Bạn đang xem: Solution: if tan x + cot x = 2, then find the value tan^17 x + cot^17 x
left( an(x)-cot(x)-left( an(x)+cot(x) ight) ight)left( an(x)-cot(x)+ an(x)+cot(x) ight)
displaystylex=1. Explanation:We know that,displaystyle an^-1x+cot^-1x=fracpi2ldotsldotsldotsldotsldotsleft(ast ight).displaystyle herefore5 an^-1x+3cot^-1x=2pi. ...
https://math.stackexchange.com/questions/2417589/solve-this-trigonometric-equation-tan-2x-cot-2x-3-tan-x-cot-x
Let anx-cotx=t. Thus, t^2+2-3t=0. For t=1 we obtain an^2x- anx-1=0, which gives x=arctanfrac1pmsqrt52+pi k, where kinmathbb Z. For t=2 we obtain an^2x-2 anx-1=0, ...
Prove the trigonometric identity? displaystyle anxleft(1-cot^2x ight)+cotxleft(1- an^2x ight)=0
Prove trig identity.Explanation:Replace in the equationdisplaystyle anx=fracsinxcosx , anddisplaystylecotx=fraccosxsinxdisplaystylefracsinxcosxleft(1-fraccos^2xsin^2x ight)+fraccosxsinxleft(1-fracsin^2xcos^2x ight)= ...
displaystyle3 an^-1left(x ight)+cot^-1left(x ight)=pi . Solve for x?
Douglas K. Feb 27, 2018 Given:displaystyle3 an^-1left(x ight)+cot^-1left(x ight)=pi Use the cosine function on both sides: displaystylecosleft(3 an^-1left(x ight)+cot^-1left(x ight) ight)=cosleft(pi ight) ...
https://math.stackexchange.com/questions/2436442/how-do-i-find-the-primitive-of-fx-arctanx2arccotanx2
Yes, you are correct. From, an(x^2) = operatornamecotleft(frac pi 2 - x^2 ight) We have, arctan(x^2) = frac pi 2 - operatornamearccot(x^2) Hence, int (arctan(x^2) + operatornamearccot(x^2)) mathrm dx = int left(frac pi 2 - operatornamearccot(x^2) + operatornamearccot(x^2) ight)mathrm dx ...
cot heta - an heta = dfraccos hetasin heta - dfracsin hetacos heta = 2 dfraccos^2 heta - sin^2 heta2,cos heta,sin heta = 2 dfraccos 2, hetasin 2, heta ...
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left( an(x)-cot(x)-left( an(x)+cot(x) ight) ight)left( an(x)-cot(x)+ an(x)+cot(x) ight)
The difference of squares can be factored using the rule: a^2-b^2=left(a-b ight)left(a+b ight).
left< eginarray l l 2 và 3 \ 5 & 4 endarray ight> left< eginarray l l l 2 và 0 và 3 \ -1 & 1 và 5 endarray ight>
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