Question

if angle A = 45° prove that sin2A = 2sinAXcosA

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Answer 1

Step-by-step explanation:

Solution

verified

A=45

∘

then 2A=90

∘

sin2A=sin90

∘

RHS

2sinAcosB=2sin45

∘

cos45

∘

=2×

2

1

×

2

1

2sinAcosB=1

LHS:

sin90

∘

=1

L.H.S.=R.H.S.

Answer 2

Correct Question:

If angle A = 45°. Prove that sin2A = 2sinA × cosA

Step-by-step explanation:

$\begin{array}{ |c |c|c|c|c|c|} \sf\angle A & \sf{0}^{ \circ} & \sf{30}^{ \circ} & \sf{45}^{ \circ} & \sf{60}^{ \circ} & \sf{90}^{ \circ} \\ \\ \sf sin \: A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} & 1 \\ \\ \sf cos \: A & 1 & \dfrac{ \sqrt{3} }{2} & \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} & 0 \\ \\ \sf \: tan \: A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \sf Not \: de fined \\ \\ \sf cosec \: A & \sf Not \: de fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } & 1 \\ \\ \sf sec \: A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \sf Not \: de fined \\ \\ \sf cot \: A & \sf Not \: de fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}$

If angle A is 45°. We have to prove that sin2(45)° = 2sin(45)° × cos(45)°

LHS,

$\longrightarrow$ sin2(45)°

$\longrightarrow$ sin(90)°

From the table of trigonometric values,

$\longrightarrow$ sin(90)° = 1

LHS = 1

___________________

RHS,

$\longrightarrow$ 2sin(45)° × cos(45)°

From the table of trigonometric values,

$\longrightarrow$ 2 × (1/√2) × 1/√2

$\longrightarrow$ √2 × 1 × 1/√2

$\longrightarrow$ √2/√2

$\longrightarrow$ 1

RHS = 1

LHS = RHS

Hence proved that, sin2A = 2sinA × cosA.