Question

PROVE the below IDENTITY :—
${\underline{\boxed{ \frac{1}{2} \sin(A) \cos(B) = \sin(A + B) }}}$

For example :—
$\red{ \dfrac{1}{2} } \sin( \red{20 \degree}) \cos( \red{20 \degree}) = \sin( \green{40 \degree})$

☺ Give step-by-step Explanation !!

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Answer 1

Answer:

Verify the fundamental trigonometric identities

Pythagorean Identities

sin2θ+cos2θ=1 1+cot2θ=csc2θ 1+tan2θ=sec2θ

Answer 2

Step-by-step explanation:

cos(A+B)=1/2 

cos(A+B)=cos 60 deg since cos 60 deg=1/2 

cancelling cos on both side we get 

A+B=60 deg.................................(1) 

similarly, 

sin(A-B)=1/2 

sin(A-B)=sin 30 deg 

A-B=30 deg....................................(... 

from 1 and 2 

A+B+A-B=90 deg 

2A=90 deg 

A=45 deg 

thus B= 15 deg 

answers A=45 AND B=15