If A = 30° and B = 60°, verify that
(i)sin (A + B) = sin A cos B + cos A sin B
(ii)cos (A + B) = cos A cos B − sin A sin B
Answers
SOLUTION :
(i) Given : A = 30°, B = 60°
Verify : Sin (A + B) = Sin A Cos B + Cos A Sin B
LHS = Sin (A + B)
On Substituting A = 30° , B = 60° in LHS
= Sin (30° + 60°)
= Sin 90°
= 1
[ sin 90°= 1]
LHS = 1
RHS = Sin A Cos B + Cos A Sin B
On Substituting A = 30° , B = 60° in RHS
= Sin 30° Cos 60° + Cos 30° Sin 60°
= ½ × ½ + √3/2× √3/2
[Sin 30°= ½ , Cos 60° = ½ , Cos 30° = √3/2, Sin 60°= √3/2]
= ¼ + ¾
= (1+3)/4
= 4/4 = 1
= 1
RHS = 1
LHS = RHS
Hence, Sin (A + B) = Sin A Cos B + Cos A Sin B
(ii) Given : A = 30°, B = 60°
Verify : Cos (A + B) = Cos A Cos B – Sin A Sin B
LHS = Cos (A + B)
On Substituting A = 30° , B = 60° in LHS
= Cos (30° + 60°)
= Cos 90° = 0
[ Cos 90° = 0]
LHS = 0
RHS = Cos A Cos B – Sin A Sin B
= Cos 30° Cos 60° – Sin 30° Sin 60°
= √3/2 × ½ - ½ × √3/2
[Cos 30° = √3/2, Cos 60° = ½, Sin 30°= ½ ,Sin 60°= √3/2]
= √3/4 - √3/4 = 0
RHS = 0
LHS = RHS = 0
Hence, Cos (A + B) = Cos A Cos B – Sin A Sin B
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Solution :-
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☆ Given,
A = 60° , B = 30°
● Now
sin(A - B) = sinA cos B - cos A sin B.
:- puting A = 60 , B = 30°
sin(60 - 30 ) = sin60cos30- cos 60sin 30
●We know, Trigonometry ratios of particular angles : sin90 = 1 , sin30 = 1/2 , sin60 = √3/2 , cos30 = √3/2 , cos60 = 1/2
sin30 = √3/2 ( √3/2 ) - 1/2 ( 1/2 )
1/2 = 3/4 - 1/4
1 /2= 2/4
1/2 = 1/2
Both Sides of the equation are equal.
Hence, We proved and verified that sin(A - B) = sinA cos B - cos A sin B. holds good for A = 60, B = 30°
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