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
"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"




Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS