Question

If sin ( A-B ) = sinA cosB - cosA sinB and cos ( A-B ) = cosA cosB + sinA sin B. Find the value of sin 15 degrees and cos 15 degrees.

Answer 1

It can be done easily by

Sin(45-30)

=√3/2 - 1/√2

=(√3-1)/2√2

And

CosA=√(1+sin^2A)

CosA=√(1+sin^2A)

Similarly

We would get the values

Answer 2

Given:

"sin (A-B) = sinA cosB - cosA sinB"

"cos (A-B) = cosA cosB + sinA sin B"

To find:

The value of $Sin15^{\circ} and Cos15^{\circ}$

Answer:

Given that

Sin (A-B) =SinA CosB - CosA SinB

Cos (A-B) =CosA CosB + SinA SinB

We need to find the $Sin15^{\circ} and Cos15^{\circ}$

$Sin15^{\circ}=Sin (45^{\circ}-15^{\circ})$

$=Sin45^{\circ} Cos15^{\circ} - Cos45^{\circ} Sin15^{\circ}$

$= \frac {1}{\sqrt {2}} \times \frac {\sqrt {3}}{2} - \frac {1}{\sqrt {2}} \times \frac {1}{2}$

$=\frac {\sqrt {3}}{2 \times {\sqrt {2}}} - \frac {1}{2 \times {\sqrt {2}}}$

$=\frac {\sqrt {3}-1}{2 \times \sqrt {2}}$

$Cos15^{\circ} = Cos (45^{\circ} - 15^{\circ})$

$= Cos45^{\circ} Cos15^{\circ} + Sin45^{\circ} Sin15^{\circ}$

$= \frac {1} {\sqrt {2}} \times \frac {\sqrt {3}}{ 2} + \frac {1}{\sqrt {2}} \times \frac {1}{2}$

$=\frac {\sqrt {3}}{2 \times \sqrt {2}} + \frac {1}{2 \times \sqrt {2}}$

$= \frac {\sqrt {3} + 1}{2 \times \sqrt {2}}$

$Sin15^{\circ}=\frac {\sqrt {3}-1}{2 \times \sqrt {2}}$

and

$Cos15^{\circ} = \frac {\sqrt {3} + 1}{2 \times \sqrt {2}}$