Question

Select the expression below that is identical to
\[\frac{ 1 }{ 2 }\sec ^{2}\alpha \cot \alpha\]

A.) csc2 alpha
B.)1/2
C.)cos 2 alpha
D.) sin alpha/2
E.) cos alpha/2

Answer 1

Use these formulas and than simplify the expression.
\[ \sec \alpha = \frac{1}{\cos \alpha}\]
\[ \cot \alpha = \frac{\cos \alpha}{\sin \alpha}\]

Answer 2

\[\frac{ 1 }{ 2 }(\frac{ 1 }{ \cos \alpha })(\frac{ \cos \alpha }{ \sin \alpha })\]

is that right?

Answer 3

except the 1/cos a should be squared...

Answer 4

Yes that's right.

Answer 5

Now you can simplify a little bit.

Answer 6

\[(\frac{ 1 }{ 2\cos ^{2}\alpha } )(\frac{ \cos \alpha }{ \sin \alpha})\]

Answer 7

You can cancel the \(\cos \alpha\).

Answer 8

so now its \[(\frac{ 1 }{ 2 \cos \alpha })(\frac{ 1 }{ \sin \alpha })\]

Answer 9

OK, now you can use the fact that \(\sin(2\alpha) = 2\cos \alpha\sin\alpha \) and simplify the denominator.

Answer 10

oh oh okay so then we get 1/sin 2 alpha

Answer 11

That's right. But it can be written as \(\csc 2\alpha\).

Answer 12

Because \(\frac{1}{\sin \alpha} = \csc \alpha\).

Answer 13

Ugh i'm looking at my list of identities and I just like dont see the ones i need they you point them out and i feel like facepalming
I'm so sorry.

Answer 14

which woul dmake it answer C.)

Answer 15

No, the correct answer is A. It's \(\csc 2\alpha\) not \(\cos 2\alpha\)

Answer 16

Annnnd you're correct. :) thank you oodles. I'll have to look at my list more closely. I really appreciate the patience and help. I'm not exactly the brightest crayon in the box.

Answer 17

Hope it helped. These trigonometric identities can be somehow tricky to remember.

Answer 18

Just a little bit. Doesn't help that I dont care a bloody lick about 'em. Just got accept to the college I want to go to so I've hit the wall.