Answer to Question #147260 in Trigonometry for Akrix Salram

Question

Let a_1,a_2,a_3,a_4,a_5 be an arithmetic progression with common difference d such that cos d=sqrt(0.2).Find cos^2(a_3) given that tg(a_1)tg(a_2) + tg(a_2)tg(a_3) + tg(a_3)tg(a_4) + tg(a_4)tg(a_5) = 16.

Answer 1

$$Solution$$

Let $a_1,a_2,a_3,a_4,a_5$ be an arithmetic progression with common difference d such that cos $d=\sqrt{0.2}$

$$a_2=a_1+d, a_3=a_2+d=a_1+2d,$$ $$a_4=a_3+d=a_1+3d, a_5=a_4+d=a_1+4d$$ $$\tan (a_2-a_1)=\dfrac{\tan a_2-\tan a_1}{1+\tan a_1\tan a_2}$$ $$\tan a_1\tan a_2=\dfrac{\tan a_2-\tan a_1}{\tan(d)}-1$$ $$\tan a_2\tan a_3=\dfrac{\tan a_3-\tan a_2}{\tan(d)}-1$$ $$\tan a_3\tan a_4=\dfrac{\tan a_4-\tan a_3}{\tan(d)}-1$$ $$\tan a_4\tan a_5=\dfrac{\tan a_5-\tan a_4}{\tan(d)}-1$$

$$\tan a_1\tan a_2+\tan a_2\tan a_3+$$ $$+\tan a_3\tan a_4+\tan a_4\tan a_5=$$

$$=\dfrac{\tan a_2-\tan a_1}{\tan(d)}-1+\dfrac{\tan a_3-\tan a_2}{\tan(d)}-1+$$ $$+\dfrac{\tan a_4-\tan a_3}{\tan(d)}-1+\dfrac{\tan a_5-\tan a_4}{\tan(d)}-1$$

Then

$$\dfrac{\tan a_2-\tan a_1}{\tan(d)}-1+\dfrac{\tan a_3-\tan a_2}{\tan(d)}-1+$$ $$+\dfrac{\tan a_4-\tan a_3}{\tan(d)}-1+\dfrac{\tan a_5-\tan a_4}{\tan(d)}-1=16$$ $$\tan a_5-\tan a_1=15\tan(d)$$ $$\dfrac{\sin(a_3+2d)}{\cos(a_3+2d)}-\dfrac{\sin(a_3-2d)}{\cos(a_3-2d)}=15\tan(d)$$ $$\dfrac{\sin(a_3+2d)\cos(a_3-2d)-\cos(a_3+2d)\sin(a_3-2d)}{\cos(a_3+2d)\cos(a_3-2d)}=$$ $$=15\tan(d)$$ $$\dfrac{\sin(a_3+2d-a_3+2d)}{\dfrac{1}{2}\big(\cos(4d)+\cos(2a_3)\big)}=15\tan(d)$$ $$\cos(4d)+\cos(2a_3)=\dfrac{2\sin(4d)}{15\tan(d)}$$ $$\cos(2a_3)=\dfrac{2\sin(4d)}{15\tan(d)}-\cos(4d)$$ $$2\cos^2(a_3)-1=\dfrac{2\sin(4d)}{15\tan(d)}-\cos(4d)$$ $$2\cos^2(a_3)=\dfrac{2\sin(4d)}{15\tan(d)}+(1-\cos(4d))$$ $$2\cos^2(a_3)=\dfrac{2\sin(4d)}{15\tan(d)}+2\sin^2(2d)$$ $$\cos^2(a_3)=\dfrac{\sin(4d)}{15\tan(d)}+\sin^2(2d)$$

$\cos(d)=\sqrt{0.2}=>\sin(d)=\pm\sqrt{1-(\sqrt{0.2})^2}=\pm\sqrt{0.8}$

Let $\sin(d)=\sqrt{0.8}.$

$\sin(d)=\sqrt{0.8}.$ Then

$$\tan(d)=\dfrac{\sin(d)}{\cos(d)}=\sqrt{\dfrac{0.8}{0.2}}=\sqrt{\dfrac{8}{2}}=2$$ $$\cos(2d)=2\cos^2(2d)-1=2(\sqrt{0.2})^2-1=-0.6$$ $$\sin(2d)=2\sin(d)\cos(d)=2\sqrt{0.8}\sqrt{0.2}=0.8$$ $$\sin(4d)=2\sin(2d)\cos(2d)=2(0.8)(2)=3.2$$ $$\sin^2(2d)=(0.8)^2=0.64$$ $$\cos^2(a_3)=\dfrac{3.2}{15\sqrt{4}}+0.64=0.6933$$

Let $\sin(d)=-\sqrt{0.8}.$

Then

$$\tan(d)=\dfrac{\sin(d)}{\cos(d)}=-\sqrt{\dfrac{0.8}{0.2}}=-\sqrt{\dfrac{8}{2}}=-2$$ $$\cos(2d)=2\cos^2(2d)-1=2(\sqrt{0.2})^2-1=-0.6$$ $$\sin(2d)=2\sin(d)\cos(d)=-2\sqrt{0.8}\sqrt{0.2}=-0.8$$ $$\sin(4d)=2\sin(2d)\cos(2d)=-2(0.8)(2)=-3.2$$ $$\sin^2(2d)=(-0.8)^2=0.64$$ $$\cos^2(a_3)=\dfrac{3.2}{-15\sqrt4{}}+0.64=0.6933$$

$\cos^2(a_3)=0.6933$