C3 Maths, writing an equation cos 2θ + sin 2θ in the form Rsin(2θ+α)?
Could anyone help me out with this question, I normally know how to go about them but I never had to deal with a double angle before.
a) Show that cos 2θ + sin 2θ can be written in the form R sin(2θ + α), where R>0 and 0 <α < π/2
b) State the maximum and minimum values of cos 2θ + sin 2θ and the angles the occur
from 0≤ θ ≤ 2π
Thanks! :)
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a)
cos2Ө + sin2Ө = R sin (2Ө + α)
cos2Ө + sin2Ө = R cosα (sin2Ө) + R sinα (cos2Ө)
1 = R sinα
1 = R cosα
tanα = 1 and α lies in 1st quadrant.
α = π/4
2 = R²
R = √2
cos2Ө + sin2Ө = √2 sin (2Ө + π/4)
b)
Max. value = √2
when
2Ө = π/4
Ө = π/8
cos(2t) + sin(2t) =>
sqrt(2) * (cos(2t) * sqrt(2)/2 + sin(2t) * sqrt(2)/2) =>
sqrt(2) * (cos(2t) * sin(pi/4) + sin(2t) * cos(pi/4)) =>
sqrt(2) * sin(pi/4 + 2t)
Maximum value occurs when sin(pi/4 + 2t) = 1
sin(pi/4 + 2t) = sin(pi/2 + 2pi * k)
pi/4 + 2t = pi/2 + 2pi * k
2t = pi/4 + 2pi * k
2t = (pi/4) * (1 + 8k)
t = (pi/8) * (1 + 8k)
k is an integer
0 < t < 2pi
0 < (pi/8) * (1 + 8k) < 2pi
0 < 1 + 8k < 16
-1 < 8k < 15
-1/8 < k < 15/8
k = 0 , 1
t = (pi/8) * (1 + 8k)
t = pi/8 , 9pi/8
Minimum occurs when 2t + pi/4 = 3pi/2 + 2pi * k
2t = 5pi/4 + 2pi * k
t = 5pi/8 + pi * k
t = (pi/8) * (5 + 8k)
t = 5pi/8 , 13pi/8
cos(2@)+sin(2@)=(1/(2)^0.5)*sin(2@+45)