Question

If sinA=1/2,find the value of cos 2A.​

Answer 1

Answer:

given : sin A = 1/2 so sin A = perpendicular / hypotenious (1)eq

Step-by-step explanation:

using phytagorus therom

hypotension square = perpendicular square + base square

2 square = 1 square + base square from eq 1

4 = 1 + base square

root 3 = base .

as cos A = base / hypotenious

cos A = root 3 / 2

so cos 2A = root 3/2 ×2 = root3

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Answer 2

$\maltese$Given :-

$sinA= \dfrac{1}{2}$

$\maltese$To find:-

$cos2A$

$\maltese$SOLUTION:-

$sinA = \dfrac{1}{2}$

As we know that $sin30^{\circ}$ is 1/2 So,

$sinA = sin30^{\circ}$

Removing sin on both sides So,

$A= 30^{\circ}$

We need cos2A substitute value of A

$cos2A = cos2(30^{\circ})$

$cos2A = cos60^{\circ}$

From trigonometric table we know that ,

$cos 60^{\circ}$ is 1/2 So,

$cos2A = \dfrac{1}{2}$

$\maltese$Know more :-

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

$\maltese$ Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

$\maltese$Trigonometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

$\maltese$Trigonometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj