So recently had a quiz in my math class, which I flunked. I had been gone for a while, so I was wondering if anyone would be willing to take me through the steps to all the problems. I'd like to get a better grasp on it.

http://i96.photobucket.com/albums/l194/anti-you92/Problem.png

Hopefully someone can help. Thanks in advance!

I'll pick #4. Ans: is (link) cos(theta)=-sqrt(2)/2 - Wolfram|Alpha (http://www.wolframalpha.com/input/?i=cos%28theta%29%3D-sqrt%282%29%2F2)

Multiply top and bottom by sqrt(2)
To get 3*PI/4 you just have to remember it. It's been way too long. That's why it too hard for me. Too much memorization.

1. sin^{-1}(-\frac12) = sin^{-1}(\frac12)

Do you know the shape of the graph of y = sin x?

I always refer to it (in my head) to find the exact values. The graph starts from (0,0), then to (30 degrees, 0.5) then to (60 degrees, \frac{\sqrt3}{2} and finally to (90 degrees, 1) for the first quadrant.

Well, sin^{-1}(\frac12) = 30^{\small{o}}

2. You should know that tan\ 60 = \sqrt3 (and you should know all the trig ratio of sin, cos and tan for 0, 30, 45, 60 and 90 degrees in exact form).

So, tan^{-1}(\sqrt3) = 60\ degrees.

Then, cos 60 = 0.5

3. I think that one must have a mistake somewhere. There is no simple exact form for that (perhaps none)

4. cos\theta = -\frac{\sqrt2}{2}

You should know that cos\ 45 = \frac{1}{\sqrt2}

From that, you'll see that you need to make the denominator with the square root of 2.

-\frac{\sqrt2}{2} = -\frac{\sqrt2}{2} \times \frac{\sqrt2}{\sqrt2} = -\frac{2}{2\sqrt2} = -\frac{1}{\sqrt2}

From there, you 'see' your y = cos x graph and you'll see that the value of theta has to be 135 and 225 degrees. (for angles from 0 to 360 degrees inclusive)

5. sin2\theta = \frac{\sqrt3}{2}

Remember what angle gives square root 3 by 2? That angle is 60 degrees.

So, 2\theta = 60\ degrees

You can solve from there.

6. A simple way to do that is to replace cos theta by another variable, like for example x.

cos^2\theta +2 cos \theta +1 = 0

x^2 + 2x + 1 = 0

Solve, you should have
x = -1
and
x = -1

So, substitute back to the trigonometric ratio.

cos\theta = -1

Recall that the angle giving -1 with the cos ratio is 180 degrees.

Hence, theta = 180 degrees.

(2) Memorizing was a pain.

Not quite, if you know how to tackle it :)

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Do you see here? The first one is an equilateral. The angles are all 60 degrees. The diagram tells you everything about the ratios. And the second one is a square.

sin\ 60 = \frac{\sqrt3}{2} for example :)

Jiminy Christmas. That makes it a snap.

With Oscar Had a Headache Over Algebra, the pythagorean theorm, 1,2 and the def of sec and csc and cot (the latter 3 are easy). That I could have handled.

I know all of the other stuff in your picture. Like the angles.

Where were you 36 years ago?

Waw. That makes it easy.

See? Just need to remember two triangles. The square roots come in naturally after you use them a couple of times. :)

In the left bbisect triangle isn't the hypotenuse the sqrt(3)


e.g. should the sqrt(3) and the 2 reversed?

Because 2^2 + 1^2 does not equal the sqrt(3)

You are missing a simple point KISS.

Pythagoras' Theorem:
The sum of the squares of two sides of a right angled triangle which are forming the right angle is equal to the square of the hypotenuse.

So,

2^2 = 1^2 + (\sqrt3)^2

Yep. Your right. BTW. I had all the way to differential equations and whatever dealt with lots of matrices.

I also had dynamics which was a real pain. I also had a structures course (basically statics for small buildings).

I am quite at ease with dynamics, linear dynamics I mean. Circular motion... not so much at ease...

Well, the most difficult things for me to remember are the formulae in simple harmonic motion. I always seem to mess up the variables :(

The goofy stuff like the velocity of a fly on a propeller of an airplane relative to the ground, Forget doing it now. I could barely do it then.

Well, I'm not versed in relativity yet :) I hope I'll get the thing quick! ;)

Did that too. 3rd semester Physics.