Express `sqrt3sin(x)-3cos(x)` in the form `R sin(x+alpha)`. (3 marks)
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Express `sqrt3sin(x)-3cos(x)` in the form `R sin(x+alpha)`. (3 marks)
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`2sqrt3 sin(x-pi/3)`
`R sin(x+alpha)=sqrt3sin(x)-3cos(x)`
`R sin(x)cos alpha +Rcos(x)sin alpha=sqrt3sin(x)-3cos(x)`
`text{Equating coefficients:}`
`R cos alpha=sqrt3, \ \ R sin alpha=-3`
`R^2` | `=(sqrt3)^2+(-3)^2=12` | |
`:.R` | `=sqrt12=2sqrt3` |
`text{Given}\ \ sin alpha<0\ \ text{and}\ \ cos alpha>0`
`=> alpha\ \ text{is in the 4th quadrant}`
`sqrt12 cos alpha` | `=sqrt3` | |
`cos alpha` | `=sqrt3/sqrt12=1/2` | |
`:.alpha` | `=-pi/3\ \ (-pi<=alpha<=pi)` |
`:.sqrt3sin(x)-3cos(x)=2sqrt3 sin(x-pi/3)`
By expressing `sqrt3 sin x + 3 cos x` in the form `A sin (x + a)`, solve `sqrt3 sin x + 3 cos x = sqrt3`, for `0 <= x <= 2pi`. (4 marks)
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`x = pi/2, \ (11pi)/6`
`A sin(x + alpha) = sqrt3 sinx + 3cosx`
`Asinx cos alpha + Acosx sin alpha = sqrt3 sinx + 3cosx`
`text(Equating co-efficients:)`
`=> Acos alpha = sqrt3`
`=> Asin alpha = 3`
`A^2` | `= (sqrt3)^2 + 3^2 = 12` |
`:. A` | `= sqrt12` |
`(Asinalpha)/(Acosalpha)` | `= 3/sqrt3` |
`tanalpha` | `= sqrt3` |
`alpha` | `= pi/3` |
`sqrt3 sinx + 3cosx` | `= sqrt3` |
`sqrt12 sin(x + pi/3)` | `= sqrt3` |
`sin(x + pi/3)` | `= 1/2` |
`x + pi/3` | `= (5pi)/6, (13pi)/6` |
`:. x` | `= pi/2, (11pi)/6\ \ \ (0 <= x <= 2pi)` |
θθ
A particular energy wave can be modelled by the function
`f(t) = sqrt5 sin 0.2t + 2 cos 0.2t, \ \ t ∈ [0, 50]`
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i. `f(t) = sqrt5 sin 0.2t + 2cos 0.2t`
`Rsin(nt – alpha)` | `= Rsin(0.2t – alpha)` |
`= Rsin 0.2t cosalpha – Rcos 0.2t sinalpha` |
`=> Rcosalpha = sqrt5,\ \ R sinalpha = −2`
`R^2` | `= (sqrt5)^2 + (−2)^2 = 9` |
`R` | `= 3` |
`cosalpha = sqrt5/3, sinalpha = −2/3`
`=> alpha\ text(is in 4th quadrant)`
`text(Base angle) = cos^(−1)(sqrt5/3) = 0.7297`
`:.alpha` | `= 2pi – 0.7297` |
`= 5.553…` |
`:. f(t) = 3sin(0.2t – 5.553)`
ii. `text(Max value occurs when)\ sin(0.2t – 5.553) = 1`
`0.2t – 5.553` | `= pi/2` |
`0.2t` | `= 7.124…` |
`t` | `= 35.62…` |
`text(Test if)\ \ t > 0\ \ text(for:)`
`0.2t – 5.553` | `= -(3pi)/2` |
`0.2t` | `= 0.8406` |
`t` | `= 4.20…` |
`:. text(Time of 1st maximum:)\ \ t = 4.2`
The current flowing through an electrical circuit can be modelled by the function
`qquad f(t) = 6sin 0.05t + 8cos 0.05t, \ \ t >= 0`
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i. `f(t) = 6sin 0.05t + 8cos 0.05t`
`Asin(at + b)` | `=Asin(0.05t + b)` | |
`= Asin(0.05t)\ cos(b) + Acos(0.05t)\ sin(b)` |
`=> Acos(b) = 6, \ Asin(b) = 8`
`A^2` | `= 6^2 + 8^2` |
`A` | `= 10` |
`=> 10cos(b)` | `= 6` |
`cos(b)` | `= 6/10` |
`b` | `= cos^(−1) 0.06 ~~ 0.927\ text(radians)` |
`:. f(t) = 10sin(0.05t + 0.927)`
ii. `text(Max occurs at)\ \ sin(0.05t + 0.927)= sin\ pi/2`
`0.05t + 0.927` | `= pi/2` |
`0.05t` | `= 0.643…` |
`:.t` | `= 12.87` |
`= 12.9\ \ (text(to 1 d.p.))` |
iii. |
Write `sqrt 3 sin x + cos x` in the form `R sin (x + alpha)` where `R > 0` and `0 <= alpha <= pi/2.` (2 marks)
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`text(See Worked Solution)`
`R sin (x + alpha)` | `=sqrt 3 sin x + cos x` |
`R sin x cos alpha + R cos x sin alpha` | `=sqrt 3 sin x + cos x` |
`text(Equating coefficients:)`
`=> R cos alpha` | `= sqrt 3` | |
`=> R sin alpha` | `= 1` |
`R^2` | `= (sqrt 3)^2 + 1^2=4` |
`:.R` | `= 2` |
`=> 2sin alpha` | `=1` |
`sin alpha` | `= 1/2` |
`alpha` | `= pi/6 qquad (0 <= alpha <= pi/2)` |
`:. sqrt 3 sin x + cos x = 2 sin (x + pi/6).`
What is the value of `tan alpha` when the expression `2sinx - cosx` is written in the form `sqrt5 sin(x - alpha)`?
A. `−2`
B. `−1/2`
C. `1/2`
D. `2`
`C`
`text(Using)\ \ sqrt5 sin(x – alpha) = sqrt5 sinx cosalpha – sqrt5cosxsinalpha,`
`=>sqrt5 sinx cosalpha – sqrt5 cosx sinalpha = 2sinx – cosx`
`=> sqrt5 cosalpha = 2, sqrt5 sinalpha = 1`
`(sinalpha)/(cosalpha)` | `= 1/2` |
`:. tan alpha` | `= 1/2` |
`⇒C`
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Give your answer, or answers, correct to two decimal places. (2 marks)
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`5 sin (x+0.93)`
i. | `A sin (x + alpha) = 3 sin x + 4 cos x` |
`A sinx cos alpha + A cos x sin alpha = 3 sin x + 4 cos x` |
`=> A cos alpha = 3\ \ \ \ \ A sin alpha = 4`
`A^2` | `= 3^2 + 4^2 = 25` |
`:. A` | `= 5` |
`=> 5 cos alpha` | `= 3` |
`cos alpha` | `= 3/5` |
`alpha` | `= cos^(-1) (3/5)= 0.9272… \ \ text(radians)` |
`:.\ 3 sin x + 4 cos x = 5 sin (x + cos^-1 (3/5)) `
ii. | `3 sin x + 4 cos x` | `= 5` |
`5 sin (x + alpha)` | `= 5` | |
`sin (x + alpha)` | `= 1` | |
`x + alpha` | `= pi/2, (5pi)/2, …` | |
`x` | `= pi/2\ – 0.9272…\ \ \ \ \ (0 <= x <= 2 pi)` | |
`= 0.6435…` | ||
`= 0.64\ text(radians)\ text{(2 d.p.)}` |