tan thita =4/5 find 4cos2thita- 5sin 2thita
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Nileshshambharkar:
ans -4/7. and -1/8. hai
ye ans hai
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Tan θ = 4 / 5
Tan θ = Perpendicular / Base
By pythagoras theorem,
H² = P² + B ²
H² = 4² + 5²
H² = 16 + 25
H² = 41
H = √ 41
Sin θ = Perpendicular / Hypotenuse
Sin θ = 4 / √ 41
Cos θ = Base / Hypotenuse
Cos θ = 5 / √ 41
ACCORDING TO THE QUESTION,
4 cos 2 θ - 5 sin 2 θ
=> 4 ( 1 - 2 sin² θ ) - 5 sin θ cos θ
=> 4 [ 1 - 2 ( 4 / √ 41 ) ²] - 5 * ( 4 / √ 41 ) * ( 5 / √ 41 )
=> 4 [ 1 - 2 * 16 / 41 ] - 100 / 41
=> 4 [ 1 - 32 / 41 ] - 100 / 41
=> 4 ( 41 - 32 / 41 ) - 100 / 41
=> ( 4 * 9 / 41 ) - 100 / 41
=> ( 36 / 41 ) - 100 / 41
=> -64 / 41.
Tan θ = Perpendicular / Base
By pythagoras theorem,
H² = P² + B ²
H² = 4² + 5²
H² = 16 + 25
H² = 41
H = √ 41
Sin θ = Perpendicular / Hypotenuse
Sin θ = 4 / √ 41
Cos θ = Base / Hypotenuse
Cos θ = 5 / √ 41
ACCORDING TO THE QUESTION,
4 cos 2 θ - 5 sin 2 θ
=> 4 ( 1 - 2 sin² θ ) - 5 sin θ cos θ
=> 4 [ 1 - 2 ( 4 / √ 41 ) ²] - 5 * ( 4 / √ 41 ) * ( 5 / √ 41 )
=> 4 [ 1 - 2 * 16 / 41 ] - 100 / 41
=> 4 [ 1 - 32 / 41 ] - 100 / 41
=> 4 ( 41 - 32 / 41 ) - 100 / 41
=> ( 4 * 9 / 41 ) - 100 / 41
=> ( 36 / 41 ) - 100 / 41
=> -64 / 41.
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