Question

"Question23
If A = 45° , verify that : sin 2A = 2 sin A cos A
Chapter6,T-Ratios of particular angles Exercise -6 ,Page number 289"

Answer 1

Hey!

Given :; A = 45°

LHS = Sin2A

Sin2 × 45

= Sin90°

= 1

RHS :- 2SinA CosA

2 × Sin45° × Cos45°

= 2 × 1/√2 × 1/√2

( √2 × √2 = 2 will cancel 2 )

= 1

LHS = RHS

Hence , Verified !!

Answer 2

Hey there!

Given to verify : sin 2A = 2 sin A cos A

If A = 45°

sin2A = sin(2*45) = sin90 = 1 [ T - Ratio of 90° , π/2 ]

sinA = sin45 = 1/√2 [ T - Ratio of 45° , π/4 ]

cosA = cos45 = 1/√2 [ T - Ratio of 45° , π/4 ]

Now,
Given equation to verify : sin 2A = 2 sin A cos A

Finding the value of Left Hand Side ( L. H. S)

= sin2A

= sin90

= 1

Finding the value of Right Hand side [ R. H. S ]

= 2 sinA * cosA

= 2 ( 1/√2)(1/√2)

= √2/√2

= 1 .

Here, We observe that L. H. S = R. H. S.

Hence proved that, sin2A = 2sinA cosA for A = 45°