac{\sin^2{\theta} + \cos^2{\theta} + \cos^2{\theta} - \sin^2{\theta}}{\cos^2{\theta} - \sin^2{\theta}} \\ &= \frac{2\cos^2{\theta}}{\cos^2{\theta} - \sin^2{\theta}}\end{align*}

Therefore

\displaystyle \begin{align*} \frac{ 2\sec{2\theta} }{ \sec{2\theta} + 1 } &= \frac{ \frac{2}{ \cos^2{\theta} - \sin^2{\theta} } }{ \frac{ 2\cos^2{\theta} }{ \cos^2{\theta} - \sin^2{\theta} } } \\ &= \frac{2}{2\cos^2{\theta}} \\ &= \frac{1}{\cos^2{\theta}} \\ &= \sec^2{\theta} \end{align*}

So we have $\displaystyle \sec^2{\theta} = \frac{ 2\sec{2\theta} }{ \sec{2\theta} + 1 }$, and if you let $\displaystyle \theta = \frac{U}{2}$ your proof is complete.

6. ## Re: Trigonometric Identities

Hello, trighelp!

$\cos 4\theta \:=\:\cos^4\!\theta - 6\sin^2\!\theta\cos^2\!\theta+\sin^4\!\theta$

$\cos4\theta \;=\;(\cos 2\theta)^2 - (\sin2\theta)^2$

. . . . . $=\;(\cos^2\!\theta - \sin^2\!\theta)^2 - (2\sin\theta\cos\theta)^2$

. . . . . $=\;\cos^4\!\theta - 2\sin^2\!\theta\cos^2\!\theta + \sin^4\!\theta - 4\sin^2\!\theta\cos^2\!\theta$

. . . . . $=\;\cos^4\!\theta - 6\sin^2\!\theta\cos^2\!\theta + \sin^4\!\theta$

$\cos^4x \:=\:\tfrac{1}{8}(3 + 4\cos2x + \cos 4x)$

$\cos^4\!x \;=\;\left(\cos^2\!x\right)^2 \;=\;\left(\dfrac{1+\cos2x}{2}\right)^2 \;=\;\tfrac{1}{4}\left(1 + 2\cos2x + \cos^2\!2x\right)$

. . . . . $=\;\tfrac{1}{4}\left(1 + 2\cos 2x + \dfrac{1+\cos4x}{2}\right) \;=\;\tfrac{1}{8}\left(2 + 4\cos2x + 1 + \cos4x\right)$

. . . . . $=\;\tfrac{1}{8}\left(3 + 4\cos2x + \cos4x\right)$

$\sec^2\tfrac{u}{2} \:=\: \dfrac{2\sec u}{\sec u+1}$

$\sec^2\tfrac{u}{2} \;=\;\frac{1}{\cos^2\frac{u}{2}} \;=\;\dfrac{1}{\frac{1+\cos u}{2}} \;=\;\frac{2}{1+\cos u}$

Multiply by $\frac{\sec u}{\sec u}:\;\;\frac{\sec u}{\sec u}\cdot\frac{2}{1 + \cos u} \;=\;\frac{2\sec u}{\sec u + 1}$

7. ## Re: Trigonometric Identities

Hey!

Thanks so much Prove It and Soroban.

It clicks now and makes perfect sense. I think I can only stare at so many identities before my brain melts. Brain melt point is 38 trig identities.

Also, TKHunny, that's a super good point. My professor told us to only use simplification, or rather did not present expansion as an option. So while solving these identities I've been doing half expansion... half simplification until I meet somewhere in the middle... if that makes sense. :P Then rewriting the whole thing to look like pure simplification to mesh with the professors teaching style as that is how she wants it on the test.

I attempted to edit my initially post. I saw a prefix "Solved" option when I first made the post but I couldn't find it, if anyone knows how to do that. This is the first time I've ever asked for math help online.

I'll tag you all with a thanks.

8. ## Re: Trigonometric Identities

Also, Prove It, what is this?

$\displaystyle b = \frac{\sum(x - \bar{x})(y - \bar{y})}{\sum(x - \bar{x})^2}$

I figured out what the ba ba ro ma ma bit is. However I don't recognize those symbols. Trig is the highest I've gone in math so far! The fact that I can't decipher this is maddening :P

9. ## Re: Trigonometric Identities

Originally Posted by trighelp
Also, Prove It, what is this?

$\displaystyle b = \frac{\sum(x - \bar{x})(y - \bar{y})}{\sum(x - \bar{x})^2}$

I figured out what the ba ba ro ma ma bit is. However I don't recognize those symbols. Trig is the highest I've gone in math so far! The fact that I can't decipher this is maddening :P
It's a Least Squares Regression Coefficient.

I suggest you go to http://www.mathhelpforum.com/math-he...-a-138041.html and download Issue 1 of the MHF e-zine, and read the article I wrote on Multiple Variable Regression.