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sin x-sin 3x / cos x+cos 3x =tan 2x
Class 11
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Answers
Step-by-step explanation:
Given :-
(Sinx-Sin 3x) / (Cos x+Cos 3x)
Correction :-
(Sinx+Sin 3x) / (Cos x+Cos 3x)
To find :-
Prove that (Sinx+Sin 3x) / (Cos x+Cos 3x) = Tan 2x.
Solution :-
On taking LHS
Given that (Sinx+Sin 3x) / (Cos x+Cos 3x)
Numerator = Sin x+ Sin 3x
This is in the form of Sin C+ Sin D
We know that
SinC + SinD = 2Sin[(C+D)/2]Cos[(C-D)/2]
=> Sin x+ Sin 3x
=> 2 Sin (x+3x)/2 Cos (x-3x)/2
=> 2 Sin (4x/2) Cos (-2x/2)
=> 2 Sin 2x Cos (-x) -----------(1)
and
The denominator = Cos x + Cos 3x
This is in the form of Cos C + Cos D
We know that
CosC + CosD=2Cos[(C+D)/2]Cos[(C-D)/2]
=> Cos x + Cos 3x
=> 2 Cos (x+3x)/2 Cos (x-3x)/2
=> 2 Cos (4x/2) Cos (-2x/2)
=> 2 Cos 2x Cos (-x) ----------(2)
Now
(Sinx+Sin 3x) / (Cos x+Cos 3x)
=> (2 Sin 2x Cos (-x))/(2 Cos 2x Cos (-x))
On cancelling 2 Cos (-x) in both numerator and the denominator
=> Sin 2x / Cos 2x
=> Tan 2x
LHS = RHS
Hence, Proved
Answer:-
(Sinx+Sin 3x) / (Cos x+Cos 3x) = Tan 2x.
Used formulae:-
SinC + SinD = 2Sin[(C+D)/2]Cos[(C-D)/2]
CosC + CosD=2Cos[(C+D)/2]Cos[(C-D)/2]
Tan X = Sin X/ Cos X
Answer:
LHS=
cosx+cos3x
sinx+sin3x
=
2cos(
2
x+3x
)cos(
2
x−3x
)
−2sin(
2
x+3x
)cos(
2
x−3x
)
=tan2x=RHS
Hence, proved