Math help.
Trig is kicking me in balls, maybe Veeky Forums can help me more than khan academy is right now.
Right now I'm working on Analytic Trigonometry
Here are the problems I'm working on. I'd like help understanding them as I have the answer sheet in the back of the book, thanks!
1. cos(20+A)
2. sec x cos (-x) + tan^2x
Numbers 61 and 67 in pic provided.
Links to videos will be helpful, too.
Nobody wants to do your dumbass grade school hw for you fuck right off m8
Read your fucking textbook. Why the fuck do kids not read their goddamn textbooks that literally provide them with examples and show how to do these? Fuck you OP and fuck your 7th grade math homework
Lets be real if you were doing trig in 7th grade you were a turbo autist with no friends
You're not going to let /b/ be fore helpful than you, are you?
Text book does jack, so do my professors.
Let's be real, we're all turbo autists. I'm just a turbo autist whose in college
We aren't doing your HW OP, it's not my fault you can't do basic simplification.
I'm not asking for you to solve it, I'm asking for help understanding it.
You know what they say about assuming, user
HS fag from third world nation here.
All these questions (except that cos(20+x), if you have to only use trigonometric identities) are easy.
61 is a rephrasal of sin^2x + cos^2x =1
You need to know that cscx = 1/sinx and secx=1/cosx for 67.
Thanks for actually trying to answer the question.
Can I eliminate both Cox^2x because they're above each other in the fraction or can I not because they have addition tagged to them?
What do after last step?
Why have you not cross multiplied?
Just cross multiply and expand.
Convert all cos to sin
Check pic related
Guys, I'm working through 67 and I simplified it down to cot x = cot^3 x
The text book says that the solution is an identity, or that they're equal to each other. Did I do something wrong or does cot^3 x simply to just cot x?
got it. Thanks
Nope, you did it wrong. I guess you cancelled stuff where you aren't supposed to. Look at it carefully.
Holy shit you guys this is simple as fuck
Factor out the numerator
OP just sub. cosec with 1/sin and sec with 1/cos, same for tan and cos.
You will get answer literally in 2 lines
Exactly.
So I converted (csc x)(cot x)/(sec x)(tanx) into their inverse (leaving the top cot as it is) and got
(1/sin x)(cot x) / (1/cos x) (1/cot x)
I eliminated the cot till I was left with (1/sin x)/(1/cos x)
then that equals cot x. I'm where did I go wrong?
(cotx)/(1/cotx) is the same as cotx * cotx/1, which is the same as cot^2x.
The rest becomes another cotx
(1/sinx)(cotx)/(1/cosx)(1/cotx) = (cosx/sinx).(cotx)^2 = cotx.(cotx)^2= RHS
Yeah, I figured that out right before you posted.
Dividing fractions always confuses me for some reason.
Hijacking thread because OP probably got the answer. How do you do pic related?
You'll get it right with practice
Nvm this shit is easy
(sin(x) + cos(x))^2 + (sin(x) - cos(x))^2 = 2
Result: True
Solution: x = π + 2 n π
Go to wolframalpha and put in the equations.
That good feel when I couldn't do this in high school, but now I can do them all in my head
Somebody plox help
lol
I'm in high school and this is shit tier
Somebody pls help me solve my stupid problem
JEE ?
Yes.
Will someone help me solve the problem?
Do you by chance have an equation sheet that came with your HW?
>equation sheet
It's DC Pandey. Answers are at the back. I just don't want to check it out. I want Veeky Forums's answers, could give me some insight.
>Textbook
>One short paragraph explanation
>Two examples that are nothing like the practice questions
Textbooks are fucking trash they always have been
Solve it for n = 2, that's sufficient to eliminate the wrong answers.
Thanks.
Are you retarded?
Have of you guys are fucking tsunfags
>bitch but still help
largely a waste of time, in quantum mechanics and physics etc. youj are never going to use trig identities but eulers equation instead