Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    If cos^2 x + sin^2 x = 1, does cos x + sin x = 1? I'm not sure because, cos^2 x = (cosx)^2 therefore when you take the square root you get cos x. So if you take the square root of everything in the trig identity cos^2 x + sin^2 x = 1 you get cos x + sin x = 1. I dont think this is right but i dont know what i'm doing wrong.
    Offline

    1
    ReputationRep:
    The square root of cos^2 + sin^2 isn't cos + sin. Think about this: 3^2 + 4^2 = 5^2 (9+16=25). Does 3 + 4 = 5?
    Offline

    14
    ReputationRep:
    nope doesnt work

    Sin45 = 1/r2

    co45 = 1/r2

    1/r2 + 1/r2 =/= 1

    this is because the square root of sin^2x+cos^2x =/= sinx+cosx
    Offline

    0
    ReputationRep:
    Your fundamentally abusing the laws of powers in this case.
    When you sqrt (sin^2+cos^2 = 1) you square root one and get +/- 1 and when you square root the other side, because its + and not times or divide, you can't just split up the two terms. Immagine that (cos^2 (x) and sin^2 (x)) are a bracket, you can't just square root both sides.

    Immagine if you could. Then you are saying that sin (x) + cos (x) = sqrt (sin^2 (x) + cos^2 (x)) so if we square both sides, we'd get the same value on either side right? then you get sin^2 (x) + cos^2(x) + 2sin(x)cos(x) = sin^2 (x) + cos^2(x) thus 2sin(x)cos(x) = 0. so it only works if sin (x) or cos (x) are 0 and for no other value.
    Good enough
    Offline

    10
    ReputationRep:
    if you're taking the square root, you take the WHOLE equation. you can't just square root cosx^2 and sinx^2 separately. that's only when the variable are multiplied, not added.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Concept186)
    The square root of cos^2 + sin^2 isn't cos + sin. Think about this: 3^2 + 4^2 = 5^2 (9+16=25). Does 3 + 4 = 5?
    Thanks man, i feel like such an idiot!
    • Thread Starter
    Offline

    0
    ReputationRep:
    Thanks everyone who replied, i'm sorry you had to witness my momentary retardedness.
    Offline

    0
    ReputationRep:
    Although it's not true generally, there are values of x where it does work, those being in any multiple of pi/2. Here's how it works:
    cos(x)+sin(x)=1

(cos(x)+sin(x))^2=1^2=1

cos^2(x)+sin^2(x)+2cos(x)sin(x)=  1

cos^2(x)+sin^2(x)=1-2cos(x)sin(x).
    So what you're saying only works when:
    2cos(x)sin(x)=0

cos(x)=0 or sin(x)=0
    x=\frac{\pi}{2} ,\frac{3\pi}{2} ,etc...
    OR
    x=0, \pi, 2\pi, etc..
    i.e. x=\frac{n \pi}{2} where n=0,1,2,...
    Offline

    2
    ReputationRep:
    depends i think what x is
    cosX +sinX = 1
    where X=0
    Offline

    17
    ReputationRep:
    If that were the case then why would anyone bother quoting the squared version?
    Offline

    0
    ReputationRep:
    Generally- no.

    Also the square root of cos^2x is not equal to cosx.
    Offline

    0
    ReputationRep:
    No
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 4, 2011
Poll
Is the Big Bang theory correct?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.