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We find max and min value of
asin@+bcos@ by +-( underroot ( a power2 + b power2))
Can we find the max and min value of asin(2@)+bcos(2@) using the same formula as +-( underroot ( a power2 + b power2)).
Please explain in detail. Urgently..
c/under root(a^2+b^2)= (a /root under(a^2+b^2)) sin@+ (b/root under(a^2+b^2) cos@
If u take a right angled triangle witrh sides a, b then its hypotenues will be under root(a^2+b^2)
therefore its in the form of:
c/root(a^2+b^2) = cos $ sin @ + sin$ cos @
c= root(a^2+ b^2) sin($+@)
-1<=sin(@+$)<= 1
therefore c lies between - root( a^2+b^2) and root(a^2+ b^2)
thus u can see even if u replace @ by 2@ u get the same result..
Plzzzzzzzzzzz approve my answer!!!!!!!!!
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