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If cosx + 2cosy + 3cosz = sin x + 2siny + 3sinz = 0 then the value of sin 3x + 8sin 3y+ 27 sin 3z is
- a)sin(x+y+z)
- b)3sin(x+y+z)
- c)18 sin (x+y+z)
- d)sin(x+2y+z)
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
If cosx + 2cosy + 3cosz = sin x + 2siny + 3sinz = 0 then the value of ...
α = cisx, β = cisy, γ = cisz
Given α + 2β + 3γ = 0 ⇒ α3 + 8β3 + 27 γ3 = 18αβγ
⇒ sin 3x + 8sin 3y + 27 sin 3z = 18sin(x + y + z)
Given α + 2β + 3γ = 0 ⇒ α3 + 8β3 + 27 γ3 = 18αβγ
⇒ sin 3x + 8sin 3y + 27 sin 3z = 18sin(x + y + z)
Most Upvoted Answer
If cosx + 2cosy + 3cosz = sin x + 2siny + 3sinz = 0 then the value of ...
Given: cosx 2cosy 3cosz = sin x 2siny 3sinz = 0
To find: The value of sin 3x 8sin 3y 27 sin 3z
Let's solve this step by step:
1. First, let's rewrite the given equation in terms of sin and cos functions:
cosx 2cosy 3cosz = 0
sin x 2siny 3sinz = 0
2. Now, let's square both equations:
(cosx 2cosy 3cosz)^2 = 0^2
(sin x 2siny 3sinz)^2 = 0^2
Expanding and simplifying, we get:
cos^2x 4cos^2y 9cos^2z + 4cosx*cosy + 6cosx*cosz + 12cosy*cosz = 0
sin^2x 4sin^2y 9sin^2z + 4sinx*siny + 6sinx*sinz + 12siny*sinz = 0
3. Since sin^2x + cos^2x = 1, we can rewrite the above equations as:
4cos^2y 9cos^2z + 4cosx*cosy + 6cosx*cosz + 12cosy*cosz = -cos^2x
4sin^2y 9sin^2z + 4sinx*siny + 6sinx*sinz + 12siny*sinz = -sin^2x
4. Rearranging the equations, we get:
cos^2x + 4cosx*cosy + 6cosx*cosz + 4cos^2y + 12cosy*cosz + 9cos^2z = 0
sin^2x + 4sinx*siny + 6sinx*sinz + 4sin^2y + 12siny*sinz + 9sin^2z = 0
5. Now, let's add the above two equations:
cos^2x + sin^2x + 4cosx*cosy + 6cosx*cosz + 4cos^2y + 12cosy*cosz + 9cos^2z + 4sinx*siny + 6sinx*sinz + 4sin^2y + 12siny*sinz + 9sin^2z = 0 + 0
Using sin^2x + cos^2x = 1, the equation simplifies to:
1 + 4(cosx*cosy + cos^2y) + 6(cosx*cosz + cosy*cosz) + 4(sin^2y + sinx*siny)
To find: The value of sin 3x 8sin 3y 27 sin 3z
Let's solve this step by step:
1. First, let's rewrite the given equation in terms of sin and cos functions:
cosx 2cosy 3cosz = 0
sin x 2siny 3sinz = 0
2. Now, let's square both equations:
(cosx 2cosy 3cosz)^2 = 0^2
(sin x 2siny 3sinz)^2 = 0^2
Expanding and simplifying, we get:
cos^2x 4cos^2y 9cos^2z + 4cosx*cosy + 6cosx*cosz + 12cosy*cosz = 0
sin^2x 4sin^2y 9sin^2z + 4sinx*siny + 6sinx*sinz + 12siny*sinz = 0
3. Since sin^2x + cos^2x = 1, we can rewrite the above equations as:
4cos^2y 9cos^2z + 4cosx*cosy + 6cosx*cosz + 12cosy*cosz = -cos^2x
4sin^2y 9sin^2z + 4sinx*siny + 6sinx*sinz + 12siny*sinz = -sin^2x
4. Rearranging the equations, we get:
cos^2x + 4cosx*cosy + 6cosx*cosz + 4cos^2y + 12cosy*cosz + 9cos^2z = 0
sin^2x + 4sinx*siny + 6sinx*sinz + 4sin^2y + 12siny*sinz + 9sin^2z = 0
5. Now, let's add the above two equations:
cos^2x + sin^2x + 4cosx*cosy + 6cosx*cosz + 4cos^2y + 12cosy*cosz + 9cos^2z + 4sinx*siny + 6sinx*sinz + 4sin^2y + 12siny*sinz + 9sin^2z = 0 + 0
Using sin^2x + cos^2x = 1, the equation simplifies to:
1 + 4(cosx*cosy + cos^2y) + 6(cosx*cosz + cosy*cosz) + 4(sin^2y + sinx*siny)
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If cosx + 2cosy + 3cosz = sin x + 2siny + 3sinz = 0 then the value of sin 3x + 8sin 3y+ 27 sin 3z isa)sin(x+y+z)b)3sin(x+y+z)c)18 sin (x+y+z)d)sin(x+2y+z)Correct answer is option 'C'. Can you explain this answer?
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