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and rewrite this in terms of only ssinx and cossx
[cosx/(1-tanx)] +[ sinx/(1-cotx)] = cosx + sinx Is this the problem? I wanted to be sure which terms are in a denominator.
Let's try some things. What is [(1-tanx)] *[(1-cotx)] =
So, 2 - tanx - cotx ? Yes?
Some people begin these proofs by changing terms to be all sines and cosines. Others look for a way to get a common denominator and hope terms will add out. Do you have a preference for one way or another?
I'm going to the DRAW and convert to sines and cosines and see what happens. I need you to check as we go along. This approach may work and may not work in terms of producing what we want but that's the way it goes with trig identities. Looking at what we are trying to show: cosx + sinx on the right side makes me think sines and cosines is the way to go.
and quick side question how woulod u solve sin(2x)= 0.5 i got to sinxcosx=0.25
Please check that and see if it makes sense.
ye it makes ense
I can only do one problem at the time without losing focus. That problem you wrote looks like a double angle problem of some sort.
ye it is but for some reason i cant simplify that one
I'm back on the original problem. I'll begin to clear fractions. I hope you looked for errors because if we miss any, it's back to scratch but that's the way math goes.
u mesed up when u combined the denominators the left ide wa lackiong a negatigve sign so u can combine the fractrion
u cant combine
Good catch! We should factor.
See if you agree.
what did u factor
Look at the .jpg upload and then what I just wrote. The key is factoring -1 from the second denominator so as to get the denominators the same after moving the negative one to the numerator with the sine squared x.
kk i get this one but how do u olve the sin (2x)=0.5
I don't know right out of my head. What's the double angle formula for sine?
Are we supposed to be proving some sort of identity or solving a trigonometric equation over some interval?
What is (square root of 2 all over 2)^2 ? I think we're going to have some 45 degree angles or pi/4 here and multiples of those.
its 2sincos= in(2x)
Look, the original problem is sin(2x) = 1/2. So where does sine x = 1/2. Why not just take half of that angle?
sin (pi/6) = 1/2. 2x = pi/6 x = pi/12. There's one angle. Are you solving over restricted interval or over the set of reals?
set ofd real
So, we need to write pi/12 in such a way to reflect the infinitely many solutions.
What is the next angle after pi/12 that makes the equation true?
um anything with a full revolution added to it
Hint: Look at this set of solutions for a set interval: sin 2x = 0.5 2x = π/6 , 5π/6 , 13π/6 , 17π/6 x = π/12 , 5π/12 , 13π/12 , 17π/12 (for 0 ≤ x ≤ 2π)
Now, that has to be expanded. It would be cool to think of a way to use n to write the answers in a shorter way.
Could we take these: x = π/12 , 5π/12 , 13π/12 , 17π/12 and add multiples of 2 pi to them?
hey could u also help me this last problem ssin3x=sinx
We do not have an answer to sin(2x) = .5 yet.
we do thouigh it would be pi/12 + n(pi)
Post each of those other problems in separate threads.
I have an idea on cos(2x) = cos(x)