Solve 4cos(2x )+ 2sin(2x) = 1 given -90° < x < 90°. Write 4cos(2x )+ 2sin(2x) in the form Rcos(2x - a), where R and a are constants.

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Step 1: Recognise 4cos(2x )+ 2sin(2x) must be put into the form given in the question form i.e. Rcos(2x - a) as it contains only one trigonometric function

Step 2: Calculate values of R and a

a. Expand Rcos(2x - a) using the suitable compound angle formulae: cos(A - B) = cosAcosB + sinAsinB, in our case A = 2x and B = a
Hence Rcos(2x - a) = R[cos(a)cos(2x) + sin(a)sin(2x)]

b. By identifying Rcos(a) and Rsin(a) as effective constants and matching these to the original form we can obtain two trigonometric equations in R and a: Rcos(a) = 4 {eq.1} & Rsin(a) = 2 {eq.2}

c. Dividing {eq.1} by {eq.2} & canceling R gives tan(a) = 1/2, thus arctan(1/2) = a, yielding a = 26.6°

d. Squaring {eq.1} & {eq.2} and taking the sum gives: R^2(cos^2(a) + sin^2(a)) = 4^2 + 2^2 = 20
Noting that this step has yielded cos^2(a) + sin^2(a) and can be replaced by 1 using the standard trigonometric identity cos^2(y) + sin^2(y) = 1, this leads to the result R = sqrt(20)


Step 3: Solve the equation given for x

a. From the result in step 2, 4cos(2x )+ 2sin(2x) = 1 can be written as sqrt(20)*cos(2x - 26.6) = 1

b. Dividing by sqrt(20) and taking the inverse cosine yields: arccos[1/sqrt(20)] = 2x - 26.6 From your calculator arccos[1/sqrt(20)] yields 77.08° but recognising -77.08° (282.92°, -282.92° etc.) are also solutions crucial(important)
This can be seen by looking at a sketch of y = cos(t)

c. Solving for x in the interval given can be done by rearrangement
2x - 26.6° = 77.08°, -77.08° (282.92°, -282.92°)
Hence x = 51.84° -25.24°
Note that (282.92°, -282.92°) give values out of the range defined for x

Step 4: Check on your calculator that the calculated values of x solve the original equation 4cos(2x )+ 2sin(2x) = 1

Prannay K. A Level Chemistry tutor, GCSE Chemistry tutor, A Level Phy...

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