Question

"Question19
Chapter6,T-Ratios of particular angles Exercise -6 ,Page number 288
Verify, sin60° cos30° - cos60° sin30° = sin30°"

Answer 1

HELLO DEAR,

we know that:-

sin(A - B) = sinAcosB - cosAcosB------(1)

sin60° cos30° - cos60° sin30° -----(2)

now on comparing ----(1) & -----(2)

we get,

sin(60° - 30°) = sin30°

sin30° = 1/2

another method,

sin60° cos30° - cos60° sin30°

= √3/2 × √3/2 - 1/2 × 1/2

= 3/4 - 1/4

= (3 - 1)/4

= 2/4

= 1/2

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hey there!
Many of us think that, We should apply the formula of sin(A - B) here, But as we are verifying the property. We should substitute values of Trigonometry ratios.


We know that,
sin60 = √3/2
cos30 = √3/2
cos60 = 1/2
sin30 = 1/2

Now,
Taking The Left Hand side,

= sin60° cos30° - cos60° sin30°

= √3/2 ( √3/2 ) - ( 1/2 ) ( 1/2 )

= 3/4 - 1/4

= 2/4

= 1/2

Taking R. H. S,
sin30

= 1/2

Therefore ,We see that L. H. S = R. H. S

Hence We proved that, sin60° cos30° - cos60° sin30° = sin30°