How do I write the function 3cosθ+4sinθ in the form Rsin(θ + α), where R and α are positive constants?

Split it up into steps:

Step 1: Use compound angle trigonometric formula to re-write Rsin(θ + α)

Since generally sin(θ ± φ) = sinθcosφ ± cosθsinφ

Rsin(θ + α) = R(sinθcosα + cosθsinα) = Rsinθcosα + Rcosθsinα

Step 2: Let Rsin(θ + α) = 3cosθ+4sinθ = Rsinθcosα + Rcosθsinα

We let them equal each other as we are trying to make them equivalent, just in different forms.

Step 3: Compare both sides of the equals (comparing coefficients of sinθ and cosθ)

Hence we can form two equations,

Comparing sinθ: 4=Rcosα

Comparing cosθ: 3=Rsinα

Step 3: Finding R (by squaring both sides of our equations)

16=R2cos2α and 9=R2sin2α

By adding these equations together we can show: 25=R2cos2α+R2sin2α

Hence, 25=R2(cos2α+sin2α)

We know that (cos2α+sin2α)=1, so 25=R2

Therefore R=5 (we take the positive root as R is positive)

Step 4: Finding α (Divide the equations to get in terms of tanα)

Since, 3=Rsinα and 4=Rcosα, we can divide the equations in such a way to get sinα/cosα

Hence, 3/4 = Rsinα/Rcosα

We can cancel down the R's in the fraction since R/R=1.

Since sinα/cosα = tanα, we can write the equation as:

tanα = 3/4

Therefore α=0.644rads (3 significant figures)

Step 5: Conclusion

As R=5 and α=0.644rads

Rsin(θ + α) = 5 sin(θ + 0.644)

Therefore 3cosθ+4sinθ = 5 sin(θ + 0.644)

Answered by Oliver G. Maths tutor

9889 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Prove by induction that, for n ∈ Z⁺ , [3 , -2 ; 2 , -1]ⁿ = [2n+1 , -2n ; 2n , 1-2n]


I always mix up my integration and differentiation. How do i stop this?


Find the derivative, dy/dx, of y = 8xcos(3x).


Solve the complex equation z^3 + 32 + 32i(sqrt(3)) = 0


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences