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)