Question

Express 2cos$\theta$ +sin$\theta$ in the form of Rcos($\theta$ −$\alpha$) where r > 0 and 0^o< $\alpha$ <90^o.

Hence,

(a) find its maximum and minimum values.

(b) solve 2cos$\theta$ +sin$\theta$ = 2 for 0^o less than or equal to $\theta$, $\theta$ is less than or equal 360^o. Give your answer to the nearest 0.1^o.

Answer 1

$$Rcos(\theta-\alpha) =Rcos\theta cos\alpha + Rsin\theta sin\alpha.$$

$2cos\theta + sin\theta=Rcosθcosα+Rsinθsinα.$

Rcosθcosα=2cosθ

Rsinθsinα=sinθ

Rcosα=2... eqtn 1

Rsinα=1...eqtn 2

Divide 2 by 1

Sinα/Cosα=1/2

Tanα=1/2

α=26.3^o

Using SOH,CAH, TOA

Sinα=1/R

Cosα=2/R

Therefore R can be the hypotenuse

Hence, R=√ (1²+2²)

R=√5

2cosθ+sinθ=(√5)Cos(θ-26.3).

a)Using Rattafication formula for 2cosθ+sinθ

±√ (a²+ b²), where a =1, b =2, max is +and minimum is -.

Max is =+√(1²+2²)=√(-5).

Min is= - √(1²+2²)= -√(-5).

b) 2cosθ+sinθ=2

2cosθ-2=-sinθ

Square both sides

(2cosθ-2)² =(-sinθ)²

4cos²θ -8cosθ+4=sin²θ

Recall that sin²θ=1-cos²θ

4cos²θ-8cosθ+4=1+cos²θ

4cos²θ-8cosθ+4-1-cos²θ=0

3cos2θ-8cosθ+3=0.

Using quadratic formula,

Cosθ=(-b±√(b²−4ac)) /2a

Where a=3,b=-8 and c=3.

Cosθ=(-(-8)±√((-8)2-(4x3x3))/(2x3)

Cosθ=(8±√(64-36))/6

Cosθ=(8±√(28)) /6

Cosθ=(8±5.29)/6

Cosθ=(8+5.29)/6 and (8-5.29)/6

Cosθ=(13.29/6) and (2.71/6)

Cosθ=2.22 and 0.45

Cosθ=2.22,Cosθ=0.45

Cosθ cannot be equal to 2.22.

Therefore Cosθ =0.45,θ=63.3^o