anonymous
  • anonymous
prove the following identities cos(x-pi) = sin(x+3pi/2) tan(2x) = 2tan(x)/1-tan^2(x) (hint switch to sine and cosine)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
matlee
  • matlee
why is yours red
anonymous
  • anonymous
use these identities for the 1st one \[\cos(A-B)=cosAcosB+sinAsinB\] \[\sin(A+B)=sinAcosB+sinBcosA\]
matlee
  • matlee
Qualified helpers dont come , i just bought some owl bucks for them and they didnt com for 30 minutes and still ahvent come anyways. I ased for a qualified helper too

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
for that second one use this identity-\[\tan(A+B)=\frac{ tanA+tanB }{ 1-tanAtanB}\] try putting A=B=x in this question :)
anonymous
  • anonymous
@lisamath12
anonymous
  • anonymous
ok so for the second one that leaves me with tan(x+x) = tanx+tanx/1-(tanx)(tanx)
anonymous
  • anonymous
I'm not sure where to go from there though
anonymous
  • anonymous
yes :) ur correct now try to simplify further
anonymous
  • anonymous
I am not sure how to do that I have all the formulas for these questions what I don't know is how to solve it after that
TheSmartOne
  • TheSmartOne
Do we have to prove one or two identities for this question?
triciaal
  • triciaal
cos(A-B) = Cos (A + B) + 2 Sin A Sin B
TheSmartOne
  • TheSmartOne
For the second identity, \(\sf\Large tan(2x) = \frac{sin(2x)}{cos(2x)}\)
TheSmartOne
  • TheSmartOne
Another hint for the second one: \(\sf\Large sin(2x) =2sin(x)cos(x)\) and \(\sf\Large cos(2x) =cos^2(x) -sin^2(x)\)
TheSmartOne
  • TheSmartOne
@lisamath12 You will get your answer faster if you cooperate :)
anonymous
  • anonymous
sorry I was busy for a short while just got back
TheSmartOne
  • TheSmartOne
ok :)
anonymous
  • anonymous
so now we have: 2sin(x) cos(x)/ Cos^2(x) - sin^2(x) is equal too 2tan(x)/1-tan^2(x)
anonymous
  • anonymous
is the next step to change the tangents on the other side too sine and cosine as well?
TheSmartOne
  • TheSmartOne
we only deal with one side at a time :)
TheSmartOne
  • TheSmartOne
so we have \(\sf\Large \frac{2sin(x)cos(x)}{cos^2(x) -sin^2(x)}\) divide both the numerator and denominator by cos^2(x)
TheSmartOne
  • TheSmartOne
\(\sf\Huge \frac{\frac{2sin(x)cos(x)}{cos^2(x)}}{\frac{cos^2(x) -sin^2(x)}{cos^2(x)}}=??\)
TheSmartOne
  • TheSmartOne
So lets deal with the numerator first :) \(\sf\Large \frac{2sin(x)cos(x)}{cos^2(x)} = ?\) Simplify it :)
anonymous
  • anonymous
alright so now we got 2sin(x)/cos(x)
anonymous
  • anonymous
and the denominator is -sin^2(x)?
TheSmartOne
  • TheSmartOne
simplify 2sin(x)/cos(x) hint: tan(x) = sin(x)/cos(x)
anonymous
  • anonymous
2 tan(x)
TheSmartOne
  • TheSmartOne
exactly! :D
TheSmartOne
  • TheSmartOne
ok, now we simplify the denominator part :)
TheSmartOne
  • TheSmartOne
\(\sf\Huge \frac{cos^2(x) -sin^2(x)}{cos^2(x)}=??\)
TheSmartOne
  • TheSmartOne
Hint: (a-b)/c = a/c - b/c
anonymous
  • anonymous
-sin^2(x)/cos^2(x)
TheSmartOne
  • TheSmartOne
Not sure how you got that, but :P \(\sf\Large \frac{cos^2(x) -sin^2(x)}{cos^2(x)}=\frac{cos^2(x)}{cos^2(x)} - \frac{sin^2(x)}{cos^2(x)} = ?\)
anonymous
  • anonymous
oh is it - tan(x). I had cos^2(x)/cos^2(x) equal 0 and that lefted me with -sin^2(x)/cos^2(x)
TheSmartOne
  • TheSmartOne
a/a = 1 \(\sf\Large \frac{cos^2(x)}{cos^2(x)} = 1\) so we are left with 1 - tan^2(x)
TheSmartOne
  • TheSmartOne
And therefore we have proved the identity
TheSmartOne
  • TheSmartOne
\(\sf\Huge \frac{\frac{2sin(x)cos(x)}{cos^2(x)}}{\frac{cos^2(x) -sin^2(x)}{cos^2(x)}}=\frac{2tan(x)}{1-tan^2(x)}\)
anonymous
  • anonymous
ohhh, yes I see it now thank you!
TheSmartOne
  • TheSmartOne
\(\sf\Large tan(2x) = \frac{2tan(x)}{1-tan^2(x)}\) LHS = RHS
TheSmartOne
  • TheSmartOne
Anytime! :D
TheSmartOne
  • TheSmartOne
Now we have this question, right? cos(x-pi) = sin(x+3pi/2)
anonymous
  • anonymous
yes exactly
TheSmartOne
  • TheSmartOne
Hold on, I'm trying to see how to solve this one. :)
TheSmartOne
  • TheSmartOne
So, I got it now :)
TheSmartOne
  • TheSmartOne
we need to use these two formulas cos(α – β) = cos(α)cos(β) + sin(α)sin(β) sin(α + β) = sin(α)cos(β) + cos(α)sin(β)
TheSmartOne
  • TheSmartOne
We shall seperately open up the LHS and RHS :)
TheSmartOne
  • TheSmartOne
Or we could use this actually: cos(α – β) = cos(α + β) + 2sin(α)sin(β)
TheSmartOne
  • TheSmartOne
So @lisamath12 lets plug it into the last formula I gave :)
anonymous
  • anonymous
alright so then we would have: cos(x-pi) = cos(x+pi) + 2sin(x)sin(pi)
TheSmartOne
  • TheSmartOne
and lets first expand cos(x+pi) using this formula: cos(α + β) = cos(α)cos(β) – sin(α)sin(β)
anonymous
  • anonymous
cos(x+pi) = cos(x)cos(pi) - sin(x)sin(pi)
TheSmartOne
  • TheSmartOne
lets simplify that hint: cos(pi) = -1 sin(pi) = 0
anonymous
  • anonymous
alright so cos(x+pi) = cos(x)(1) - sin(x)(0)
TheSmartOne
  • TheSmartOne
cos(x)(-1) ** so simplifying that gets us -cos(x) right?
TheSmartOne
  • TheSmartOne
and let's simplify the other half 2sin(x)sin(pi)
TheSmartOne
  • TheSmartOne
remember sin(pi) = 0
anonymous
  • anonymous
2sin(x)(0)=0
TheSmartOne
  • TheSmartOne
correct so we got the LHS as -cos(x) lets expand the RHS too
TheSmartOne
  • TheSmartOne
sin(x+3pi/2) using the formula: sin(α + β) = sin(α)cos(β) + cos(α)sin(β)
anonymous
  • anonymous
sin(x+3pi/2) = sin(x)cos(3pi/2) + cos(x)sin(3pi/2)
anonymous
  • anonymous
sin(x)(0) + cos(x)(-1)
TheSmartOne
  • TheSmartOne
and simplifying it more you get -cos(x)
TheSmartOne
  • TheSmartOne
therefore -cos(x) = -cos(x) and LHS = RHS
anonymous
  • anonymous
once again big help
TheSmartOne
  • TheSmartOne
anytime :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.