IIT JEE , AIEEE Forum - Mathematics , Trignometry

ranjana_25

if

a sin@ +b cos@ = c

prove that

a cos@ - b sin@ = the whole root of

thedumbheadwithnobrain

a²+b²-c²=a²+b²-a²sin²@+b²cos²@-2absin@cos@=a²cos²@+b²sin²@-2absin@cos@=(acos@-bsin@)²

hence (a²+b²-c²)^(1/2) =acos@-bsin@

Prajju

Taking
asin@ + bcos@ = c
squaring
==> a^2sin^2@ + b^2cos^@ + 2abcos@sin@ = c^2
==> a^2(1 - cos^2@) + b^2(1 - sin^2@) + 2absin@cos@ =c^2
==>a^2cos^2@ + b^2sin^2@ - 2abcos@sin@ = a^2 + b^2 - c^2
==> acos@ - bsin@ = sqrt ( a^2 + b^2 - c^2 )

mukulaish

yaar the answer is root(a^2+b^2+c^2)

the solution is as follows

asin@ + bcos@=c
let acos@-bsin@=x

squarin and addin the 2 equations we get
a^2+b^2=c^2+x^2

hence x=root(a^2+b^2-c^2)

vnkt.swaroop

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a2+b2-c2=a2+b2 -a2sin2@+b2cos2@-2absin@cos@=a2cos2@+ b2sin2@-2absin@cos@=(acos@-bsin@)2

hence (a2+b2-c2)(1/2) =acos@-bsin@

a²+b²-c²=a²+b²-a²sin²@+b²cos²@-2absin@cos@=a²cos²@+b²sin²@-2absin@cos@=(acos@-bsin@)²

hence (a²+b²-c²)^(1/2) =acos@-bsin@