If sinA+sinB+sinC=3 then cosA+CosB+cosC= ?

Question

Answer

Sin90°+ sin90°+sin90°= 3so,cos90°+cos90°+cos90°=0

Answer

0

Answer

Maximum value of sin Ф is 1.
So sinA+sinB+sinC=3 means
sin A = 1
sin B = 1
sin C = 1
Thus, A = B = C = 90 degree

cos A = cos 90 = 0
cos B = cos 90 = 0
cos C = cos 90 = 0

cos A + cos B +cos C = 0...

Hope it helps ....

Answer

The maximum value of sin is 1and minimum value is -1  that means here all the values of sins are 1 . if one sin is not 1 then you cannot make such values for other two sins to satisfy the equation(keeping in mind the minimum and maximum value of sin).hence here all the sins are 1.which means that all the cos will be 0.so cosA+cosB+cosC is 0

Answer

Solution:

Given:

sinA sinB sinC=3

To Find:

cosA CosB cosC= ?

Solution:

1. Using Trigonometric Identities:

We know that:

sin²A + cos²A = 1

sin²B + cos²B = 1

sin²C + cos²C = 1

Multiplying these equations, we get:

sin²A sin²B sin²C + cos²A sin²B sin²C + sin²A cos²B sin²C + sin²A sin²B cos²C = 1

Using the identity,

sin²A sin²B sin²C = (sinA sinB sinC)²

Substituting the given value of sinA sinB sinC, we get:

(sinA sinB sinC)² + cos²A sin²B sin²C + sin²A cos²B sin²C + sin²A sin²B cos²C = 1

3² + cos²A sin²B sin²C + sin²A cos²B sin²C + sin²A sin²B cos²C = 1

9 + cos²A sin²B sin²C + sin²A cos²B sin²C + sin²A sin²B cos²C = 1

cos²A sin²B sin²C + sin²A cos²B sin²C + sin²A sin²B cos²C = -8

Dividing both sides by sinA sinB sinC, we get:

cos²A cos²B cos²C + sin²A sin²B sin²C + sin²A cos²B sin²C + cos²A sin²B sin²C + sin²A sin²B cos²C + cos²A cos²B sin²C = -8 / (sinA sinB sinC)

cos²A cos²B cos²C + sin²A sin²B sin²C + sin²A cos²B sin²C + cos²A sin²B sin²C + sin²A sin²B cos²C + cos²A cos²B sin²C = -8 / 3

Now, adding and subtracting sin²A cos²B cos²C, cos²A sin²B cos²C, and cos²A cos²B sin²C on both sides, we get:

cos²A cos²B cos²C + sin²A sin²B sin²C + sin²A cos²B sin²C + cos²A sin²B sin²C + sin²A sin²B cos²C + cos²A cos²B sin

Answer

Ans will be 0
because sinA+sinB+sinC=3
which means,
=sin90+sin90+sin90=1+1+1=3
Now for,
=cosA+cosB+cosC
as @=90
=cos90+cos90+cos90=0+0+0=0
therefore 0 is the ans.

Answer

Maximum value of sinx is 1 so A.B.C all angle sould be 90 hence  0 is right answer

Answer

Maximum value of sinx is 1 the angle is 90' then all cos is zero the sum is zero

Answer

Answer is 0

Answer

In this querstion 3 is only possible condition when all three sin are in their maximum value which is 1 so the A=B=C=90 degree so the sum of cosines will be zero

Answer

Since, -1≤sinX≤1.
We have,
sinA+sinB+sinC=3
So, the value if each of the sines must be 1. Thus,
sinA=sinB=sinC=1
From general equations,
A=B=C= π/2
Thus, for the cosines part,
cosA+cosB+cosC= cosπ/2+cosπ/2+cosπ/2
=0
Hence, proved!

Answer

0

Answer

0

Answer

Bhai it is possible only if angles are equal and equal to pie/2 so cosa + cosb+ cosc= 0

Answer

Cos90+cos90+cos90=0...

Answer

4

Answer

0

Answer

3

Answer

Since maximum value of sin can be 1, and given SinA+SinB+SinC=3 so when sin is 1, cos is 0. From this we get CosA+CosB+CosC=0

Answer

As sin90=1
then sin90+sin90+sin90=3
so,cos90+cos90+cos90=0
As,cos90=0

Answer

Its 0.
as sin a has maximum value of 1.
so all sin a,sin b,sin c are equal.
therefore Q is 90
and cos 90 is 0.
so answer is 0

Answer

0

Answer

0 is the only correct answer for this question

Answer

Put the angle as 90 and you will get the answer as 0

Answer

Put A=B=C=90 which satisfies the equation

Answer

Ans is 0

Answer

0

Answer

0000000

Answer

Sin function having maximum value is equal to 1.
so angle A, B, C having maximum value 90 for making
sinA+sinB+sinC is equal tu 3 and on these value answer is eqal tu 0

Answer

0 is the right answer.

Answer

Bhai ,,, maximum value of Sina/sinb/sinc is 1,,,so ,in this equation if one less than one other two must be greater than 1 ,that's not possible .so Sina=sinb=sinc=1
then cosa+cosb+cosc=0.

Answer

SinA+sinB+sinC=3..it is only possible when value of sin is maximum..we know -1《sinA《1...so sinA=1.sinB=.sinC=1.
A,B,C=90.. so cosA+cosB+cosC=cos90+cos90+cos90=0

Answer

The maximum value for sin is 1 and since you have three sin's adding up to 3, each of the sin's must be 1!
So A = B = C = 90 degrees or any angle coterminal with 90 degrees.

And the cos of 90 degrees is zero. So cos(A)+cos(B)+cos(C) = 0+0+0 = 0

Answer

Answer is zero because putting value A,B,C, 90 we get  3 and when we put that value in cosA+ cosB +cosC we get 0 ok u hope you understand

Answer

0 is correct  answer  put a,b,c=90

Answer

0        because sinA +sinB+sinC =3 for maximum value of sinA=1 sinB=1 sinC =1 and for maximum value of sin.  cos will be zero for example at 90 so cosA+cosB+cosC=0

Answer

Max value of sin(x) =1 so the sum is 3 only if sinA, sinB,SinC=1 each for which angle =90degree and cos 90 degree = 0 so cosA+ cosB+ cosC=0 hope u got it

Answer

Diff. Is not apropiate By which u will diff. A, B, C then there will be another type of variable in the question.Correct ans is 0 since A,B,C=π÷2

Answer

It's an extreme value prblm. ans=0

Answer

SinA+sinB+sinC=3
sinA=sinB=sinC=1
A=B=C=90
cosA+cosB+cosC
=cos90+cos90+cos90
=0 [All angles are measured in degrees ]

Answer

If sin A + sinB + sinC =3 then cos A + cos B +cosC =0

Answer

SinA+sinB+sinC=0 it is only possible when A=B=C=90
hence cosA+cosB+cosC=0

Answer

If SinA+SinB+SinC=3 then each angle A=B=C=90 degree
now putting value in
CosA + cosB+cosC=cos90+cos90+cos90
=0 Answer

Answer

We know that sinx≤1.Here sinA +sinB + sinC =>
A=B=C=π/2.Hence cosA+cosB+cosC=0

Answer

0

Answer

3 cos

Answer

0

Answer

Sin90+sin90+sin90=3
cos90+cos90+cos90=0

Answer

Zero

Answer

Ans is 3 or zero

Answer

0

Answer

Substitute A=B=C=90
SO cosA+CosB+CosC=0

Answer

0 because sin has maximum value us equal to 1 this condition satisfies when SinA =SinB =SinC =1 implies A,B,C are 90degree so CosA = 0

Answer

0

Answer

SinA Sin B SonC equal to one then this is possible so Cosine is zero

Answer

0

Answer

Maximum value of sin is 1, which means sinA = sinB= sinC =1, so 1+1+1=3 i.e, A=B=C=π/2
thus, cosA + cosB + cosC= 0

Answer

00

Answer

Pi

Answer

From given A=B=C=90•  then the answer is '0'.

Answer

If A+B+C=π => A=π - ( B+C). Therefore, (cos A+cos B)+cos C= 2 cos(A+B/2) cos(A-B/2)+(1-2 sin^2 C/2)= 2 cos(π/2-C/2) cos(A-B/2)+ 1- 2 sin^2C/2= 2 sin C/2{ cos(A-B/2)+ sin C/2} +1 = 2 sin C/2{ cos(A-B/2) + sin( π/2 - (A+B)/2)} +1= 2 sin C/2 { cos(A-B)/2 + cos (A+B)/2 } +1 =2 sin C/2 ( 2 cos A/2• cosB/2 )+1 = 4 cos A/2 • cos B/2 • sin C/2 + 1. Answer.

Answer

Max. value of sin 'theta'= 1 .
then sinA =SinB=SinC=90 in that case ,
cosA=cosB=cosC=0.
hence, cosA+cosB+cosC=0

Answer

Since maximum value of sine function is 1 hence each sinA sinB and sinC values are 1
and hence value of their cos is 0
therefore cosA + cosB+ cosC = 0

Answer

Easy h yr ,0

Answer

Let A=B=C=90 DEGREES
then sinA+sinB+sinC=1+1+1=3
cosA+cosB+cosC=0

Answer

Boundary value problem
a,b,c equals pi/2
so ans is zero

Answer

0

Answer

Maxmium value of sine x is 1 so sinA +sineB +sineC = 3 and sin×= 1 at x = 90•  so for cosA +cosB+cosC= 0 because at A,B,C =90  their sum equal to 0

Answer

0

Answer

Cos90 =0 then answer is 0

Answer

As we know max{sin(x)} is '1'.
and it is given in the question that, sinA + sinB + sinC = 3 that means every term is at Maxx and equal to '1'.
hence, A = B = C = π/2

using this you can easily find value of the asked in question
ANS = 0

Answer

The answer is -3.

Answer

As we know sin90 =1 because sin 90 maximum value is 1 and cos 90 =0
so sin90+ sin90+sin90=3
cos90+cos90+cos90=3

Answer

000

Answer

Max value of sin(theta) can be 1 so here sinA = sinB = sinC=1, so that there sum is 3 (given in question).ThereforeA = B = C = 90 degrees.

Answer

Differentiate it

Answer

Yubraj

Answer

0

Answer

A=B=C=90
cos90+cos90+cos90=<0>

Answer

SinA=1 ,A=90 answer is 0

Answer

SinA=sinB=sinC=1 => A=B=C=90 degrees
thus cosA=cosB=cosC=0

Answer

0

Answer

Answer will be 0

replace A, B, C = 90

sin90 + sin90 + sin90 = 1

=> cos90 +cos90+cos90 = 0

Answer : 0

Answer

A=B=C=90Cos90+cos90+cos90=0

Answer

It's answer is 0 because
sin 90+sin 90+sin90=3
Therefore, sin 90=1
1+1+1=3
Again cos 90+cos90+ cos90
Therefore, cos 90=0
0+0+0=0

Answer

Sin a+sinb+sinc=3 is only when sin value takes place 1 nd then cos takes value 0. so cosa+cosb+cosc=0

Answer

The answer is 0.

since |sinø|

That's why,
cosA+cosB+cosC = 0.

Answer

0 obviously FOR SINA+SINB+SINC=3 ALL ANGLES SHOULD BE 90 I.E A=B=C=90

Answer

0

Answer

It is (20+40-60)

Answer

3 terms of sine when added gives 3 means A, B, C are 90 degrees.  means cosA,  cosB,  cosC are all 0. Therefore sum is zero(ans)

Answer

0

Answer

O

Answer

SinA + sinB+sinC=3 it is only possible when
A=B=C=90. But in a triangle sum of all angles is 180.
here is 270. so this is not possible.

Answer

This condition is only possible when all the angles are π/2, so when we put the all the angles π/2 in CosA + CosB + CosC, the answer will be 0.

Answer

Look sina can have maximum value of one and when sina + sinb + sinc comes out to be 3 hence a,b,c =π/2. now just put all values in cosa +cosb + cosc and get answer which comes out to be Zero

Answer

0

Answer

As value of sine function(sin) ranges from -1 to 1 so for sinA + sinB + sinC =3 all of them should be equal to 1. If sinA = 1 then cosA = 0 similarly cosB and cosC will be 0 . Therefore cosA + cosB + cosC will eqaul to 0.
cos A + cos B + cos C = 0

Answer

3 only can happen when all the sine is maximum
this means that all angles A,B,C are of 90
so cosA+cosB+cosC= 0

Answer

CosA+cosB+cosC=0

Answer

As we all know the value of sin(90)=1 .
And we all know the range of sin is [-1,1]
Here the angle A,B,C are same equal to 90.
sinA+sinB+sinC=sin90+sin90+sin90=3.
substituting the values of A,B and C the value of
cosA+cosB+cosC=cos90+cos90+cos90=0.
because the value of Cos90 is 0.

Answer

0 is right

Answer

0

Answer

CosA+cosB+cosC=0

Answer

Given,
sin a+sin b+sin c=3
or,(sin a-1)+(sin b-1)+(sin c-1)=0
Now, comparing both sides of the equations:-
sin a-1=0
or, sin a=1
or, a=90
similarly,
b=90
and, c=90
Hence,
cos a +cos b+cos c
=cos 90 +cos 90 +cos 90
=0

Answer

CosA+cosB+cosC=0
because, if we replace A,B,C. of sinA+sinB+sinC with 90` ,then
answer will be 3, so , similarly,
if A,B,C of cosA+cosB+cosC is equal to 90`, then answer
will be 0

Answer

If we take A=B=C=90 degree then ,
firstly we have to take the sin term sin90+sin90+sin90=1+1+1=3 so it is correct as 3.
now in cos,
cos90+cos90+cos90=0+0+0=0
and the correct answer is O.

Answer

2

Answer

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