Question

If sin(A-B)= 1/2 , cos(A+B)= 1/2;and 0<(A+B)<90° and A>B . then find the value of sinB​

Answer 1

Step-by-step explanation:

sin 15°=√6-√2/4.

Refer the above attachment.

Answer 2

Given:

• Sin(A-B) = 1/2.

• Cos(A+B) = 1/2.

• 0°<(A+B)<90°.

Find:

The Value of SinB.

Solution:

We know the angle value of sin and cos which give us the value 1/2.

They are 30° and 60° respectively,i.e.,

Sin30° = 1/2 and cos60° = 1/2.

So, we can say that:

Sin(A-B) = Sin30° = 1/2 and Cos(A+B) = Cos60° = 1/2.

From this:

A-B = 30. → Equation 1.

A+B = 60. → Equation 2.

Equation 1 + Equation 2:

2A = 90°.

A = 45°.

A-B = 30°

45°-B = 30°.

B = 45° - 30°.

B = 15°.

Now:

SinB = Sin15°

The value of Sin15°:

Sin15° = Sin(45°-30°)

By applying Sin(α-β) = sinα.cosβ - cosα.sinβ:

Sin15° = sin45°.cos30° - cos45°.sin30°.

$Sin15 = \frac{1}{\sqrt{2} }.\frac{\sqrt{3} }{2} -\frac{1}{\sqrt{2} } .\frac{1}{2}\\\\Sin15 = \frac{\sqrt{3} }{2\sqrt{2} } - \frac{1}{2\sqrt{2} } \\\\Sin15 = \frac{\sqrt{3}-1 }{2\sqrt{2} }$

$\therefore SinB = \frac{\sqrt{3} - 1}{2\sqrt{2} }$