Question

show that
4sin∅cos∅=2sin²∅​

Answer 1

Topic:

Trigonometry

Question:

Prove that

$4 \sin\alpha \cos\alpha = 2 { \sin }^{2} \alpha$

Solution:

$4 \sin \alpha \cos \alpha = 2 \times { \sin}^{2} \alpha$

$take \: 2 \: common \: \\ 2(2 \sin\alpha \cos \alpha ) \\ = 2 \sin2 \alpha$

More Information :

Trignometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigonometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj

Multiples:

sin2θ= 2 sinθ cosθ

cos2θ= cos²θ-sin²θ

tan2A= 2tanA/1-tan²A