Answer to Question #142697 in Math for mkami

Question

Let a1, a2, a3, a4, a5 be an arithmetic progression with common difference d such that cosd = √0,6. Find cos^2(a3) given that tga1tga2 + tga2tga3 + tga3tga4 + tga4tg5 = 11.

Answer 1

$$a_2=a_1+d, a_3=a_2+d=a_1+2d,$$

$$a_4=a_3+d=a_1+3d, a_5=a_4+d=a_1+4d$$

$$\tan (a_2-a_1)=\dfrac{\tan a_2-\tan a_1}{1+\tan a_1\tan a_2}$$

$$\tan a_1\tan a_2=\dfrac{\tan a_2-\tan a_1}{\tan(d)}-1$$

$$\tan a_2\tan a_3=\dfrac{\tan a_3-\tan a_2}{\tan(d)}-1$$

$$\tan a_3\tan a_4=\dfrac{\tan a_4-\tan a_3}{\tan(d)}-1$$

$$\tan a_4\tan a_5=\dfrac{\tan a_5-\tan a_4}{\tan(d)}-1$$

$$\tan a_1\tan a_2+\tan a_2\tan a_3+$$

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

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

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

Then

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

$$+\dfrac{\tan a_4-\tan a_3}{\tan(d)}-1+\dfrac{\tan a_5-\tan a_4}{\tan(d)}-1=11$$

$$\tan a_5-\tan a_1=15\tan(d)$$

$$\dfrac{\sin(a_3+2d)}{\cos(a_3+2d)}-\dfrac{\sin(a_3-2d)}{\cos(a_3-2d)}=15\tan(d)$$

$$\dfrac{\sin(a_3+2d)\cos(a_3-2d)-\cos(a_3+2d)\sin(a_3-2d)}{\cos(a_3+2d)\cos(a_3-2d)}=$$

$$=15\tan(d)$$

$$\dfrac{\sin(a_3+2d-a_3+2d)}{\dfrac{1}{2}\big(\cos(4d)+\cos(2a_3)\big)}=15\tan(d)$$

$$\cos(4d)+\cos(2a_3)=\dfrac{2\sin(4d)}{15\tan(d)}$$

$$\cos(2a_3)=\dfrac{2\sin(4d)}{15\tan(d)}-\cos(4d)$$

$$2\cos^2(a_3)-1=\dfrac{2\sin(4d)}{15\tan(d)}-\cos(4d)$$

$$2\cos^2(a_3)=\dfrac{2\sin(4d)}{15\tan(d)}+(1-\cos(4d))$$

$$2\cos^2(a_3)=\dfrac{2\sin(4d)}{15\tan(d)}+2\sin^2(2d)$$

$$\cos^2(a_3)=\dfrac{\sin(4d)}{15\tan(d)}+\sin^2(2d)$$

$\cos(d)=\sqrt{0.6}=>\sin(d)=\pm\sqrt{1-(\sqrt{0.6})^2}=\pm\sqrt{0.4}$

Let $\sin(d)=\sqrt{0.4}.$ Then

$$\tan(d)=\dfrac{\sin(d)}{\cos(d)}=\sqrt{\dfrac{0.4}{0.6}}=\sqrt{\dfrac{2}{3}}$$

$$\cos(2d)=2\cos^2(2d)-1=2(\sqrt{0.6})^2-1=0.2$$

$$\sin(2d)=2\sin(d)\cos(d)=2\sqrt{0.4}\sqrt{0.6}=0.4\sqrt{6}$$

$$\sin(4d)=2\sin(2d)\cos(2d)=2(0.4\sqrt{6})(0.2)=0.16\sqrt{6}$$

$$\sin^2(2d)=(0.4\sqrt{6})^2=0.96$$

$$\cos^2(a_3)=\dfrac{0.16\sqrt{6}}{15\sqrt\dfrac{2}{3}{}}+0.96=0.992$$

Let $\sin(d)=-\sqrt{0.4}.$ Then

$$\tan(d)=\dfrac{\sin(d)}{\cos(d)}=-\sqrt{\dfrac{0.4}{0.6}}=-\sqrt{\dfrac{2}{3}}$$

$$\cos(2d)=2\cos^2(2d)-1=2(\sqrt{0.6})^2-1=0.2$$

$$\sin(2d)=2\sin(d)\cos(d)=-2\sqrt{0.4}\sqrt{0.6}=-0.4\sqrt{6}$$

$$\sin(4d)=2\sin(2d)\cos(2d)=-2(0.4\sqrt{6})(0.2)=-0.16\sqrt{6}$$

$$\sin^2(2d)=(-0.4\sqrt{6})^2=0.96$$

$$\cos^2(a_3)=\dfrac{-0.16\sqrt{6}}{-15\sqrt\dfrac{2}{3}{}}+0.96=0.992$$

$\cos^2(a_3)=0.992$