Question Number 193381 by Mingma last updated on 12/Jun/23
Answered by som(math1967) last updated on 12/Jun/23
![2sin^2 4θ +2sin^2 2θ−2sin^2 θ=1 1−cos8θ+1−cos4θ−1+cos2θ=1 cos2θ−(cos4θ+cos8θ)=0 ⇒cos2θ−2cos6θcos2θ=0 ⇒cos2θ{1−2cos6θ}=0 ⇒(1−2cos6θ)=0 [ θ<90] cos6θ=(1/2) 6θ=60 ∴θ=10](https://www.tinkutara.com/question/Q193383.png)
$$\:\mathrm{2}{sin}^{\mathrm{2}} \mathrm{4}\theta\:+\mathrm{2}{sin}^{\mathrm{2}} \mathrm{2}\theta−\mathrm{2}{sin}^{\mathrm{2}} \theta=\mathrm{1} \\ $$$$\mathrm{1}−{cos}\mathrm{8}\theta+\mathrm{1}−{cos}\mathrm{4}\theta−\mathrm{1}+{cos}\mathrm{2}\theta=\mathrm{1} \\ $$$${cos}\mathrm{2}\theta−\left({cos}\mathrm{4}\theta+{cos}\mathrm{8}\theta\right)=\mathrm{0} \\ $$$$\Rightarrow{cos}\mathrm{2}\theta−\mathrm{2}{cos}\mathrm{6}\theta{cos}\mathrm{2}\theta=\mathrm{0} \\ $$$$\Rightarrow{cos}\mathrm{2}\theta\left\{\mathrm{1}−\mathrm{2}{cos}\mathrm{6}\theta\right\}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{2}{cos}\mathrm{6}\theta\right)=\mathrm{0}\:\:\left[\:\theta<\mathrm{90}\right] \\ $$$${cos}\mathrm{6}\theta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\mathrm{6}\theta=\mathrm{60}\:\:\therefore\theta=\mathrm{10} \\ $$$$ \\ $$
Commented by Mingma last updated on 12/Jun/23
Perfect ��
Commented by mnjuly1970 last updated on 13/Jun/23
$$\:\:\:\theta\:=\mathrm{10}^{\:°} \:\:,\:\:\theta\:=\:\mathrm{45}^{°} \\ $$
Commented by som(math1967) last updated on 13/Jun/23
$${yes},\:{if}\:{co}\mathrm{2}\theta=\mathrm{0}\Rightarrow\theta=\mathrm{45} \\ $$