Question

sin47° + sin 61º- sin 11º-sin2° = cos7°

Answer 1

Answer:

sin47 ∘ +sin61 ∘ −(sin11 ∘ +sin25 ∘ )

Using the transformation angle formula sinC+sinD=2sin( C+D / 2)cos( C−D/ 2)

=(2sin( 47+61 / 2 )cos( 47−61 / 2))−(2sin( 11+25 /2)cos( 11−25 / 2))

=(2sin( 108 / 2 )cos( −14 / 2) ))−(2sin( 36 /2 )cos( −14 / 2 ))

=(2sin54∘ cos−7 ∘ )−(2sin18 ∘cos−7 ∘ )

=2cos7 ∘ (sin54 ∘−sin18 ∘ ) using cos(−θ)=cosθ

=2cos7 ∘ (2sin( 54−18 / 2 )cos( 54+18 / 2 ))

using the transformation angle formula sinC−sinD=2sin(

C−D / 2 )cos( C+D / 2 )

=4cos7∘sin18∘ cos36∘

We know that sin18∘ = √5 −1 / 4

and

cos36∘ = √5+1 / 4

=4cos7∘(√5−1 /4) × (√5+1 /4)

=4cos7∘ 4 / 4×4 =cos7∘

Answer 2

Answer:

using the transformation angle formula sin C+ sin D =2sin(C+D/2) cos(C-D/2)

=[2sin(47+61/2)cos(47-61/2)]

(2sin[11+25/2]cos[11-25/2])

={2si (108/2)cos(-14/2)-(2sin(36/2)cos(-14/2)

we know that sin18°=√5-1/4 andcos 36°=√5+1/4

=4 cos 7°√5-1/4×√5+1/4

=4cos7°4/4×4=cos7°