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Simplifying a cos x + b sin x

Simplifying a cos x + b sin x

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Explainer Video

Tutor: Toby

Summary

Simplifying acos⁡x+bsin⁡xa \cos x + b \sin xacosx+bsinx​​

In a nutshell

You need to know how to write expressions of the form acos⁡x±bsin⁡xa \cos x \pm b \sin xacosx±bsinx in the form Rcos⁡(x±α)R \cos (x \pm \alpha)Rcos(x±α) or Rsin⁡(x±β)R \sin (x \pm \beta)Rsin(x±β). You should also be able to solve equations of the form acos⁡x±bsin⁡x=ca \cos x \pm b \sin x = cacosx±bsinx=c. The addition formulae are used to help derive these expressions.



​Rcos⁡(x±α)R \cos (x \pm \alpha)Rcos(x±α) and Rsin⁡(x±β)R \sin(x \pm \beta)Rsin(x±β)​​

Use the addition formulae to write expressions of the form acos⁡x±bsin⁡xa \cos x \pm b \sin xacosx±bsinx in the form Rcos⁡(x±α)R \cos (x \pm \alpha)Rcos(x±α) or Rsin⁡(x±β)R \sin(x \pm \beta)Rsin(x±β). Once the addition formulae has been applied, sketch a right-angled triangle to find RRR and α\alphaα or β\betaβ.


Example 1

Write the expression 4sin⁡x+2cos⁡x4 \sin x + 2 \cos x4sinx+2cosx in the form Rsin⁡(x+α)R \sin (x+ \alpha)Rsin(x+α).


Use the addition formulae on Rsin⁡(x+α)R \sin (x+ \alpha)Rsin(x+α):

4sin⁡x+2cos⁡x=Rsin⁡(x+α)4sin⁡x+2cos⁡x=R[sin⁡xcos⁡α+cos⁡xsin⁡α]4sin⁡x+2cos⁡x=Rsin⁡xcos⁡α+Rcos⁡xsin⁡α\begin{aligned}4 \sin x + 2 \cos x &= R \sin (x+\alpha) \\4 \sin x + 2 \cos x &= R [\sin x \cos \alpha+ \cos x \sin \alpha] \\4 \sin x + 2 \cos x &= R \sin x \cos \alpha+ R\cos x \sin \alpha \\\end{aligned}4sinx+2cosx4sinx+2cosx4sinx+2cosx​=Rsin(x+α)=R[sinxcosα+cosxsinα]=Rsinxcosα+Rcosxsinα​​​


Equate the coefficients of sin⁡x\sin xsinx and cos⁡x\cos xcosx. Then draw a sketch to find RRR and α\alphaα using trigonometric rules:

​4=Rcos⁡α2=Rsin⁡α4 = R \cos \alpha \\2 = R \sin \alpha4=Rcosα2=Rsinα​​
Maths; Trigonometry and modelling; KS5 Year 13; Simplifying a cos x + b sin x
​R=42+22=25R = \sqrt{4^2+2^2}= 2\sqrt5R=42+22​=25​​​
​tan⁡α=24α=26.6°\begin{aligned}\tan \alpha &= \dfrac{2}{4}\\ \\\alpha& = 26.6\degree \end{aligned}tanαα​=42​=26.6°​​​


Therefore, 4sin⁡x+2cos⁡x=25sin⁡(x+26.6°)‾\underline{4 \sin x + 2 \cos x = 2 \sqrt5 \sin (x+26.6\degree)}4sinx+2cosx=25​sin(x+26.6°)​.


Solving equations with acos⁡x+bsin⁡xa\cos x + b \sin xacosx+bsinx​​

You should be able to solve equations of the form acos⁡x+bsin⁡x=ca\cos x + b \sin x=cacosx+bsinx=c. First use the above method to write the equation in the form Rsin⁡(x+α)R \sin (x+\alpha)Rsin(x+α) or Rcos⁡(x+α)R \cos (x+\alpha)Rcos(x+α) before solving the trigonometric equation. A sketch of the graph may help.

 

Example 2

The expression 4cos⁡x+7sin⁡x4 \cos x + 7 \sin x4cosx+7sinx can be written in the form Rcos⁡(x−α)R \cos (x-\alpha)Rcos(x−α).

a) Find RRR and α\alphaα.

b) Find the maximum value of the expression Rcos⁡(x−α)R \cos (x-\alpha)Rcos(x−α) and the minimum value of xxx for which this value occurs in the range 0≤x≤360°0 \le x \le 360\degree0≤x≤360°.

c) Solve 4cos⁡x+7sin⁡x=24 \cos x + 7 \sin x=24cosx+7sinx=2 in the range 0≤x≤360°0 \le x \le 360\degree0≤x≤360°.


a) Use the addition formula on Rcos⁡(x−α)R \cos (x-\alpha)Rcos(x−α):

4cos⁡x+7sin⁡x=Rcos⁡(x−α)4cos⁡x+7sin⁡x=R[cos⁡xcos⁡α+sin⁡xsin⁡α]4cos⁡x+7sin⁡x=Rcos⁡xcos⁡α+Rsin⁡xsin⁡α\begin{aligned}4 \cos x + 7 \sin x &= R \cos (x-\alpha) \\4 \cos x + 7 \sin x &= R [\cos x \cos \alpha + \sin x \sin \alpha] \\4 \cos x + 7 \sin x &= R \cos x \cos \alpha + R\sin x \sin \alpha \\\end{aligned}4cosx+7sinx4cosx+7sinx4cosx+7sinx​=Rcos(x−α)=R[cosxcosα+sinxsinα]=Rcosxcosα+Rsinxsinα​​​


Equate coefficients to find RRR and α\alphaα:

4=Rcos⁡α7=Rsin⁡αR=42+72R=65‾tan⁡α=74α=60.3°‾ (3 s.f.)\begin{aligned}4 &= R \cos \alpha \\7 &= R \sin \alpha \\\\R &= \sqrt{4^2+7^2} \\R &= \underline{\sqrt{65}}\\\\\tan \alpha&= \dfrac 7 4 \\ \\\alpha&= \underline{60.3 \degree} \ (3 \ s.f.)\end{aligned}47RRtanαα​=Rcosα=Rsinα=42+72​=65​​=47​=60.3°​ (3 s.f.)​​​


b) The expression is therefore 65cos⁡(x−60.3°)\sqrt{65} \cos (x-60.3\degree)65​cos(x−60.3°) and the maximum value occurs when cos⁡(x−60.3°)=1\cos(x-60.3\degree) = 1cos(x−60.3°)=1, so is 65×1\sqrt{65} \times 165​×1.​

Therefore, the maximum value of the expression is 65‾\underline{\sqrt{65}}65​​.


To find the minimum value of xxx for which this occurs, equate the expression to 65\sqrt{65}65​ and solve:

65cos⁡(x−60.3°)=65cos⁡(x−60.3°)=1x−60.3°=cos⁡−1(1)x−60.3°=0x=60.3°‾\begin{aligned}\sqrt{65} \cos (x-60.3\degree) &= \sqrt{65} \\\cos (x-60.3\degree) &= 1 \\x-60.3\degree &= \cos^{-1} (1) \\x-60.3\degree &= 0 \\x&= \underline{60.3\degree}\end{aligned}65​cos(x−60.3°)cos(x−60.3°)x−60.3°x−60.3°x​=65​=1=cos−1(1)=0=60.3°​​​​


c) Substitute the expression 65cos⁡(x−60.3°)\sqrt{65} \cos (x-60.3\degree)65​cos(x−60.3°) in and solve:

4cos⁡x+7sin⁡x=265cos⁡(x−60.3°)=2cos⁡(x−60.3°)=265x−60.3°=cos⁡−1(265)x−60.3°=75.64,284.36x=135.9°,344.7° (1 d.p.)‾\begin{aligned}4 \cos x + 7 \sin x&=2 \\\sqrt{65} \cos (x-60.3\degree)& = 2 \\\cos (x-60.3\degree) &= \dfrac{2}{\sqrt{65}} \\x-60.3\degree &= \cos^{-1} \Big(\dfrac{2}{\sqrt{65}}\Big) \\x-60.3\degree &= 75.64 , 284.36 \\x &= \underline{135.9 \degree, 344.7 \degree \ (1 \ d.p.)}\end{aligned}4cosx+7sinx65​cos(x−60.3°)cos(x−60.3°)x−60.3°x−60.3°x​=2=2=65​2​=cos−1(65​2​)=75.64,284.36=135.9°,344.7° (1 d.p.)​​



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FAQs - Frequently Asked Questions

How do you write expressions of the form acosx ± bsinx in the form Rcos(x±α) or Rsin(x±β)?

Use the addition formulae to write expressions of the form acosx ± bsinx in the form Rcos(x±α) or Rsin(x±β). Equate the coefficients of cos x and sin x, draw a right-angled triangle and use Pythagoras' theorem to find R and trigonometry to find α.

How do you solve equations of the form acosx±bsinx=c?

Write the expression acosx±bsinx in the form Rcos(x±α) or Rsin(x±β) and solve as a normal trigonometric equation.

What is the maximum value of an expression in the form Rcos(x±α) or Rsin(x±β)?

Maximum values occur when cos(x±α) or sin(x±β) are equal to 1. Therefore the maximum value is equal to R.

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