MHB Establishing Identities - Possible Misprint on Worksheet

[Dundee3]

Hey guys, I'm just going over my work files for this week and I've noticed 2 problems that have fooled me, and seem a little bizarre for my current mathematical level. They do involve trig, and a part of me is hoping for a misprint on the sheet, but I'm more sure that it's my own inadequacies that are hurting me here.

The first problem is listed as follows:

sin((\pi/4) + x) = sqrt2/2(cosxx +sinx)

My objective there is to "Establish the Identity", and the 2 'x's right next to each other on the right side of the equal sign is confusing to me.

Secondly:

cos(\alpha+\beta)/cos\alphacos\beta = cot\beta - tan\alpha

That second problem is also under the objective "Establish the Identity". For problem 2, I feel like I've solved it as far as I can, but the answer I keep leading back to once I've gone through standard replacement procedures for identity establishment I keep getting:

cos(\alpha + \beta)/cos/alphacos\beta = (cos\beta - sin\alpha)/sin\betacos\alpha

As you guys might have guessed, our previous lesson was over the sum and difference identities and using them to simplify expressions. Any help would be incredible.

Thank you guys.

 

[skeeter]

extra x must be a typo ...

$$\sin \left(\frac{\pi}{4} + x\right) =$$

$$\sin \left(\frac{\pi}{4}\right) \cos{x} + \cos\left(\frac{\pi}{4}\right) \sin{x} =$$

$$\frac{\sqrt{2}}{2} \cos{x} + \frac{\sqrt{2}}{2} \sin{x} =$$

$$\frac{\sqrt{2}}{2} \left(\cos{x}+\sin{x}\right)$$

 

[skeeter]

I disagree with the outcome of your 2nd identity ...

$$\frac{\cos(\alpha+\beta)}{\cos{\alpha}\cos{\beta}} =$$

$$\frac{\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}} =$$

$$\frac{\cos{\alpha}\cos{\beta}}{\cos{\alpha}\cos{\beta}}-\frac{\sin{\alpha}\sin{\beta}}{\cos{\alpha}\cos{\beta}} =$$

$$1 - \tan{\alpha}\tan{\beta} =$$

$$\cot{\beta}\tan{\beta} - \tan{\alpha}\tan{\beta} =$$

$$\tan{\beta}(\cot{\beta} - \tan{\alpha})$$

 

[Dundee3]

Thank you so much!

Correcting the typo in the first problem made everything fall perfectly into place.

You did an incredible job and I thank you for it.

My only concern was on the second expression. Doing the computation we're left with:

tanB(cotB - tanA)

And, I've noticed on my original worksheet that the outcome we're "supposed" to get is simply:

CotB - TanA (without the term on the outside of the parentheses)

Is this another typo? Or another of my delusions?

 

[skeeter]

My only concern was on the second expression. Doing the computation we're left with:

tanB(cotB - tanA)

And, I've noticed on my original worksheet that the outcome we're "supposed" to get is simply:

CotB - TanA (without the term on the outside of the parentheses)

Is this another typo? Or another of my delusions?

as I stated, I disagree with what's on the WS ... if my work has an error (and it may), then one of the other helpers will point it out.

 

[skeeter]

you can check both ... pick two random angles for alpha & beta whose cosine is not zero.

 

[Dundee3]

Thank you so much =) Is there a way I can repay you?

 

[skeeter]

Thank you so much =) Is there a way I can repay you?

$ 10 million would be nice (just kidding) ... just pay it forward to someone else.

 

[topsquark]

$ 10 million would be nice (just kidding) ... just pay it forward to someone else.

skeeter old pal. You really don't understand the whole reward thing, do you?

Take a chess board. Place one penny on the first square, two pennies on the next, four pennies on the next, 8, 16, etc. until the whole board is full. All you want is all the money on the board. Say you will accept payment plans for merely 0.01% yearly. His scions will take the remainder of the loan and pay it to your scions for however many generations it will take. :)

That's my standard bribe for changing grades.

-Dan

 

[skeeter]

I wouldn't live long enough to spend it all ... but I'd have a hell of a lot of fun trying. ;)