1. ## Trig theory help...

Take home stuff due tomorrow...the applied trig I understand...heights of mountains and such, but the theory is lost on me. The following are all questions from the assignment to which I do not know the answers. Any help would be appreciated. Thank you in advance.

Prove the identity.

1) ([5(csc^2)x] + 4cscx - 1)/[(cot^2)x] = [5cscx-1]/[cscx-1]

2) cosx/secx-1 - cosx/secx+1 = 2cosx/tan^2x

3) csc^4 - cot^4x = csc^2x + cot^2x

4) cos[x+[y-(pi/2)]] = sin(x+y)

2. next set is

1) 4 cot 4x = 2 cot 2u - 2 tan 2u

2) cos 4u = cos^2 2u - sin^2 2u

3) 4csc2x = 2csc^2x tan x

edit: I got these worked out, thanks for the help!

3. I'm just going to post them all even while I still work to figure this out.

write the expression as the sine, cosine, or tan of an angle...(multiple choice)
cos(pi/2)cos(pi/11)+sin(pi/2)sin(pi/11)
^This one I actually tried to solve using the calculator, but I couldn't get it to work out for me.

Find all solutions to the given equation in interval [0,2pi]

A) cosx = sin2x

B) cos^2[x/2] =cos^2x

Again, any help would be appreciated...I'm LOST.

edit: last problem, I promise...this honestly seems like it should be basic stuff, but my brain is fried.

position of a weight on a spring is s(t) = -9cos 8 pi t inches after t seconds. What is max height the weight reaches above the equilibrium position and when does it reach said max height?

Thanks,

- JM

4. Hello, HeLlIoN!

You are expected to know these identities:

. . $\sec^2\!\theta \;=\;\tan^2\!\theta + 1 \quad\Longleftrightarrow\quad \sec^2\1x - 1 \:=\:\tan^2\!\theta$

. . $\csc^2\theta \:=\:\cot^2\!\theta + 1 \quad\Longleftrightarrow\quad \csc^2\!\theta - \cot^2\!\theta \:=\:1$

$1)\;\dfrac{5\csc^2\!x + 4\csc x - 1}{\cot^2\!x} \;=\;\dfrac{5\csc x-1}{\csc x-1}$

The left side is:

. . $\displaystyle \frac{(5\csc x - 1)(\csc x + 1)}{\csc^2\!x - 1} \;=\;\frac{(5\csc x - 1)(\csc x + 1)}{(\csc x - 1)(\csc x + 1)} \;=\;\frac{5\csc x - 1}{\csc x - 1}$

$2)\; \dfrac{\cos x}{\sec x-1} - \dfrac{\cos x}{\sec x+1} \;=\; \dfrac{2\cos x}{\tan^2\!x}$

On the left side, get a common denominator:

$\displaystyle \frac{\cos x}{\sec x-1}\cdot\frac{\sec x+1}{\sec x +1} - \frac{\cos x}{\sec x + 1}\cdot\frac{\sec x - 1}{\sec x - 1} \;=\;\frac{\cos x(\sec x + 1) - \cos x(\sec sx - 1)}{(\sec x - 1)(\sec x + 1)}$

. . $\displaystyle =\;\frac{(1 + \cos x) - (1 - \cos x)}{\sec^2\!x-1} \;=\;\frac{2\cos x}{\tan^2\!x}$

$3)\; \csc^4\!x - \cot^4\!x \;=\; \csc^2\!x + \cot^2\!x$

$\text{Factor the left side: }\:\underbrace{(\csc^2\!x - \cot^2\!x)}_{\text{This is 1}}(\csc^2\1x + \cot^2\1x) \;=\;\csc^2\!x + \cot^2\!x$

$4)\; \cos\left[x+(y-\frac{\pi}{2})\right] \;=\; \sin(x+y)$

We need these compound angle identities:

. . $\sin(A \pm B) \;=\;\sin A\cos B \pm \cos A\sin B$

. . $\cos(A \pm B) \;=\;\cos A\cos B \mp \sin A\sin B$

The left side is:

$\cos[x + (y - \frac{\pi}{2})] \;=\;\cos x\cos(y-\frac{\pi}{2}) - \sin x\sin(y-\frac{\pi}{2})$

. . $=\; \cos x\left(\cos y\cos\frac{\pi}{2} + \sin y\sin\frac{\pi}{2}\right) - \sin x\left(\sin y\cos\frac{\pi}{2} - \cos y\sin\frac{\pi}{2}\right)$

. . $=\; \cos x(\cos y\cdot 0 + \sin y\cdot1) - \sin x(\sin y\cdot 0 - \cos y\cdot 1)$

. . $=\;\cos x\sin y -+ \sin x\cos y$

. . $=\; \sin(x+y)$

5. Thank you! I'm going to try the ones in my second post based on what you showed, does anyone have any thoughts on the word problems in post #3?

Also, it wasn't really explained to me, but what is the math formula represented by tan/sin/cos? What I mean is; when I hit one of those keys, what is the calculator actually doing? I think that will help me understand. Thanks again!

- JM

6. Originally Posted by HeLlIoN
write the expression as the sine, cosine, or tan of an angle...(multiple choice)
cos(pi/2)cos(pi/11)+sin(pi/2)sin(pi/11)
This one, I came out with:

$sin(13pi/22)$

Would that be correct?

7. Originally Posted by HeLlIoN
position of a weight on a spring is s(t) = -9cos 8 pi t inches after t seconds. What is max height the weight reaches above the equilibrium position and when does it reach said max height?
I got a height of 9 inches after .13 seconds.

8. Alright...I think I have all the rest figured out, just can't get this last one...can someone walk me through the steps?

Find all solutions to the given equation in interval [0,2pi]
cos^2[x/2] =cos^2x

Thanks

9. Originally Posted by HeLlIoN
Alright...I think I have all the rest figured out, just can't get this last one...can someone walk me through the steps?

Find all solutions to the given equation in interval [0,2pi]
cos^2[x/2] =cos^2x

Thanks

$\displaystyle \cos\left({x\over2}\right)=\pm \sqrt{{1+\cos x}\over2}\ \ \to\ \ \cos^2\left({x\over2}\right)={{1+\cos x}\over2}$

$\displaystyle {{1+\cos x}\over2}=\cos^2 x$
Therefore: $\displaystyle 0=2\cos^2 x-\cos x -1$.