1. ## Trig identity question

Prove that

4 cos (θ + 60) cos (θ + 30) = √3 - 2sinθ

I have said

4 cos (θ + 60) cos (θ + 30)

= 4(cosθ cos60 − sinθ sin 60) (cosθ cos30 − sinθ sin 30)

= 4(cosθ 1/2 - sinθ √3/2) (cosθ √3/2 − sinθ 1/2 )

not sure where to go now

thanks

2. Originally Posted by gracey
Prove that

4 cos (θ + 60) cos (θ + 30) = √3 - 2sinθ

I have said

4 cos (θ + 60) cos (θ + 30)

= 4(cosθ cos60 − sinθ sin 60) (cosθ cos30 − sinθ sin 30)

= 4(cosθ 1/2 - sinθ √3/2) (cosθ √3/2 − sinθ 1/2 )

not sure where to go now

thanks
Hi gracey,

Are you sure the right side of the equation is not $\displaystyle \sqrt{3}-2 \sin {\color{red}2}\theta$?

3. Hello, Gracey!

masters is right . . . you left out a "2" . . .

Prove that: .$\displaystyle 4\cos(\theta + 60)\cos(\theta+30) \:=\:\sqrt{3} - 2\sin{\color{red}2}\theta$

I have said:

$\displaystyle 4\cos(\theta+60)\cos(\theta+30) \;=\; 4(\cos\theta\cos60 - \sin\theta\sin60) (\cos\theta\cos30 - \sin\theta\sin30)$

. . . . . $\displaystyle = \;4\left(\cos\theta\cdot\tfrac{1}{2} - \sin\theta\cdot\tfrac{\sqrt{3}}{2}\right)\left(\co s\theta\cdot\tfrac{\sqrt{3}}{2} - \sin\theta\cdot \tfrac{1}{2}\right)$

not sure where to go now . . . Really?

There is only one thing you can do ... multiply it out!

You have: .$\displaystyle 4\left(\frac{\cos\theta - \sqrt{3}\sin\theta}{2}\right)\left(\frac{\sqrt{3}\ cos\theta - \sin\theta}{2}\right)$

. . . . . . $\displaystyle =\;(\cos\theta - \sqrt{3}\sin\theta)(\sqrt{3}\cos\theta - \sin\theta)$

. . . . . . $\displaystyle = \;\sqrt{3}\cos^2\!\theta - \sin\theta\cos\theta - 3\sin\theta\cos\theta + \sqrt{3}\sin^2\!\theta$

. . . . . . $\displaystyle = \;\sqrt{3}\underbrace{(\cos^2\!\theta + \sin^2\!\theta)}_{\text{This is 1}} - 4\sin\theta\cos\theta$

. . . . . . $\displaystyle = \;\sqrt{3} - 2\underbrace{(2\sin\theta\cos\theta)}_{\text{This is }\sin2\theta}$

. . . . . . $\displaystyle =\;\sqrt{3}-2\sin2\theta$