Find the value of $\cos (15^{ \circ }).$

Solution:

We know that

$\cos (x- y) = \cos x \ \cos y + \sin x \ \sin y$

$\cos (x+ y) = \cos x \ \cos y - \sin x \ \sin y$

$\therefore \cos (15^{ \circ })= \cos (45^{ \circ } -30^{ \circ }) = \cos \biggl (\frac {\pi}{4}- \frac {\pi}{6}\biggl)$

$= \cos 45^{ \circ } \cos30^{ \circ } + \sin 45^\circ \sin 30^\circ$

$\frac {1}{\sqrt2} \times \frac {\sqrt3}{2} + \frac {1}{\sqrt2} \times \frac {1}{2}= \frac {(1+ \sqrt3)} {2\sqrt2}$

$2\cos x \cos y = \cos (x -y) + \cos (x +y)$

$2\sin x \sin y = \cos (x -y) - \cos (x +y)$

Find the value of $\cos (7.5 ^ \circ ) = \cos \biggl (\frac {15 ^ \circ}{2} \biggl).$

Solution:
We know that

$\cos (2x) = \cos (x+x)$

$\cos (2x) = \cos^2x - \sin^2x$

$\cos (2x) = \cos^2x - (1-\cos^2x)$

$\cos (2x) = 2\cos^2x - 1$

$\cos (2x) =1- 2\sin^2x$

$2\cos^2x =1+ \cos2x$

$\cos x = \pm \sqrt{\frac{1+ \cos2x}{2}}$

$\cos\ 7.5 ^ \circ = \sqrt{\frac{1+ \cos15 ^ \circ}{2}}$

• $\cos \biggl(x- \frac {\pi}{2}\biggl) = \cos x \ \cos \frac {\pi}{2} + \sin x \ \sin \frac {\pi}{2}$

$\quad \quad = \sin x \quad \quad \biggl(x=y+ \frac {\pi}{2} \biggl)$

• $\cos \biggl(x+ \frac {\pi}{2}\biggl) = \cos x \ \cos \frac {\pi}{2} - \sin x \ \sin \frac {\pi}{2}$

$\quad \quad = -\sin x \quad \quad \biggl(x=y- \frac {\pi}{2} \biggl)$

• $\sin \biggl(y+ \frac {\pi}{2}\biggl) = \cos \biggl(y+\frac {\pi}{2}- \frac {\pi}{2} \biggl) =\cos(y)$

• $\sin \biggl(y- \frac {\pi}{2}\biggl) = -\cos \biggl(y-\frac {\pi}{2}+ \frac {\pi}{2} \biggl) = \cos (y)$

$\sin z = \cos \biggl(z - \frac {\pi} {2}\biggl)$

$\therefore \sin (x- y) = \cos \biggl(x - y- \frac {\pi} {2}\biggl)$

$\sin (x- y) = \cos \biggl(x - \biggl(y +\frac {\pi} {2}\biggl)\biggl)$
We know that,
$\cos (x- y) = \cos x \ \cos y + \sin x \ \sin y$

$\therefore \sin (x- y) = \cos x \ \cos \biggl(y +\frac {\pi} {2}\biggl) + \sin x \ \sin \biggl(y +\frac {\pi} {2}\biggl)$

$\sin (x- y) = \cos x \ (-\sin y) + \sin x \ \cos y$

$\sin (x- y) = \sin x \ \cos y - \cos x \ \sin y$

$\sin (x+ y) =\sin (x -(-y))$

$\because \sin (x- y) = \sin x \ \cos y - \cos x \ \sin y$

$\therefore \sin (x+y) = \sin x \ \cos (-y) - \cos x \ \sin (-y)$

$\sin (x+ y) = \sin x \ \cos y + \cos x \ \sin y$

$\sin (x- y) = \sin x \ \cos y - \cos x \ \sin y \cdots (i)$

$\sin (x+ y) = \sin x \ \cos y + \cos x \ \sin y \cdots (ii)$
Add $(i) \ \& \ (ii)$ ,

$2\ \sin x \ \cos y = \sin (x +y) + \sin (x-y)$

Subtract $(ii) \ \& \ (i)$ ,

$2\ \cos x \ \sin y = \sin (x +y) - \sin (x-y)$

$\sin (x+ y) = \sin x \ \cos y + \cos x \ \sin y$
$x=y$

$\sin 2x = \sin (x +x)$

$\sin 2x = 2 \sin x \cos x$

$\cos 2x = 2 \cos^2x -1$

$\cos 2x = 2(1-\sin^2x) -1$

$\cos 2x = 1-2\sin^2x$

$\therefore \sin^2x = \frac {1-\cos 2x} {2}$

$\sin x = \pm \sqrt \frac {1-\cos 2x} {2}$

$x =7.5 ^{\circ}$

$\sin\ 7.5 ^{\circ} = \sqrt \frac {1-\cos 15^{\circ}} {2}$

We know that

$2 \cos x \ \cos y = \cos (x + y) + \cos (x - y)$

Let, $x = \frac {(A + B)} {2}, \quad y = \frac {(A - B)} {2}$

$\therefore 2 \cos \frac {(A + B)} {2} \times \cos \frac {(A - B)} {2} = \cos A + \cos B$

We know that
$2 \sin x \ \sin y = \cos (x-y) - \cos (x+y)$

Let, $x = \frac {(A + B)} {2}, \quad y = \frac {(A - B)} {2}$

$\therefore 2 \sin \frac {(A + B)} {2} \times \sin \frac {(A - B)} {2} = \cos (B) - \cos (A)$

We know that
$2 \sin(x) \ \cos (y) = \sin (x+y) + \sin (x-y)$

Let $x = \frac {(A + B)} {2}, \quad y = \frac {(A - B)} {2}$

$\sin (A) + \sin (B) = 2 \sin \frac {(A + B)} {2} \times \cos \frac {(A - B)} {2}$

$\because \sin (x +y) = \sin x \cos y + \cos x \sin y$

$\& \ \sin (x -y) = \sin x \cos y - \cos x \sin y$
$\therefore 2 \cos(x) \ \sin (y) = \sin (x+y) - \sin (x-y)$
Let, $x = \frac {(A + B)} {2}, \quad y = \frac {(A - B)} {2}$

$\therefore \sin (A) - \sin (B) = 2 \cos \frac {(A + B)} {2} \times \sin \frac{(A - B)}{2}$

$\cos (3x) = \cos (2x+x)$

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

$\cos (3x) = \cos 2x\ \cos x -\sin 2x\ \sin x$

$\cos (3x) = (2\cos^ 2x - 1) \cos x \\ \quad \quad -2 \sin x\ \cos x \ \sin x$

$\cos (3x) = 2\cos^ 3x - \cos x \\ \quad \quad -2 \cos x (1- \cos^2 x )$

$\cos (3x) = 2\cos^ 3x - \cos x -2 \cos x + 2 \cos^3 x$

$\cos (3x) = 4\ \cos^ 3x - 3\ \cos x$

$\sin (3x) = \sin (2x +x)$

$\sin (A+B) = \sin A \cos B + \cos A \sin B$

$\sin (3x) = \sin 2x \ \cos x + \cos 2x \ \sin x$

$\sin (3x) = 2\sin x \ \cos x \cos x + (1- 2\sin^2 x) \sin x$

$\sin (3x) = 2\sin x (1- \sin^2 x) + \sin x -2 \sin ^3x$

$\sin (3x) = 3\sin x - 4 \sin ^3x$

$\tan (x) = \frac {b} {a}$

$\sin (x) = b$

$\cos (x) = a$

$\tan (x) = \frac {\sin(x)} {\cos(x)}$

$\tan (x) = \frac {b} {a} =\frac{\sin x}{\cos x}$

$\tan (-x) = \frac {\sin (-x)} {\cos (-x)} =\frac {-\sin x} {\cos x}$

$\tan (-x) = -\ \tan x$

$\tan (x+ {\pi}) = \frac {\sin (x+ {\pi})} {\cos (x+ {\pi})} = \frac {\sin x \cos {\pi} + \cos x \sin {\pi}} {\cos x \cos {\pi} - \sin x \sin {\pi}}\\ \quad \quad = \tan x$

Similarly, $\tan (x- {\pi}) = \tan x$

$\tan \biggl(\frac {\pi} {2} -x\biggl) = \frac {\sin (\frac {\pi} {2} -x )} {\cos (\frac {\pi} {2} -x)} = \frac {\cos x} {\sin x}$

$\tan \biggl(\frac {\pi} {2} -x\biggl) = \frac {1} {\tan x} \\= \cot x \ (\text{co- tangent})$