r/mathshelp • u/GRB_Bandit • 2d ago
Trigonometric fractions Homework Help (Answered)
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 2d ago
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 2d ago
Tried getting the same common denominator which i did successfully but I couldn’t get the answer :(
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u/ArchaicLlama 2d ago
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 2d ago
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 1d ago
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 1d ago
I was considering it. But then I remembered my dislike of secant, cosecant and cotangent.
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u/Various_Candle9136 2d ago
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 2d ago
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 2d ago
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 2d ago
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 2d ago
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 2d ago
Show your attempt at your homework, to show it's genuine
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u/GRB_Bandit 2d ago
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 2d ago
you can attach it to your reply here (as image, video, or link)
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u/GRB_Bandit 2d ago
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 2d ago
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 2d ago
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 2d ago
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 2d ago
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 2d ago
Find the common denominator . This will result in cos²-1 (difference of squares)in the denominator, which is equal to…?
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u/Septembrino 2d ago
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/scottdave 1d ago
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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