Consider ?ACB, right-angled at C, in which AB = 29 units, BC = 21 units and ?ABC = ?. Determine the values of i. cos 2? + sin2?, ii. cos2? – sin2?.

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Vivek Mishra

Chemistry Teacher

For right angle triangle use pythagorous theorem (AC)^2 = (AB)^2 + (BC)2 PUT value of all find answer
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1, sqrt(41/841)
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Science Teacher

first find the value of AC by Pythagoras theorem then use the formula for cos2x and sin2x
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Use Pythagoras theorem 29sq-21sq= xsq X=20 Area= 21*10=210
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In Δ ACB, we have AC = √(AB2 − BC2) = √ = √(8)(50) = √400 = 20 units So, sin θ = AC/AB = 20/29, cos θ = BC/AB = 21/29. Now, (i) cos2θ + sin2θ = (29/21)2 + 21/292 = (202 + 212)/292 = (400 + 441)/841 = 1, and (ii) cos2θ -- sin2θ = (29/21)2 -- 21/292 = / 292 = 41/841
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In Δ ACB, we have AC = √(AB2 − BC2) = √[(29)2 − (21) 2) = √[(29 − 21)(29 + 21)] = √(8)(50) = √400 = 20 units So, sin θ = AC/AB = 20/29, cos θ = BC/AB = 21/29. Now, (i) cos2θ + sin2θ = (29/21)2 + 21/292 = (202 + 212)/292 = (400 + 441)/841 = 1, and (ii) cos2θ – sin2θ = (29/21)2 – 21/292 = [(21 + 20)(21 – 20)] / 292 = 41/841 read less
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Mathematics Tutor

Ab=29 bc=21 ac^2=ab^2-bc^2 Ac=20 Angleabc=cos^-1 (21/29)=sin^-1 (20/29) cos2?+sin2?=2cos^2?-1+2sin?cos? =2 (21/29)^2-1+2 (21*20/29*29)=881/841 cos2?-sin2?=2 (21/29)^2-1-2 (21*20/29*29)=-799/841
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Right triangle 1 1.034 2 0.964
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41/29 and1/29
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Using Pythagoras’s theorem; Third side = (29^2-21^2)^(1/2)=20 Now angle B=Cos^-1 (20/21) I assume next part asks cos^2B+Sin^2B, this will be 1 irrespective of what is B Second with - sign this equals to cos 2B or 2cos^2(B)-1 Just put cos B value here and get the result.
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Using Pythagoras’s theorem; Third side = (29^2-21^2)^(1/2)=20 Now angle B=Cos^-1 (20/21) I assume next part asks cos^2B+Sin^2B, this will be 1 irrespective of what is B Second with - sign this equals to cos 2B or 2cos^2(B)-1 Just put cos B value here and get the result. read less
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Maths Magic

1) 41/29. 2) 1/29.
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