Verifying Trigonometric Identities.

M

mrtrixta

New member
Joined
Feb 16, 2008
Messages
12
Ok So I'm stuck on another problem. Gee, when I think that I'm getting it I run into something else.

Verify that:

\(\displaystyle {{{\rm cos }\alpha } \over {\sec \alpha {\rm }}}{\rm } + {\rm }{{\sin {\rm }\alpha } \over {csc\alpha }} = {\rm sec}^{\rm 2} - {\rm tan}^{\rm 2}\)

I'm not sure what to do, according to my text I'd have to first work on the side that is more difficult, which is the left side. My thought was to make a common denominator then add both sides:

\(\displaystyle {{\cos \alpha \csc \alpha } \over {\sec \alpha \csc \alpha }} + {{\sin \alpha \sec \alpha } \over {\sec \alpha \csc \alpha }} = {{\cos \alpha \csc \alpha + \sin \alpha \sec \alpha } \over {\sec \alpha \csc \alpha }}{\rm ,}\)

From there I'm stuck. I was thinking that I change all that into terms of sin and cos. The thing is that I'm not sure how to proceed from there. It all looks a little messy to me. Please help! Thanks! Wouls I have to work on both sides??
 
For trig identities it is very tempting to make them more difficult than they really are.

I would suggest changing everything to sines and cosines: (forgive me, I don't know LaTex yet)

cos/(1/cos) + sin/(1/sin)

multiply by the reciprocal:

cos^2 + sin^2 = 1
1 = sec^2 - tan^2
 
That's amazing!! I mean I did get to that point but I didn't even think about looking at it like that. Just a bit too much to take in. :D Thanks a lot!!
 
Here's another one that I need some help with:

Verify that:

\(\displaystyle (1 - \cos ^2 \alpha )(1 + \cos ^2 \alpha ) = 2\sin ^2 \alpha - \sin ^4 \alpha\)

My next step was work on the left side and for the first part of the equation I substitued for sin so I got this:

\(\displaystyle (\sin ^2 \alpha )(1 + \cos ^2 \alpha )\)

I guess I can work on the right side to so I just factored sin out. I got this.

\(\displaystyle \sin \alpha (2\sin \alpha - \sin ^3 \alpha )\)

Maybe I'm looking too much into it. Is there any way to look at this so that it all makes sense. Thanks!!
 
Ok so I did that and I got:

\(\displaystyle \sin ^2 \alpha (2 - \sin ^2 \alpha )\)

I still can't see it. :?:
 
mrtrixta said:
Here's another one that I need some help with:

Verify that:

\(\displaystyle (1 - \cos ^2 \alpha )(1 + \cos ^2 \alpha ) = 2\sin ^2 \alpha - \sin ^4 \alpha\)

My next step was work on the left side and for the first part of the equation I substitued for sin so I got this:

\(\displaystyle (\sin ^2 \alpha )(1 + \cos ^2 \alpha )\)

sin[sup:raejfl7a]2[/sup:raejfl7a]a(1 + 1 - sin[sup:raejfl7a]2[/sup:raejfl7a]a)

2sin[sup:raejfl7a]2[/sup:raejfl7a]a - sin[sup:raejfl7a]4[/sup:raejfl7a]a
Click to expand...
 
Geez, I really need to pay more attention to details with all these formulas. I didn't think I could do something like that. Thanks ever so much.
 
Verify that:

\(\displaystyle {{\cot \alpha + 1} \over {\cot \alpha - 1}} = {{1 + \tan \alpha } \over {1 - \tan \alpha }}\)

I usually have somewhat an idea but for this one I'm so lost. :?: I have no idea where to start. Maybe I'm burnt out. :?

Please help!
 
try multiplying the left side by

\(\displaystyle \frac{\tan{\alpha}}{\tan{\alpha}}\)
 
Great thanks!! Is there some way to look at these problems and know how to "attack" them? :lol: I mean, everytime I look at these problems, it's like I'm looking at gibberish. :(
 
mrtrixta said:
Verify that:

\(\displaystyle {{\cot \alpha + 1} \over {\cot \alpha - 1}} = {{1 + \tan \alpha } \over {1 - \tan \alpha }}\)

I usually have somewhat an idea but for this one I'm so lost. :?: I have no idea where to start. Maybe I'm burnt out. :?

Please help!
Click to expand...

Since left-hand-side(LHS) is all "cot" and RHS is all "tan" - can we write "tan in terms of "cot" (or vise-versa).

Sure we can

cot (a) = 1/ tan (a)

then LHS

\(\displaystyle {{\cot \alpha + 1} \over {\cot \alpha - 1}}\)

\(\displaystyle ={{{1\over {\tan \alpha}} + 1} \over {1\over {\tan \alpha}} - 1}}\)

and continue...
 
You must log in or register to reply here.
Top