If A = 60° and B = 30°, prove 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.
(iii) cos (A - B) = cos A cos B + sin A sin B.
tan A-tan B
(iv) tan (A - B) =tanA -tan B /
1+tan A tan B
Answers
Step-by-step explanation:
Hey mate !!
Here's your answer !!
Given :
A = 60° , B = 30°
To verify :
Sin ( A - B ) = Sin A . Cos B - Cos A . Sin B
Proof :
Let's substitute the values of A and B directly in the equation that is required for verification.
= Sin ( 60 - 30 ) = Sin 60 . Cos 30 - Cos 60 . Sin 30
= Sin 30 = Sin 60 . Cos 30 - Cos 60 . Sin 30 -----( Eqn. 1 )
We know that,
Sin 30 = 1 / 2
Sin 60 = √ 3 / 2
Cos 30 = √ 3 / 2
Cos 60 = 1 / 2
Now let's substitute the values in Equation 1.
=> 1 / 2 = √ 3 / 2 * √ 3 / 2 - 1 / 2 * 1 / 2
=> 1 / 2 = ( √ 3 / 2 )² - ( 1 / 2 )²
=> 1 / 2 = ( 3 / 4 - 1 / 4 )
=> 1 / 2 = ( 2 / 4 )
=> 1 / 2 = 1 / 2
Hence LHS = RHS
Hence verified !!
Hope my answer helped you mate !!
Cheers !!
Answer:
(i) sin(A+B)=sinA cosB+cosA sinB.
A=60° & B=30°.
sin(60°+30°)=sin60° cos30°+cos60° sin30°.
sin60°=√3/2.
cos30°=√3/2.
cos60°=1/2.
sin30°=1/2.
sin(90°)=√3/2 √3/2+ 1/2 1/2.
sin90°=1.
1=(√3/2)²+(1/2)².
1=3/4+1/4.
1=3+1/4.
1=4/4.
1=1.
(ii) cos(A+B)=cosA cosB-sinA sinB.
A=60° & B=30°.
cos(60°+30°)=cos60° cos30°-sin60° sin60°.
cos60°=1/2.
cos30°=√3/2.
sin60°=√3/2.
sin30°=1/2.
cos(90°)=1/2 √3/2-√3/2 1/2.
cos(90°)=0.
0=√3/4-√3/4.
0=0.
Sorry I didn't understand other two questions.