##### <Success>Trigonometric functions : lecture 3</Success> <div style='float: left; text-align: left; width:100%; font-size:30px;'> </div> <p>======</p> <h3 id="successproblem-1success"><Success>Problem 1</Success></h3> <div style='float: left; text-align: left; width:100%; font-size:30px;'> <p>Find the value of $\cos (15^{ \circ }).$</p> <p><strong>Solution:</strong></p> <p>We know that</p> <p>$\cos (x- y) = \cos x \ \cos y + \sin x \ \sin y $</p> <p>$\cos (x+ y) = \cos x \ \cos y - \sin x \ \sin y $</p> <p>$\therefore \cos (15^{ \circ })= \cos (45^{ \circ } -30^{ \circ }) = \cos \biggl (\frac {\pi}{4}- \frac {\pi}{6}\biggl) $</p> <p>$= \cos 45^{ \circ } \cos30^{ \circ } + \sin 45^\circ \sin 30^\circ $</p> <p>$\frac {1}{\sqrt2} \times \frac {\sqrt3}{2} + \frac {1}{\sqrt2} \times \frac {1}{2}= \frac {(1+ \sqrt3)} {2\sqrt2}$</p> </div> <p>======</p> <h5 id="successexpression-for-cosine-and-sine-functionsuccess"><Success>Expression for cosine and sine function</Success></h5> <div style='float: left; text-align: left; width:100%; font-size:30px;'> <p>$2\cos x \cos y = \cos (x -y) + \cos (x +y) $</p> <p>$2\sin x \sin y = \cos (x -y) - \cos (x +y) $</p> </div>

### <Success>Problem 2</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Find the value of $\cos (7.5 ^ \circ ) = \cos \biggl (\frac {15 ^ \circ}{2} \biggl).$</p> <p><strong>Solution:</strong><br> We know that</p> <p>$\cos (2x) = \cos (x+x) $</p> <p>$\cos (2x) = \cos^2x - \sin^2x $</p> <p>$\cos (2x) = \cos^2x - (1-\cos^2x) $</p> <p>$\cos (2x) = 2\cos^2x - 1 $</p> <p>$\cos (2x) =1- 2\sin^2x $</p> </div> <p>======</p> <h5 id="successsolutionsuccess"><Success>Solution</Success></h5> <div style='float: left; text-align: left; width:100%; font-size:30px;'> <p>$2\cos^2x =1+ \cos2x $</p> <p>$\cos x = \pm \sqrt{\frac{1+ \cos2x}{2}} $</p> <p>$\cos\ 7.5 ^ \circ = \sqrt{\frac{1+ \cos15 ^ \circ}{2}} $</p> </div>

### <Success>Expression for cosine function</Success> <div style='float: left; text-align: left; width:100%; font-size:30px;'> <ul> <li>$\cos \biggl(x- \frac {\pi}{2}\biggl) = \cos x \ \cos \frac {\pi}{2} + \sin x \ \sin \frac {\pi}{2} $</li> </ul> <p>$\quad \quad = \sin x \quad \quad \biggl(x=y+ \frac {\pi}{2} \biggl)$</p> <ul> <li>$\cos \biggl(x+ \frac {\pi}{2}\biggl) = \cos x \ \cos \frac {\pi}{2} - \sin x \ \sin \frac {\pi}{2} $</li> </ul> <p>$\quad \quad = -\sin x \quad \quad \biggl(x=y- \frac {\pi}{2} \biggl) $</p> </div> <p>======</p> <h3 id="successexpression-for-sine-functionsuccess"><Success>Expression for sine function</Success></h3> <div style='float: left; text-align: left; width:100%; font-size:30px;'> <ul> <li> <p>$\sin \biggl(y+ \frac {\pi}{2}\biggl) = \cos \biggl(y+\frac {\pi}{2}- \frac {\pi}{2} \biggl) =\cos(y) $</p> </li> <li> <p>$\sin \biggl(y- \frac {\pi}{2}\biggl) = -\cos </Default>\biggl(y-\frac {\pi}{2}+ \frac {\pi}{2} \biggl) = \cos (y) $</p> </li> </ul> </div>

### <Success>$\text{Formula of} \ \sin (x - y)$</Success> <div style='float: left; text-align: left; width:100%; font-size:30px;'> <p>$\sin z = \cos \biggl(z - \frac {\pi} {2}\biggl) $</p> <p>$\therefore \sin (x- y) = \cos \biggl(x - y- \frac {\pi} {2}\biggl) $</p> <p>$\sin (x- y) = \cos \biggl(x - \biggl(y +\frac {\pi} {2}\biggl)\biggl) $<br> We know that,<br> $\cos (x- y) = \cos x \ \cos y + \sin x \ \sin y $</p> <p>$\therefore \sin (x- y) = \cos x \ \cos \biggl(y +\frac {\pi} {2}\biggl) + \sin x \ \sin \biggl(y +\frac {\pi} {2}\biggl) $</p> </div> <p>======</p> <h3 id="successtextformula-of--sin-x---ysuccess"><Success>$\text{Formula of} \ \sin (x - y)$</Success></h3> <div style='float: left; text-align: left; width:100%; font-size:30px;'> <p>$\sin (x- y) = \cos x \ (-\sin y) + \sin x \ \cos y $</p> <p>$\sin (x- y) = \sin x \ \cos y - \cos x \ \sin y $</p> </div>

### <Success>$\text{Formula of} \ \sin (x + y)$</Success> <div style='float: left; width:100%;font-size:30px;'> <p>$\sin (x+ y) =\sin (x -(-y)) $</p> <p>$\because \sin (x- y) = \sin x \ \cos y - \cos x \ \sin y $</p> <p>$\therefore \sin (x+y) = \sin x \ \cos (-y) - \cos x \ \sin (-y) $</p> <p>$\sin (x+ y) = \sin x \ \cos y + \cos x \ \sin y $</p> </div>

##### <Success>$\text{Formula of} \ 2 \sin x \cos y$ <div style='float: left; width:100%;font-size:30px;'> <p>$\sin (x- y) = \sin x \ \cos y - \cos x \ \sin y \cdots (i) $</p> <p>$\sin (x+ y) = \sin x \ \cos y + \cos x \ \sin y \cdots (ii) $<br> Add $(i) \ \& \ (ii)$ ,</p> <p>$2\ \sin x \ \cos y = \sin (x +y) + \sin (x-y) $</p> <p>Subtract $(ii) \ \& \ (i)$ ,</p> <p>$2\ \cos x \ \sin y = \sin (x +y) - \sin (x-y) $</p> <p>======</p> <h3 id="successtextformula-of--sin-2xsuccess"><Success>$\text{Formula of} \ \sin 2x$</Success></h3> <div style='float: left; text-align: left; width:100%; font-size:30px;'> <p>$\sin (x+ y) = \sin x \ \cos y + \cos x \ \sin y $<br> $x=y$</p> <p>$\sin 2x = \sin (x +x) $</p> <p>$\sin 2x = 2 \sin x \cos x $</p> </div>

### <Success>$\text{Value of} \ \sin 7.5^{\circ}$</Success> <div style='float: left;text-align: left; width:100%;font-size:30px;'> <p>$\cos 2x = 2 \cos^2x -1 $</p> <p>$\cos 2x = 2(1-\sin^2x) -1 $</p> <p>$\cos 2x = 1-2\sin^2x $</p> <p>$\therefore \sin^2x = \frac {1-\cos 2x} {2} $</p> <p>$\sin x = \pm \sqrt \frac {1-\cos 2x} {2} $</p> <p>$x =7.5 ^{\circ} $</p> <p>$\sin\ 7.5 ^{\circ} = \sqrt \frac {1-\cos 15^{\circ}} {2} $</p> </div>

### <Success>$\text{Formula of} \ \cos A + \cos B$</Success> <div style='float: left; text-align: left; width:100%; font-size:30px;'> <p>We know that</p> <p>$2 \cos x \ \cos y = \cos (x + y) + \cos (x - y) $</p> <p>Let, $x = \frac {(A + B)} {2}, \quad y = \frac {(A - B)} {2} $</p> <p>$\therefore 2 \cos \frac {(A + B)} {2} \times \cos \frac {(A - B)} {2} = \cos A + \cos B $</p> <p>======</p> <h3 id="successtextformula-of--cos-b---cos-asuccess"><Success>$\text{Formula of} \ \cos B - \cos A$</Success></h3> <div style='float: left; text-align: left; width:100%; font-size:30px;'> <p>We know that<br> $2 \sin x \ \sin y = \cos (x-y) - \cos (x+y) $</p> <p>Let, $x = \frac {(A + B)} {2}, \quad y = \frac {(A - B)} {2} $</p> <p>$\therefore 2 \sin \frac {(A + B)} {2} \times \sin \frac {(A - B)} {2} = \cos (B) - \cos (A) $</p> </div>

### <Success>$\text{Formula of} \ \sin A + \sin B$</Success> <div style='float: left; text-align: left; width:100%; font-size:30px;'> <p>We know that<br> $2 \sin(x) \ \cos (y) = \sin (x+y) + \sin (x-y) $</p> <p>Let $x = \frac {(A + B)} {2}, \quad y = \frac {(A - B)} {2} $</p> <p>$\sin (A) + \sin (B) = 2 \sin \frac {(A + B)} {2} \times \cos \frac {(A - B)} {2} $</p> <p>======</p> <h3 id="successtextformula-of--sin-a---sin-bsuccess"><Success>$\text{Formula of} \ \sin A - \sin B$</Success></h3> <div style='float: left; text-align: left; width:100%; font-size:30px;'> <p>$\because \sin (x +y) = \sin x \cos y + \cos x \sin y $</p> <p>$\& \ \sin (x -y) = \sin x \cos y - \cos x \sin y $<br> $\therefore 2 \cos(x) \ \sin (y) = \sin (x+y) - \sin (x-y) $<br> Let, $x = \frac {(A + B)} {2}, \quad y = \frac {(A - B)} {2} $</p> <p>$\therefore \sin (A) - \sin (B) = 2 \cos \frac {(A + B)} {2} \times \sin \frac{(A - B)}{2} $</p> </div>

### <Success>$\text{Formula of} \ \cos 3x$</Success> <div style='float: left;text-align: left; width:100%; font-size:30px;'> <p>$\cos (3x) = \cos (2x+x) $</p> <p>$\cos (A+B) = \cos A \cos B -\sin A\ \sin B $</p> <p>$\cos (3x) = \cos 2x\ \cos x -\sin 2x\ \sin x $</p> <p>$\cos (3x) = (2\cos^ 2x - 1) \cos x \\ \quad \quad -2 \sin x\ \cos x \ \sin x $</p> <p>$\cos (3x) = 2\cos^ 3x - \cos x \\ \quad \quad -2 \cos x (1- \cos^2 x ) $</p> <p>$\cos (3x) = 2\cos^ 3x - \cos x -2 \cos x + 2 \cos^3 x $</p> <p>$\cos (3x) = 4\ \cos^ 3x - 3\ \cos x $</p> </div>

### <Success>$\text{Formula of} \ \sin 3x$</Success> <div style='float: left; text-align: left; width:100%;font-size:30px;'> <p>$\sin (3x) = \sin (2x +x) $</p> <p>$\sin (A+B) = \sin A \cos B + \cos A \sin B $</p> <p>$\sin (3x) = \sin 2x \ \cos x + \cos 2x \ \sin x $</p> <p>$\sin (3x) = 2\sin x \ \cos x \cos x + (1- 2\sin^2 x) \sin x $</p> <p>$\sin (3x) = 2\sin x (1- \sin^2 x) + \sin x -2 \sin ^3x $</p> <p>$\sin (3x) = 3\sin x - 4 \sin ^3x $</p> </div>

### <Success>Tangent function</Success> <div style='float: left;text-align: left; width:50%; font-size:30px;'> <p>$\tan (x) = \frac {b} {a} $</p> <p>$\sin (x) = b $</p> <p>$\cos (x) = a $</p> <p>$\tan (x) = \frac {\sin(x)} {\cos(x)} $</p> </div> <div style='float: right; width:50%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/2fd657dc-6a93-4205-c553-283f8d9a4e00/public"></p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/f6c7b9ff-ff78-4b70-67e5-11ac24286e00/public"></p> </div>

### <Success>$\text{Graph of} \ \tan x$</Success> <div style='float: left; text-align: left; width:50%;font-size:30px;'> <p>$\tan (x) = \frac {b} {a} =\frac{\sin x}{\cos x} $</p> </div> <div style='float: right; width:40%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/060cc0b2-cbe8-427f-c930-9aa0abdd8800/public"> <img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/b233a6df-b0af-4c6b-4fa8-6fe3cc9a8000/public"></p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/d93fe119-e7dc-4670-32af-14f3f4ef7f00/public"></p> </div>

### <Success>Tangent formula</Success> <div style='float: left; text-align: left; width:100%;font-size:30px;'> <p>$\tan (-x) = \frac {\sin (-x)} {\cos (-x)} =\frac {-\sin x} {\cos x} $</p> <p>$\tan (-x) = -\ \tan x $</p> <p>$\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$</p> <p>Similarly, $\tan (x- {\pi}) = \tan x $</p> </div>

#### <Success>$Prove \tan({\pi}/{2} -x)=\cot x$</Success> <div style='float: left; text-align: left; width:100%; font-size:30px;'> <p>$\tan \biggl(\frac {\pi} {2} -x\biggl) = \frac {\sin (\frac {\pi} {2} -x )} {\cos (\frac {\pi} {2} -x)} = \frac {\cos x} {\sin x}$</p> <p>$\tan \biggl(\frac {\pi} {2} -x\biggl) = \frac {1} {\tan x} \\= \cot x \ (\text{co- tangent}) $</p> </div>

### <Success>Thank you</Success> <div> </div>