Question

Q18. If sin2A=2sinA is true then value of Ain
degree) is
0
30
Ο Ο Ο Ο
45
60​

Answer 1

Correct Question :

If sin2A=2sinA is true then value of A is ?

a) 30°

b) 0°

c) 45°

d) 60°

Solution :

We know that :- Sin 2x =2 Sin x Cos x

therefore , sin2A = 2 Sin A Cos A

➝ sin2A = 2sinA

➝ 2 Sin A Cos A = 2 Sin A

➝ 2 Sin A Cos A = 2 Sin A

➝ 2 Sin A Cos A -2 Sin A = 0

➝ 2 Sin A( Cos A - 1) = 0

➝ Sin A( Cos A - 1) = 0

Sin A = 0 or CosA - 1 = 0

➡️Sin A = 0 , then A = 0°

➡️ CosA - 1 = 0

➡️ CosA = 1 , then A = 0°

Therefore, value of A = 0°

Answer :

➝ Option (b) 0° is correct

$\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}$

Answer 2

Solution

sin 2A = 2 sin A

As a first step bring all terms to LHS.

sin 2A – 2 sin A = 0

As we know that sin 2A = 2 sin A cos A,putting the formula in above equation we get:

2 sin A cos A - 2 sin A=0 2 sin A (cos A – 1) = 0

2 sin A=0

or cos A – 1 = 0

sin A = sin 0 or cos A = cos 0

A=0 or A = 0

Therefore, the value of A=0.

Note: Whenever this type of problem appears first try to resolve the equation using trigonometric identity, after this we form an equation taking 2 sin A as common and after Remember the basic trigonometry identities that solve the equation and find the value of A.

$\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}$