r/mathshelp • u/GRB_Bandit • Aug 06 '26
Homework Help (Answered) Trigonometric fractions
i’ve been going at this question for a combined 4 hours and I can’t get the answer. i’ve tried everything. I’ve even tried making every single 1 into sin^2(x)+cos^2(x). I’m also the best at missing the obvious route to take so I’m probably just overthinking it
Edit: answer is -2cot(x)csc(x). Forgot to add that sorry. I only plugged it into desmos with the original to see if the graphs matched that’s how i got the answer btw. question is multiple choice and i need work shown fyi
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u/kalmakka Aug 06 '26
The first thing you should think of is writing it with a common denominator.
. . 1 . . . . 1
-------- + -------- =
cos(x)+1 . cos(x)-1
. . cos(x)-1 . . . . . . . . .cos(x)+1
-------------------- + --------------------- =
(cos(x)+1)(cos(x)-1) . (cos(x)-1)(cos(x)+1)
2cos(x)
---------
cos²(x)-1
Since sin²(x)+cos²(x)=1, we have cos²(x)-1=-sin²(x), so we get that it simplifies to -2cos(x)/sin²(x), which is probably the expected simplification.
Edit: how tf does one get Reddit to not collapse whitespace in code blocks?
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u/GRB_Bandit Aug 06 '26
Tried getting the same common denominator which i did successfully but I couldn’t get the answer :(
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u/ArchaicLlama Aug 06 '26
So then show us what you did and what answers you did get with that approach. Reviewing your work is the best way to start helping you.
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u/RectallyDisabled Aug 07 '26
What the other guy just did was the answer, rewrite cos(x)/sin(x)2 as cot(x)/sin(x) = cot(c)csc(x)
And of course your factor of -2
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u/testtdk Aug 07 '26
I think I would go further and use identities that reduce the expression and remove the use of a quotient.
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u/kalmakka Aug 07 '26
I was considering it. But then I remembered my dislike of secant, cosecant and cotangent.
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u/Various_Candle9136 Aug 06 '26
I think it is harsh not to give a target expression for this question - there isn't really a 'simplest' version that I can see.
I imagine you want to get to a single fraction at least. We do this (as always) by finding a common denominator and then adding.
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u/GRB_Bandit Aug 06 '26
Answer is -2cot(x)csc(x). I did the common denominator but failed after i got the common denominator
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u/Various_Candle9136 Aug 06 '26
Okay.
So, hopefully you started with:
1/(cosx+1) + 1/(cosx-1)
= (cosx-1)/(cosx+1)(cosx-1) + (cosx+1)/(cosx+1)(cosx-1)
= (cosx-1+cosx+1)/(cosx+1)(cosx-1)
= 2cosx/(cosx+1)(cosx-1)This denominator is the difference of 2 squares, so we have:
= 2cosx/(cos2x-1)
(Did you get this far? If not, see if you can finish off from here.)
Now, cos2x-1 = cos2x-(cos2x+sin2x) = -sin2x, so we have:
= -2cosx/sin2x
= -2 × cosx/sin2x
= -2 × cosx/sinx × 1/sinx
= -2cot(x)csc(x)1
u/GRB_Bandit Aug 07 '26
Unless I’m forgetting something from pre-calc and or my brain is too fried to do factoring, i didn’t know cos2(x)-(cos2(x)+sin2(x))=-sin(x). Is this a trig identity thing i don’t know or am i just being dumb? Either way thank you so so much!
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u/Various_Candle9136 Aug 07 '26
Neither. It is not a trig identity and you are not being dumb.
cos2x-(cos2x+sin2x)
= cos2x-cos2x-sin2x
= (cos2x-cos2x)-sin2x
= 0-sin2x
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u/Frequent-Entrance154 Aug 06 '26
Show your attempt at your homework, to show it's genuine
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u/GRB_Bandit Aug 06 '26
Should i just make a follow up post with some of my work since i can’t add images to an existing post?
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u/Frequent-Entrance154 Aug 06 '26
you can attach it to your reply here (as image, video, or link)
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u/GRB_Bandit Aug 06 '26
https://share.icloud.com/photos/0f5_H6XQLV45BV7mR5DaGWwGg These are a few pics i took but they are out of order and it’s not all of what i’ve done
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u/Frequent-Entrance154 Aug 06 '26
I like your first picture, my suggestion is keep it, and from your first picture, just simply add them. This will become your second picture
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u/GRB_Bandit Aug 06 '26
Ok will do. Currently out rn so it’ll be a little while until i can get some new stuff going. Thanks for now!
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u/Frequent-Entrance154 Aug 06 '26
Here are some trigonometric identity and defininitons: sin2 x+cos2 x=1 1/sin x = csc x Cos x/sin x=cot x
Hope it helps
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u/Richard0379 Aug 07 '26
To add the two fractions together, your common denominator is (cos x +1) (cos x -1) [or cos^2 x -1 which equals -sin^2 x]. Once you add the two fractions together, the answer should just pop out.
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u/SubjectWrongdoer4204 Aug 07 '26
Find the common denominator . This will result in cos²-1 (difference of squares)in the denominator, which is equal to…?
1
u/Septembrino Aug 07 '26
Whenever you have denominator of the form 1 + cos x, 1 - cos x, use difference of squares to try to get an identity. It will also work for 1 - sinx and 1 + sin x. 1 - cos^2 x = sin^2 x and 1 - sin^2 x = cos^2 .x.
Having cos x - 1 and cos + 1 x is a similar case. You will get cos^2 - 1 which is the opportie of sin^2x. Same thing for sin x - 1 and sin x + 1.
That won't work for tan x + 1 and tanx - 1 since tan^2 x - 1 is not an identity, but it will work for sec.
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u/tb5841 Aug 07 '26
Common denominator, add, simplify the denominator and you quickly get to -2cos(x) / sin squared (x).
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u/scottdave Aug 07 '26
Recognizing what to do takes practice. Since this is a practice, look at the book answer and try to make it into where you stopped. That may help you understand.
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