(sin A+sin 3A+sin 5A+sin 7A)/ (COS A+cos3A+cos 5A+cos 7A) = tan 4A
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Step-by-step explanation:
L.H.S
=
cosA+cos3A+5A+cos7A
sinA+sin3A+sin5A+sin7A
We know that
sinC+sinD=2sin
2
C+D
⋅cos
2
C−D
Therefore,
=
2cos(
2
A+3A
)⋅cos(
2
A−3A
)+2cos(
2
5A+7A
)⋅cos(
2
5A−7A
)
2sin(
2
A+3A
)⋅cos(
2
A−3A
)+2sin(
2
5A+7A
)⋅cos(
2
5A−7A
)
=
2cos(2A)⋅cos(−A)+2cos(6A)⋅cos(−A)
2sin(2A)⋅cos(−A)+2sin(6A)⋅cos(−A)
=
2cos(2A)⋅cos(A)+2cos(6A)⋅cos(A)
2sin(2A)⋅cos(A)+2sin(6A)⋅cos(A)
[cos(−θ)=cosθ]
=
cos(2A)+cos(6A)
sin(2A)+sin(6A)
=
cos(2A)+cos(6A)
sin(2A)+sin(6A)
=
2cos(
2
2A+6A
)⋅cos(
2
2A−6A
)
2sin(
2
2A+6A
)⋅cos(
2
2A−6A
)
=
cos(4A)
sin(4A)
=tan4A
Hence, proved.
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