Answer to Question #147260 in Trigonometry for Akrix Salram

Question #147260
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.
1
Expert's answer
2020-11-30T12:08:59-0500
SolutionSolution

Let a1,a2,a3,a4,a5a_1,a_2,a_3,a_4,a_5 be an arithmetic progression with common difference d such that cos d=0.2d=\sqrt{0.2}


a2=a1+d,a3=a2+d=a1+2d,a_2=a_1+d, a_3=a_2+d=a_1+2d,a4=a3+d=a1+3d,a5=a4+d=a1+4da_4=a_3+d=a_1+3d, a_5=a_4+d=a_1+4dtan(a2a1)=tana2tana11+tana1tana2\tan (a_2-a_1)=\dfrac{\tan a_2-\tan a_1}{1+\tan a_1\tan a_2}tana1tana2=tana2tana1tan(d)1\tan a_1\tan a_2=\dfrac{\tan a_2-\tan a_1}{\tan(d)}-1tana2tana3=tana3tana2tan(d)1\tan a_2\tan a_3=\dfrac{\tan a_3-\tan a_2}{\tan(d)}-1tana3tana4=tana4tana3tan(d)1\tan a_3\tan a_4=\dfrac{\tan a_4-\tan a_3}{\tan(d)}-1tana4tana5=tana5tana4tan(d)1\tan a_4\tan a_5=\dfrac{\tan a_5-\tan a_4}{\tan(d)}-1




tana1tana2+tana2tana3+\tan a_1\tan a_2+\tan a_2\tan a_3++tana3tana4+tana4tana5=+\tan a_3\tan a_4+\tan a_4\tan a_5=




=tana2tana1tan(d)1+tana3tana2tan(d)1+=\dfrac{\tan a_2-\tan a_1}{\tan(d)}-1+\dfrac{\tan a_3-\tan a_2}{\tan(d)}-1++tana4tana3tan(d)1+tana5tana4tan(d)1+\dfrac{\tan a_4-\tan a_3}{\tan(d)}-1+\dfrac{\tan a_5-\tan a_4}{\tan(d)}-1


Then



tana2tana1tan(d)1+tana3tana2tan(d)1+\dfrac{\tan a_2-\tan a_1}{\tan(d)}-1+\dfrac{\tan a_3-\tan a_2}{\tan(d)}-1++tana4tana3tan(d)1+tana5tana4tan(d)1=16+\dfrac{\tan a_4-\tan a_3}{\tan(d)}-1+\dfrac{\tan a_5-\tan a_4}{\tan(d)}-1=16tana5tana1=15tan(d)\tan a_5-\tan a_1=15\tan(d)sin(a3+2d)cos(a3+2d)sin(a32d)cos(a32d)=15tan(d)\dfrac{\sin(a_3+2d)}{\cos(a_3+2d)}-\dfrac{\sin(a_3-2d)}{\cos(a_3-2d)}=15\tan(d)sin(a3+2d)cos(a32d)cos(a3+2d)sin(a32d)cos(a3+2d)cos(a32d)=\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)}==15tan(d)=15\tan(d)sin(a3+2da3+2d)12(cos(4d)+cos(2a3))=15tan(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(2a3)=2sin(4d)15tan(d)\cos(4d)+\cos(2a_3)=\dfrac{2\sin(4d)}{15\tan(d)}cos(2a3)=2sin(4d)15tan(d)cos(4d)\cos(2a_3)=\dfrac{2\sin(4d)}{15\tan(d)}-\cos(4d)2cos2(a3)1=2sin(4d)15tan(d)cos(4d)2\cos^2(a_3)-1=\dfrac{2\sin(4d)}{15\tan(d)}-\cos(4d)2cos2(a3)=2sin(4d)15tan(d)+(1cos(4d))2\cos^2(a_3)=\dfrac{2\sin(4d)}{15\tan(d)}+(1-\cos(4d))2cos2(a3)=2sin(4d)15tan(d)+2sin2(2d)2\cos^2(a_3)=\dfrac{2\sin(4d)}{15\tan(d)}+2\sin^2(2d)cos2(a3)=sin(4d)15tan(d)+sin2(2d)\cos^2(a_3)=\dfrac{\sin(4d)}{15\tan(d)}+\sin^2(2d)

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



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

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



tan(d)=sin(d)cos(d)=0.80.2=82=2\tan(d)=\dfrac{\sin(d)}{\cos(d)}=\sqrt{\dfrac{0.8}{0.2}}=\sqrt{\dfrac{8}{2}}=2cos(2d)=2cos2(2d)1=2(0.2)21=0.6\cos(2d)=2\cos^2(2d)-1=2(\sqrt{0.2})^2-1=-0.6sin(2d)=2sin(d)cos(d)=20.80.2=0.8\sin(2d)=2\sin(d)\cos(d)=2\sqrt{0.8}\sqrt{0.2}=0.8sin(4d)=2sin(2d)cos(2d)=2(0.8)(2)=3.2\sin(4d)=2\sin(2d)\cos(2d)=2(0.8)(2)=3.2sin2(2d)=(0.8)2=0.64\sin^2(2d)=(0.8)^2=0.64cos2(a3)=3.2154+0.64=0.6933\cos^2(a_3)=\dfrac{3.2}{15\sqrt{4}}+0.64=0.6933

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

Then



tan(d)=sin(d)cos(d)=0.80.2=82=2\tan(d)=\dfrac{\sin(d)}{\cos(d)}=-\sqrt{\dfrac{0.8}{0.2}}=-\sqrt{\dfrac{8}{2}}=-2cos(2d)=2cos2(2d)1=2(0.2)21=0.6\cos(2d)=2\cos^2(2d)-1=2(\sqrt{0.2})^2-1=-0.6sin(2d)=2sin(d)cos(d)=20.80.2=0.8\sin(2d)=2\sin(d)\cos(d)=-2\sqrt{0.8}\sqrt{0.2}=-0.8sin(4d)=2sin(2d)cos(2d)=2(0.8)(2)=3.2\sin(4d)=2\sin(2d)\cos(2d)=-2(0.8)(2)=-3.2sin2(2d)=(0.8)2=0.64\sin^2(2d)=(-0.8)^2=0.64cos2(a3)=3.2154+0.64=0.6933\cos^2(a_3)=\dfrac{3.2}{-15\sqrt4{}}+0.64=0.6933

cos2(a3)=0.6933\cos^2(a_3)=0.6933




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