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calculate: sinx+sin2x+sin3x=1+cosx+cos2x ∈(pi/2;5pi/6)

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alexa0048 | Student, Grade 11 | (Level 3) eNoter

Posted May 16, 2013 at 2:57 PM via web

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calculate:

sinx+sin2x+sin3x=1+cosx+cos2x ∈(pi/2;5pi/6)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 16, 2013 at 3:45 PM (Answer #1)

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You need to group the members to both sides, such that:

`(sin x + sin 3x) + sin 2x = (1 + cos 2x) + cos x`

You need to convert the sum of sines into a product, using the following formula, such that:

`sin a + sin b = 2 sin((a+b)/2)*cos((a-b)/2)`

Reasoning by analogy, yields:

`sin x + sin 3x = 2 sin((x+3x)/2)*cos((x-3x)/2)`

`sin x + sin 3x = 2 sin(2x)*cos(-x)`

Since `cos (-a) = cos a` yields:

`sin x + sin 3x = 2 sin(2x)*cos(x)`

Replacing `2 sin(2x)*cos(x)` for `sin x + sin 3x` to the left, yields:

`2 sin(2x)*cos(x) + sin 2x = (1 + cos 2x) + cos x`

You need to use the half angle formula, to the right, such that:

`cos (a/2) = sqrt((1 + cos a)/2)`

Reasoning by analogy, yields:

`1 + cos 2x = 2 cos^2 x`

Replacing `2 cos^2 x` for` 1 + cos 2x` , to the right, yields:

`2 sin(2x)*cos(x) + sin 2x = 2 cos^2 x + cos x `

Factoring out `sin 2x` to the left side, yields:

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

Factoring out cos x to the right, yields:

`sin 2x*(2cos x + 1) = cos x*(2cos x + 1)`

`sin 2x*(2cos x + 1) - cos x*(2cos x + 1) = 0`

Factoring out `(2cos x + 1)` yields:

`(2cos x + 1)*(sin 2x - cos x) = 0`

Using zero product property yields:

`{(2cos x + 1 = 0),(sin 2x - cos x = 0):}`

`{(cos x = -1/2),(2sin x*cos x - cos x = 0):}`

`{(x = (pi - pi/3)),(cos x*(2sin x - 1) = 0):}`

`x = (2pi)/3 valid, (2pi)/3 in (pi/2, (5pi)/6)`

`cos x*(2sin x - 1) = 0`

Using again zero product property yields:

`{(cos x = 0),(2sin x - 1 = 0):} => {(x = pi/2),(sin x = 1/2):} => {(x = pi/2),(x = (5pi)/6):}`

Both values `x = pi/2` and `x = (5pi)/6` are invalid since they are not included in interval `(pi/2, (5pi)/6).`

Hence, evaluating the solution to the given equation, in `(pi/2, (5pi)/6)` , yields `x = (2pi)/3.`

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