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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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In triangle ACB, angle ACB is right angle, AB = 29 unit, it is hypotenuse, BC = 21unit, So, AC2= hypotenuse2- BC2= AB2 - BC2= 292 – 212= 202 or AC = 20 unit. angle ABC = angle β. sin β = AC/AB = 20/29 and cos β= BC/AB = 21/29 …now Question 1: Sin2 β +cos2 β... read more

In triangle ACB, angle ACB is right angle, AB = 29 unit, it is hypotenuse, BC = 21unit,

So, AC^{2}= hypotenuse^{2}- BC^{2}= AB^{2 }- BC^{2}= 29^{2 }– 21^{2}= 20^{2} or AC = 20 unit. angle ABC = angle β.

sin β = AC/AB = 20/29 and cos β= BC/AB = 21/29 …now

Question 1:

Sin^{2} β +cos^{2} β = (20/29)^{2} + (21/29)^{2} = 20^{2} /29^{2} + 21^{2} /29^{2} = 841/841 = 1 answer.

Question 2:

cos^{2} β – sin^{2} β = (21/29)^{2} –( 20/29)^{2} = 41/ 29^{2} = 41/841 = .048 answer.

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In ACB, we have AC = (AB2 BC2) = = (8)(50) = 400 = 20 units So, sin = AC/AB = 20/29, cos = BC/AB = 21/29. Now, (i) cos2 + sin2 = (29/21)2 + 21/292 = (202 + 212)/292 = (400 + 441)/841 = 1, and (ii) cos2 -- sin2 = (29/21)2 -- 21/292 = / 292 = 41/841 read more

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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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 more

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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