sinx + cosx = -1?

Question

i was gonna divide by cosx after moving the sinx to make a tan but that -1 gets in the way...grrrrr help is welcome! =)

okay...details - solve for 0deg < x <360deg...

and this WAS the question

it suggests the 't' formula which i couldn't get to work...and also the formula about asinx + bsinx = rsin (x + y)....

Answer 1

square both sides
(sin x )^2 +(cos x)^2 + 2 sin x cos x = 1

as 2sinx cos x = sin 2x

sin2x =0

As in trigonometric equations

sin x = 0, then x= n * 180deg
where n is an integer

thus
2x = n* 180 deg
x = n* 90 deg
by inspection method

put 1,2,3,4 etc as the value of n and see if the conditions are satisfied

the values satisfying the conditions are
x= 180, 270

Answer 2

sinx cosx 1

Answer 3

Yous can use the R-Formula.
Let sinx+cosx=Rsin(x+α)
sinx+cosx=Rsinxcosα+Rcosxsinα

By comparing coefficients;
Rcos α=1----------------------(1)
Rsin α=1-----------------------(2)

Divide (2) and (1) to get tan α,

Rsinα/Rcosα=tanα=1

tanα=1
α=tan^-1(1)
α=45°

R= √1^2+1^2 = √2
Form a Rsin(x+α) structure, and equate it with 1,
√2sin(x+45°)=1
sin(x+45°)=1/√2
x+45°=45
x=0

Since sinx+cosx-1=0 lies on the 1st and 2nd quadrant,
x=180°-0,360°-90°
x=180°,270°

Answer 4

From inspection x=pi gives sinx=0 and cosx=-1 and 3pi/2 gives sinx=-1 and cosx=0.

But you can derive it as follows.

sinx + cosx=-1
sinx^2-cosx^2=-1*(sinx-cosx)
sinx^2-cosx^2=(cosx-sinx)

sinx^2-cosx^2=(cosx-(-1-cosx))
sinx^2-cosx^2=(2cosx+1)
sinx^2+cosx^2=2cosx^2+2cosx+1
cosx^2+cosx=0

cosx(1+cosx)=0 so cosx=0 and cosx=-1

when cosx=0 sinx=-1 which occurs at 3pi/2
when cosx=-1 sinx=0 which occurs at pi

Answer 5

Square both sides (this will generate extra solutions which we need to double-check at the end)...
→ sin²x + 2 sinxcosx + cos²x = 1
→ 1 + sin(2x) = 1
→ sin(2x) = 0
→ 2x = k*pi
→ x = k*pi / 2

I assume you want solutions in [0, 2pi) so this would mean the possible solutions are
x=0, pi/2, pi, 3pi/2.
However, as I said, we need to check these against the original equation... it turns out that sinx + cosx = 1 (not -1) for x=0 and x=pi/2.

So the only solutions are x=pi, x=3pi/2.

A much simpler way to find the answer is to observe graphically that the line x + y = -1 crosses the unit circle at (-1,0) and (0,-1) and therefore sin t + cos t = 1 at these two points.

Answer 6

sinx + cosx = -1
sinx= -(1+cosx)
2sinx/2cosx/2= -2cosx/2cosx/2
sinx/2= -cosx/2

which means
sinx/2+cosx/2=0

if we put x=270 deg
then x/2 becomes 135(90+45)
i.e. in 2nd quadrant.........
so sin is +ve n cos is -ve
so....sin270/2+cos270/2 gives zero

im sure thr can be more angles like tht..........try 2 find out n tell me also.......take care

Answer 7

or sin^2x + cos^2x +2 sinx cosx = 1
or 1 + 2 sinx +cosx = 1
or 2 sin x cos x = 0
i.e. either cos x = 0 or sin x = 0
x = 90 or x = 0

Answer 8

This Site Might Help You.

RE:
sinx + cosx = -1?
i was gonna divide by cosx after moving the sinx to make a tan but that -1 gets in the way...grrrrr help is welcome! =)

Answer 9

sinx + cosx = -1

square both sides

sin^2x + 2sinxcosx + cos^2x = 1

since sin^2x+ cos^2x = 1 and 2sinxcosx = sin2x

1 + sin2x = 1

sin2x = 0

2x = arcsin 0

x = (arcsin 0)/2

Answer 10

There are a couple of ways to solve the question but a question of the form

A sin x + B cos x = C where A B C are constants can be solved by letting A = r cos t B = r sin t

and proceding

here A = 1 and B = 1 C = -1

1= r cost
1= r sin t

sqauare and add r = sqrt(2)
devide tan t =1 or t = pi/4

we have lhs = r cost sin x + r sin t cos x
= r( sin(x+t)) = -1

sin (x + t) = -1/sqrt(2) = sin (-pi/4)
x+t = npi -pi/4 or npi+5pi/4
x = npi-pi/4-pi/4 or npi+5pi/4-pi/4
= npi- pi/2 or npi + pi

putting n = 1 i in first expression we get 3pi/2(270 degrees and 0 in second we get pi(180 degress)