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Grade 10Trigonometry

cosA+SinA=root2cosA,show thatcosA-sinA=root2 sinA

Profile image of rachel
11 Years agoGrade 10
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3 Answers

Profile image of Sourabh Singh
11 Years ago
Hii student

Take LHS from to prove and do the squaring and try to writecosA-SinA andSin2A in terms ofcosA+SinA.

Mke certain rearrangement by using the property

cos2A+Sin2A = 1

and you will get the answer.




Thanks.
.
Profile image of Nikhil Gaddam
11 Years ago
  1. CosA+SinA=​root2CosA
  2. SinA=root2CosA-CosA
  3. SinA=CosA(root2-1)
Multiply by (root2+1) on numerator as well as denominator
 
SinA=CosA(root2-1)(root2+1)  /   (root2+1) 
SinA=CosA(2-1)  /  (root2+1) 
SinA=CosA / (root2+1) 
CosA=SinA(root2+1) 
CosA=root2SinA+SinA
CosA-SinA=root2SinA
Thus Prooved
 
Profile image of Anonymous
5 Years ago

cosA + sinA = root 2cosA [given]

( cosA + sinA)^2= 2 cos^2A

cos^2A + sin^2A + 2 sinA cosA = 2 cos^2A

1+ 2 sinA cosA = 2 cos^2A

or 2 sinA cosA = 2 cos^2A -1

now cosA - sinA = sqrt[ (cosA+ sinA)^2 - 4 sinA cosA ]

= sqrt[ 2 cos^2A -2( 2os^2A - 1)

=sqrt2 sqrt [ cos^2A - 2cos^2A +1 ]

=sqr2 sqrt[ 1- cos^2A]

= sqrt2 sinA PROVED

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