If the sides of a triangle are cos 2alpha +cos 2beta+2cos(alpha+beta) and sin2alpha +sin 2beta +2sin(alpha +beta),then the length of the hypotenuse is
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hi my name is PAVANKALYAN
Solution:
Let a = cos 2α + cos 2β + 2 cos(α+β)=
2 cos(α+β) . cos(α−β) + 2 cos(α+β)=2 cos(α+β)[cos(α−β) + 1]Let b = sin 2α + sin 2β + 2 sin(α+β)= 2 sin (α+β) . cos(α−β) + 2 sin(α+β)= 2 sin(α+β) . [cos(α−β) + 1]Now, let h be the hypotenuse.
Now, h2 = a2 + b2 [Pythagoras Theorem]
⇒h2 = 4 cos2(α+β) . [cos2(α−β)+1+2 cos(α−β)] + 4 sin2(α+β) [cos2(α−β)+1+2 cos(α−β)]⇒h2 = 4 [cos2(α−β)+1+2 cos(α−β)] [cos2(α+β) + sin2(α+β)]⇒h2 = 4[cos(α−β) + 1]2⇒h2 = 4[2 cos2(α−β2)]2⇒h2 = 16.cos4(α−β2)⇒h = 4 cos2(α−β2)
Solution:
Let a = cos 2α + cos 2β + 2 cos(α+β)=
2 cos(α+β) . cos(α−β) + 2 cos(α+β)=2 cos(α+β)[cos(α−β) + 1]Let b = sin 2α + sin 2β + 2 sin(α+β)= 2 sin (α+β) . cos(α−β) + 2 sin(α+β)= 2 sin(α+β) . [cos(α−β) + 1]Now, let h be the hypotenuse.
Now, h2 = a2 + b2 [Pythagoras Theorem]
⇒h2 = 4 cos2(α+β) . [cos2(α−β)+1+2 cos(α−β)] + 4 sin2(α+β) [cos2(α−β)+1+2 cos(α−β)]⇒h2 = 4 [cos2(α−β)+1+2 cos(α−β)] [cos2(α+β) + sin2(α+β)]⇒h2 = 4[cos(α−β) + 1]2⇒h2 = 4[2 cos2(α−β2)]2⇒h2 = 16.cos4(α−β2)⇒h = 4 cos2(α−β2)
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