Question

If,[A = π÷3 B =π÷6 ] prove that:sin(A + B) = sinAcosB + cosAsin B ​

Answer 1

Answer:

WE have to find value of sin15

o

=sin(45

o

−30

o

)

=sin45

o

cos30

o

−cos45

o

sin30

o

=

2

1

×

2

3

−

2

1

×

2

1

=

2

1

×

2

3

−

2

1

×

2

1

=

2

2

3

−1

Answer 2

Step-by-step explanation:

Given A = π÷3= 60° , B=π÷6=30°

Now , LHS

= sin(A+B)

= sin(60°+30°)

= sin90°

=1

RHS

=sinAcosB+cosAsinB

=sin60°cos30°+cos60°sin30°

$= \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2} + \frac{1}{2} \times \frac{1}{2}$

$= \frac{3}{4} + \frac{1}{4}$

$= \frac{3 + 1}{4}$

$= \frac{4}{4}$

$= 1$

Hence , LHS = RHS

i.e sin(A+B) = sinAcosB + cosAsinB