Math - help really needed
posted by Rani .
I'm sorry to double post; I don't want to seem impatient, but I really need help with this.
Prove each idenity.
I haven't even gotten 'round to sny of the quedtions because the first one is just so hard. I'm not really sure I'm uderstanding how to use the quotient and pythagorean identities. I'm so confused. I can't make sense of it and I have tried so many different ways. I must have spent over an hour on he first problem and still I can't come up with an answer. I'm just really frustrated; could someone please help me. I'd really appreciate it.
i will do the first one
LS = 1 + 1/(sin^2x/cos^2x)
= 1 + cos^2x/sin^2x
= (sin^2x + cos^2x)/sin^2x
I usually try to change all ratios to sines and cosines
the second is quite easy.
add the left side terms by taking a common denominator of cosx
you will get
sin^2 + cos^2 = 1 (a^2 + b^2 = c^2)
sec^2 - tan^2 = 1
csc^2 = cot^2 = 1 (the co-version of the above)
To subtract those two, you need a common denominator, so multiply the (cosx/1) by (cosx/cosx). When you subtract them, the new numerator is an identity, so it can be re-written. When you simplify, you're left with sinxtanx.
The rest are very similar. I hope you the idea of how to solve these types of problems - you just rearrange the complicated side to get identities that can be re-written.
oh, ok, I see.
thank you all very much. I think I get it now, so hopefully I can actually get them on my own. :)
wait, for the first one if sin^2+cos^2=1, how did you even cancel some out?
I'll give you a hint:
Try replacing the 1 with tan^2x/tan^2x. After add that to 1/tan^2x, replace the denominator (the tan^2x) with sin^2/cos^2.
okay, I'll try it out, thank you.
I'm really sorry. I don't want to sound stupid, but when I add tan^2x/tan^2x to 1/tan^2x I get 1+tan^2x/tan^2x and then the tans cancel out and I'm left with just one. Obviously, that's not correct, but I can't find my mistake.
You actually get (1/tan^2x)+1, which is cot^2 + (csc^2x-cot^2x), which is csc^2x, which is 1/sin^2x. And that's your answer.
(You could only cancel out the tan^2 if it were (Some #)(tan^2x)/tan^2x, rather than (1)+(tan^2x)/tan^2x. Dividing is the inverse of multiplying, so dividing only cancels what is multiplied, not added.)
but if I have identical denominators, why can't I just add the 1 to the tan^2X?
^ I split that fraction, making the 1/denom. a fraction, and adding it to the tan^2/denom fraction.
I haven't learned what cot and csc are. Can I still solve this with just tan, cos, and sin?
cot= 1/tan or inverse of tangent
csc= 1/sin or inverse of sine
(1-cos^2x)= sin x
(sin x)(1/sin x)=1
The first one is simpler if you do it this way:
Multiply both sides by [sin(x)]^2
sin^2x + sin^2x/tan^2x = 1
since tan^2x = sin^2x/cos^2x the equation becomes:
sin^2x + sin^2x/(sin^2x/cos^2x) = 1
sin^2x + sin^2x*cos^2x/sin^2x = 1
the two sin^2x cancels and you are left with:
sin^2x + cos^2x = 1 which is true!
Hope that helps!