Question

sin(A-B)=1/2, cos(A+B)=1/2 find A and B​

Answer 1

Step-by-step explanation:

sin(A-B) =1/2

cos(A+B) =1/2

right hand side is same

so

left side must be same

so

sin(A-B) =cos(A+B)

sin(A-B) =sin{90-(A+B) }

compare both side

A-B=90-(A+B)

A-B=90-A-B

A=90-A

2A=90

A=45

we know that A+B=90

so

45+B=90

B=45

so

A=45

B=45

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Answer 2

Topic:-

Trigonometry

Question:-

$sin(A-B)= \dfrac{1}{2} \\ cos(A+B)= \dfrac{1}{2} \\Then\: find\: the \:values\: of \:A \:and\: B$

Solution:-

$sin(A-B)= \dfrac{1}{2}$

$cos(A+B)= \dfrac{1}{2}$

$So, Sin(A-B) and cos(A+B) also be equal$

$Sin(A-B) = Cos(A+B)$

$We know that Cos( A+B) = Sin (90-(A+B))$

So,

$Sin(A-B) = Sin (90-(A+B))$

$Sin(A-B) = Sin (90-A-B)$

$By \: comparing \: both \: sides$

$A-B = 90-A-B$

$A \cancel{-B} = 90-A\cancel{-B}$

$A= 90-A$

$A+A= 90$

$2A= 90$

$A= \dfrac{90}{2}$

$A= \dfrac{\cancel{90}\:45}{\cancel{2}}$

$A= 45$

$We \:know\:that \:A+B=90°$

$A=45°$

$45+B=90°$

$B=90-45$

$B= 45°$

Answer:-

$A= 45° and B= 45°$

More Information :-

Trigon metric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj

Multiples:-

$sin2\theta= 2sin\theta cos\theta$

$sin2\theta=\dfrac{2tan\theta}{1+tan²\theta}$

$cos2\theta= cos²\theta-sin²\theta$

$cos2\theta= 1-2sin²\theta$

$cos2\theta= 2cos²\theta-1$

$cos2\theta= \dfrac{1-tan²\theta}{1+tan²\theta}$

$tan2\theta= \dfrac{2tan\theta}{1-tan²\theta}$

$cot2\theta= \dfrac{cot²\theta-1}{2cot\theta}$

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