sin x + cos x = 1

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sinx + cosx = 1
=> 1/rt2 sinx + 1/rt cosx = 1/rt2
=> cos45 sinx + sin45 cosx = 1/rt2 ............ since cos45 = sin45 = 1/rt2
=> sin(x+45) = 1/rt2...........using identity for sin(A+B)
=> x+45 = 45, 135
=> x= 0, 90

Nicer, methinks.

Reply 21

yazan_l

e-unit

that's what Fly. said....

(1/√2)sinx + (1/√2)cosx = 1/√2
cos45.sinx + sin45.cosx = 1/√2
sin(x+45) = 1/√2
x+45 = 45,135
x=0,90

besides, sea tea... why this TOO late post?

Woops.

Quite a long thread for what appeared to be a simple question.

Reply 24

Christophicus

Original post by &#925

Squaring a function can sometimes create extra roots, another solutions is to say

Rsin(x+@)=R(sinxcos@+sin@cosx)

Where in this case R=sqrt(2), cos@=sin@=1, hence tan@=1 and @=45 deg

Therefore

sqrt(2)sin(x+45)=1

t. f. x+45=arcsin((1/sqrt(2)))=45, 135

t. f. x=0, 90 Q. E. D.

Newton

Simba

There is generally quite a neat way of solving questions like this:

sin x° + cos x° = rsin(x + a)°

rsin(x + a)° = rsinx°cosa° + rcosx°sina°

Comparing co-efficients:

rcosa° = 1, rsina° = 1

Squaring and adding:

r²cos²a° + r²sin²a° = 2
r²(cos²a° + sin²a°) = 2
r² = 2
r = √2 (we take the positive root).

Substituting into one of the equations above:

√2 cosa° = 1
cos a° = √2/2
a° = 45°

Therefore sin x° + cos x° = √2sin(x + 45)°

Looking at the original equation:

sin x° + cos x° = 1
√2sin(x + 45)° = 1
sin(x + 45)° = √2/2
(x + 45)° = 45°, 135°
x° = 0°, 90°

Manifest

Hmmm thought as much....SIMBA are you a genius? You are in yr 11 aren't you?

Hmm.

Reply 25

Kolya

The quickest solution is obtained by drawing the graphs and using some common sense. :rolleyes:

Reply 26

yazan_l

Baal_k

The quickest solution is obtained by drawing the graphs and using some common sense. :rolleyes:

haha that's the quickest way to get a zero! :biggrin:

Reply 27

Kolya

yazan_l

haha that's the quickest way to get a zero! :biggrin:

It depends if he wanted the simple answer to the question or all the ticks on the "assessment objectives mark scheme" for an exam. I inferred the former.

Reply 28

yazan_l

Baal_k

It depends if he wanted the simple answer to the question or all the ticks on the "assessment objectives mark scheme" for an exam. I infered the former.

well... a more simple way is just look the answer from the back of the book!

Reply 29

Kolya

It is a crass suggestion that the ability to use common sense and a logical sketch when solving a Maths problem, reducing the amount of algebra that you have to do (reducing the chance of an error) is akin to copying from the back of the book.

Reply 30

yazan_l

reducing the chance of an error

graphs are so INACURATE to get answers! besides: why would u reduce the amount of algebra used? it's MATH.... not football!

Reply 31

Kolya

Graphs are accurate enough to see when sinx + cosx = 1, and therefore they can solve this equation incredibly quickly.

Why would you want to reduce the amount of algebra used? Look at Simba's method and then draw a 3cm x 5cm graph and decide which is easier and more concise.

Reply 32

Kolya

Or just solve the problem in your head this way, I doubt many people can do Simba's algebra in their head and even get an answer, let alone as quick.

Reply 33

yazan_l

Baal_k

is it? try to time how long it will take u to draw that 3cm x 5cm graph! AGAIN asi said: this is math not football...

Reply 34

Kolya

It took me 9.2 seconds to draw and label the graph required. (Slightly quicker in my head)

Reply 35

Malik

I cant believe you guys are even arguing over this, I mean who cares just do the method you think works best.

He started it! :p:

ok i'm gonna stop this coz i feel as if i'm in 5th grade! "he started" ... "no he did" ... "no he did"...

Reply 38

sea tea

yazan_l

besides, sea tea... why this TOO late post?

eh? your solution wasn't quite right, somebody pointed out an angle which satisfied cos x = o, sin x = 0, but not cosx + sinx = 1. i simply tidied it up for you.

Reply 39

yazan_l

--------------------------------------------------------------------------------

Quote:
Originally Posted by Neapolitan
By squaring you've introduced new solutions; for example x = 3pi/2 satisfies your last line but not the original - so you would need to check which of your final solutions satisfied the original equation.

Or there are a couple of other ways you could do this:

(i) write sinx = 2 sin(x/2) cos (x/2) and cosx = 1 - 2 sin²(x/2)

or

(ii) write sinx + cosx = √2sin(x+pi/4)

ya maybe i forgot to say that u have to check which values of x satisfy the first equation,,, but this doesnt mean that what i did was wrong...
i was going to do it as u did it in (i)... but i found it simpler to use my way...

cheers

someone tidied it up before u!