Bài tập Áp dụng công thức biến đổi tổng thành tích lớp 11 (có lời giải)

Hướng dẫn giải:

Đáp án đúng là: B

Xét đáp án A

Có sin10° + sin11° + sin15° + sin16° = (sin10° + sin16°) + (sin11° + sin15°)

= 2sin13°cos3° + 2sin13°cos2°

= 2sin13°(cos3° + cos2°)

= 2sin13°×2cos2°30'cos0°30'

= 4sin13°cos2°30'cos0°30'.

Vậy đáp án A đúng

Xét đáp án B

sina + sin2a + sin3a + sin4a = (sina + sin4a) + (sin2a + sin3a)

= $2\sin \frac{5a}{2}cos\frac{3a}{2}+2\sin \frac{5a}{2}cos\frac{a}{2}$

$=2\sin \frac{5a}{2}(cos\frac{3a}{2}+cos\frac{a}{2})$

$=2\sin \frac{5a}{2}\cdot 2\cos acos\frac{a}{2}$

$=4\sin \frac{5a}{2}\cos acos\frac{a}{2}$.

Vậy đáp án B sai

Xét đáp án C

cosa + cos2a + cos3a + cos4a = (cosa + cos4a) + (cos2a + cos3a)

= $2\cos \frac{5a}{2}cos\frac{3a}{2}+2\cos \frac{5a}{2}\cos \frac{a}{2}$

$=2\cos \frac{5a}{2}(cos\frac{3a}{2}+\cos \frac{a}{2})$

$=2\cos \frac{5a}{2}(2\cos a\cos \frac{a}{2})$

$=4\cos \frac{5a}{2}\cos a\cos \frac{a}{2}$.

Vậy đáp án C đúng

Xét đáp án D

1 + sina + cosa + tana = $1+\sin a+\cos a+\frac{\sin a}{\cos a}$

$=\frac{c\text{osa}+\sin a\cos a+{\cos }^{2}a+\sin a}{\cos a}$

$=\frac{c\text{osa}+\sin a+\cos a(\sin a+\cos a)}{\cos a}$

$=\frac{(1+\cos a)(\sin a+\cos a)}{\cos a}$

$=\frac{\sqrt{2}(1+\cos a)(\frac{\sqrt{2}}{2}\sin a+\frac{\sqrt{2}}{2}\cos a)}{\cos a}$

$=\frac{2\sqrt{2}{\cos }^{2}\frac{a}{2}(\sin acos\frac{π}{4}+\cos a\sin \frac{π}{4})}{\cos a}$

$=\frac{2\sqrt{2}{\cos }^{2}\frac{a}{2}\sin (a+\frac{π}{4})}{\cos a}$.

Vậy đáp án D đúng.

Hướng dẫn giải:

Đáp án đúng là: A

sin70° − sin20° + sin50° = (sin70° − sin20°) + sin50°

= 2cos45°sin25° + 2sin25°cos25°

= 2sin25°(cos45° + cos25°)

= 2sin25°(2cos35°cos10°)

= 4sin25°cos35°cos10°

= 4cos65°cos35°cos10°.

Hướng dẫn giải:

Đáp án đúng là: C

cosa – cos5a = $-2\sin \frac{a+5a}{2}\sin \frac{a-5a}{2}=-2\sin 3a\sin (-2a)=2\sin 3a\sin 2a$.

Hướng dẫn giải:

Đáp án đúng là: D

$\frac{\sin 7α-\sin 5α}{\sin 7α+\sin 5α}=\frac{2\cos \frac{7α+5α}{2}\sin \frac{7α-5α}{2}}{2\sin \frac{7α+5α}{2}\text{cos}\frac{7α-5α}{2}}=\frac{2\cos 6α\sin α}{2\sin 6α\text{cos}α}$= cot6$α$tan$α$.

Hướng dẫn giải:

Đáp án đúng là: C

$I=\frac{cos5x+cos3x}{\sin 5x-\sin 3x}=\frac{2cos\frac{5x+3x}{2}cos\frac{5x-3x}{2}}{2cos\frac{5x+3x}{2}\sin \frac{5x-3x}{2}}=\frac{2cos4x\cos x}{2cos4x\sin x}=\cot x=\frac{1}{\tan x}=3$.

Hướng dẫn giải:

Đáp án đúng là: B

$cos\left( {x + \frac{\pi }{4}} \right) - cos\left( {x - \frac{\pi }{4}} \right)$$= - 2\sin \frac{{\left( {x + \frac{\pi }{4}} \right) + \left( {x - \frac{\pi }{4}} \right)}}{2}\sin \frac{{\left( {x + \frac{\pi }{4}} \right) - \left( {x - \frac{\pi }{4}} \right)}}{2}$

$= - 2\sin x\sin \frac{\pi }{4}$$= - \sqrt 2 \sin x$.

Hướng dẫn giải:

Đáp án đúng là: C

P = cos(120° + x) + cos(120° − x) – cosx

= $2cos\frac{(120^{\circ}+x)+(120^{\circ}-x)}{2}cos\frac{(120^{\circ}+x)-(120^{\circ}-x)}{2}-\cos x$

$=2cos120^{\circ}\cos x-\cos x$

$=2\cdot (\frac{-1}{2})\cos x-\cos x$

$=-2\cos x$

Hướng dẫn giải:

Đáp án đúng là: D

$\frac{\sin 10^{\circ}+\sin 20^{\circ}}{cos10^{\circ}+cos20^{\circ}}=\frac{2\sin \frac{10^{\circ}+20^{\circ}}{2}cos\frac{10^{\circ}-20^{\circ}}{2}}{2cos\frac{10^{\circ}+20^{\circ}}{2}cos\frac{10^{\circ}-20^{\circ}}{2}}=\frac{2\sin 15^{\circ}cos5^{\circ}}{2cos15^{\circ}cos5^{\circ}}$ = tan15°.

Đáp án đúng là: D

Có cos18° = cos78° + cosa°

$\Leftrightarrow$ cosa° = cos18° − cos78°

$\Leftrightarrow cos\alpha ^\circ = - 2\sin \frac{{18^\circ + 78^\circ }}{2}\sin \frac{{18^\circ - 78^\circ }}{2}$

$\Leftrightarrow$ cosa° = −2sin48°sin(−30°)

$\Leftrightarrow$ cosa° = sin48°

$\Leftrightarrow$ cosa° = cos42°.

Vậy số dương nhỏ nhất của a là 42.

Hướng dẫn giải:

Đáp án đúng là: C

$\frac{1}{\sin 18^{\circ}}-\frac{1}{\sin 54^{\circ}}=\frac{\sin 54^{\circ}-\sin 18^{\circ}}{\sin 18^{\circ}\sin 54^{\circ}}=\frac{2cos\frac{54^{\circ}+18^{\circ}}{2}\sin \frac{54^{\circ}-18^{\circ}}{2}}{\sin 18^{\circ}\sin 54^{\circ}}$

$=\frac{2cos36^{\circ}\sin 18^{\circ}}{\sin 18^{\circ}\sin 54^{\circ}}=\frac{2\sin 54^{\circ}}{\sin 54^{\circ}}=2$

Đáp án đúng là: D

$\cos a + \cos b = 2\cos \frac{{a + b}}{2} \cdot \cos \frac{{a - b}}{2}$

Đáp án đúng là: A

A = cos3a – cosa = $- 2\sin \frac{{3a + a}}{2} \cdot \sin \frac{{3a - a}}{2} = 2\sin 2a \cdot \sin a$.

Đáp án đúng là: D

$\sin a - \sin b = 2\cos \frac{{a + b}}{2} \cdot \sin \frac{{a - b}}{2}$.

Đáp án đúng là: B

$\sin \alpha - 1 = \sin \alpha - \sin \frac{\pi }{2} = 2\cos \left( {\frac{{\alpha + \frac{\pi }{2}}}{2}} \right) \cdot \sin \left( {\frac{{\alpha - \frac{\pi }{2}}}{2}} \right) = 2\cos \left( {\frac{\alpha }{2} + \frac{\pi }{4}} \right) \cdot \sin \left( {\frac{\alpha }{2} - \frac{\pi }{4}} \right)$

Đáp án đúng là: D

$P = \frac{{\cos a - \cos 5a}}{{\sin 4a + \sin 2a}} = \frac{{ - 2\sin 3a \cdot \sin \left( { - 2a} \right)}}{{2\sin 3a \cdot \cos a}} = \frac{{\sin 2a}}{{\cos a}} = \frac{{2\sin a \cdot \cos a}}{{\cos a}} = 2\sin a$.

Đáp án: a) Sai. b) Sai. c) Đúng. d) Đúng.

a) Sai. $\cos 3x + \cos x = 2\cos \frac{{3x + x}}{2} \cdot \cos \frac{{3x - x}}{2} = 2\cos 2x \cdot \cos x$.

b) Sai. $\sin 3x + \sin 2x = 2\sin \frac{{3x + 2x}}{2} \cdot \cos \frac{{3x - 2x}}{2} = 2\sin \frac{{5x}}{2} \cdot \cos \frac{x}{2}$.

c) Đúng. $\cos 4x - \cos x = - 2\sin \frac{{4x + x}}{2} \cdot \sin \frac{{4x - x}}{2} = - 2\sin \frac{{5x}}{2} \cdot \sin \frac{{3x}}{2}$.

d) Đúng. $\sin 5x - {\mathop{\rm s}\nolimits} {\rm{inx}} = 2\cos \frac{{5x + x}}{2} \cdot \sin \frac{{5x - x}}{2} = 2\cos 3x \cdot \sin 2x$.

Đáp án: a) Đúng. b) Sai. c) Sai. d) Đúng.

a) Đúng. sin2x = 2sinx∙cosx.

b) Sai. $\cos a - \cos b = - 2\sin \frac{{a + b}}{2} \cdot \sin \frac{{a - b}}{2}$.

c) Sai. sina + sin2a + sin3a + sin4a + sin5a + sin6a

= (sina + sin6a) + (sin2a + sin5a) + (sin3a + sin4a)

$= 2\sin \frac{{7a}}{2} \cdot \cos \frac{{5a}}{2} + 2\sin \frac{{7a}}{2} \cdot \cos \frac{{3a}}{2} + 2\sin \frac{{7a}}{2} \cdot \cos \frac{a}{2}$

$= 2\left( {\cos \frac{{5a}}{2} + \cos \frac{{3a}}{2} + \cos \frac{a}{2}} \right) \cdot \sin \frac{{7a}}{2} = 2\left( {2\cos \frac{{3a}}{2} \cdot \cos a + \cos \frac{{3a}}{2}} \right) \cdot \sin \frac{{7a}}{2}$

$= 2\left( {2\cos a + 1} \right) \cdot \cos \frac{{3a}}{2} \cdot \sin \frac{{7a}}{2} = 4\left( {\cos a + \frac{1}{2}} \right) \cdot \cos \frac{{3a}}{2} \cdot \sin \frac{{7a}}{2}$

$= 4\left( {\cos a + \cos \frac{\pi }{3}} \right) \cdot \cos \frac{{3a}}{2} \cdot \sin \frac{{7a}}{2}$

$= 8\cos \left( {\frac{a}{2} + \frac{\pi }{6}} \right) \cdot \cos \left( {\frac{a}{2} - \frac{\pi }{6}} \right) \cdot \cos \frac{{3a}}{2} \cdot \sin \frac{{7a}}{2}$.

d) Đúng.

Vế trái:

$\frac{{\sin A + \sin B}}{{\cos A + \cos B}} = \frac{{2\sin \frac{{A + B}}{2} \cdot \cos \frac{{A - B}}{2}}}{{2\cos \frac{{A + B}}{2} \cdot \cos \frac{{A - B}}{2}}} = \frac{{2\sin \frac{{A + B}}{2}}}{{2\cos \frac{{A + B}}{2}}} = \tan \frac{{A + B}}{2} = \frac{{\sin \frac{{A + B}}{2}}}{{\cos \frac{{A + B}}{2}}}$.

Vế phải:

$\frac{1}{2}\left( {\tan A + \tan B} \right) = \frac{1}{2} \cdot \left( {\frac{{\sin A}}{{\cos A}} + \frac{{\sin B}}{{\cos B}}} \right) = \frac{1}{2}\left( {\frac{{\sin A\cos B + \cos A\sin B}}{{\cos A\cos B}}} \right)$

$= \frac{1}{2}\frac{{\sin \left( {A + B} \right)}}{{\cos A\cos B}} = \frac{{\sin \frac{{A + B}}{2}\cos \frac{{A + B}}{2}}}{{\cos A\cos B}}$.

Khi đó $\frac{{\sin \frac{{A + B}}{2}}}{{\cos \frac{{A + B}}{2}}} = \frac{{\sin \frac{{A + B}}{2}\cos \frac{{A + B}}{2}}}{{\cos A\cos B}}$$\Leftrightarrow \cos A\cos B = {\cos ^2}\frac{{A + B}}{2}$

$\Leftrightarrow \frac{1}{2}\left[ {\cos \left( {A - B} \right) + \cos \left( {A + B} \right)} \right] = \frac{{1 + \cos \left( {A + B} \right)}}{2}$$\Leftrightarrow \cos \left( {A - B} \right) = 1 \Leftrightarrow A - B = 0 \Leftrightarrow A = B$ hay $\widehat A = \widehat B$ ⇔ tam giác ABC cân tại C.

Đáp án: a) Sai. b) Đúng. c) Sai. d) Đúng.

a) Sai. $\frac{{\sin 2x + \sin 4x + \sin 6x}}{{2\left( {1 - \cos x} \right)}}$

$= \frac{{\left( {\sin 2x + \sin 6x} \right) + \sin 4x}}{{2\left( {1 - \cos x} \right)}} = \frac{{2\sin \frac{{2x + 6x}}{2} \cdot \cos \frac{{2x - 6x}}{2} + \sin 4x}}{{2\left( {1 - \cos x} \right)}}$

$= \frac{{2\sin 4x \cdot \cos 2x + \sin 4x}}{{2\left( {1 - \cos x} \right)}} = \frac{{\sin 4x\left( {2\cos 2x + 1} \right)}}{{2\left( {1 - \cos x} \right)}} = \frac{{\sin 4x\left( {2\cos 2x + 1} \right)}}{{4{{\sin }^2}\frac{x}{2}}}$.

b) Đúng. $\frac{{1 + \cos x + \cos 2x + \cos 3x}}{{2{{\cos }^2}x + \cos x - 1}}$

$= \frac{{\left( {1 + \cos 2x} \right) + \left( {\cos x + \cos 3x} \right)}}{{\left( {2{{\cos }^2}x - 1} \right) + \cos x}} = \frac{{2{{\cos }^2}x + 2\cos 2x \cdot \cos x}}{{\cos x + \cos 2x}}$

$= \frac{{2\cos x \cdot \left( {\cos x + \cos 2x} \right)}}{{\cos x + \cos 2x}} = 2\cos x$.

c) Sai. $\frac{{\cos x - \cos 2x + \cos 3x}}{{{\mathop{\rm s}\nolimits} {\rm{inx}} - \sin 2x + \sin 3x}}$

$= \frac{{\left( {\cos x + \cos 3x} \right) - \cos 2x}}{{\left( {{\mathop{\rm s}\nolimits} {\rm{inx}} + \sin 3x} \right) - \sin 2x}} = \frac{{2\cos 2x\cos x - \cos 2x}}{{2\sin 2x\cos x - \sin 2x}} = \frac{{\cos 2x\left( {2\cos x - 1} \right)}}{{\sin 2x\left( {2\sin x - 1} \right)}} = \cot 2x$

d) Đúng. $D = \sin a + \sin b + \sin c - \sin \left( {a + b + c} \right)$

$= \left( {\sin a + \sin b} \right) - \left[ {\sin \left( {a + b + c} \right) - \sin c} \right]$

$= 2\sin \frac{{a + b}}{2} \cdot \cos \frac{{a - b}}{2} - \left[ {2\cos \frac{{\left( {c + a + b + c} \right)}}{2} \cdot \sin \frac{{\left( {a + b + c - c} \right)}}{2}} \right]$

$= 2\sin \frac{{a + b}}{2} \cdot \cos \frac{{a - b}}{2} - 2\cos \frac{{a + b + 2c}}{2} \cdot \sin \frac{{a + b}}{2}$

$= 2\sin \frac{{a + b}}{2} \cdot \left( {\cos \frac{{a - b}}{2} - \cos \frac{{a + b + 2c}}{2}} \right)$

$= 2\sin \frac{{a + b}}{2}\left[ { - 2 \cdot \sin \frac{{\frac{{a - b + a + b + 2c}}{2}}}{2} \cdot \sin \frac{{\frac{{a - b - 2c - a - b}}{2}}}{2}} \right]$

$= 2\sin \frac{{a + b}}{2} \cdot \left( { - 2} \right) \cdot \sin \frac{{a + c}}{2} \cdot \sin \frac{{ - \left( {b + c} \right)}}{2}$

$= 4\sin \frac{{a + b}}{2} \cdot \sin \frac{{b + c}}{2} \cdot \sin \frac{{c + a}}{2}$.

Đáp án: a) Sai. b) Sai. c) Sai. d) Sai.

a) Sai. sina − sin2a + sin3a = (sina + sin3a) – sin2a

$= 2\sin \frac{{a + 3a}}{2} \cdot \cos \frac{{a - 3a}}{2} - \sin 2a = 2\sin 2a\cos a - \sin 2a$

$= \sin 2a \cdot \left( {2\cos a - 1} \right)$.

b) Sai. sinx + 2sin3x + sin5x = (sinx + sin5x) + 2sin3x

$= 2\sin \frac{{x + 5x}}{2} \cdot \cos \frac{{x - 5x}}{2} + 2\sin 3x = 2\sin 3x \cdot \cos 2x + 2\sin 3x = 2\sin 3x \cdot \left( {\cos 2x + 1} \right)$

$= 2\sin 3x \cdot 2{\cos ^2}x = 4\sin 3x \cdot {\cos ^2}x$.

c) Sai. 2cosx – cos3x – cos5x = 2cosx – (cos3x + cos5x)

$= 2\cos x - 2\cos \frac{{3x + 5x}}{2} \cdot \cos \frac{{3x - 5x}}{2} = 2\cos x - 2\cos 4x \cdot \cos x = 2\cos x\left( {1 - \cos 4x} \right)$

$= 2\cos x \cdot 2{\sin ^2}2x = 4\cos x \cdot {\sin ^2}2x$.

d) Sai. $\frac{{\cos 5x - \cos x}}{{\sin 4x + \sin 2x}} = \frac{{ - 2\sin \frac{{5x + x}}{2} \cdot \sin \frac{{5x - x}}{2}}}{{2\sin \frac{{4x + 2x}}{2} \cdot \cos \frac{{4x - 2x}}{2}}}$

$= \frac{{ - 2\sin 3x \cdot \sin 2x}}{{2\sin 3x \cdot \cos x}} = - \frac{{2\sin x\cos x}}{{\cos x}} = - 2\sin x$.

Đáp án: a) Đúng. b) Đúng. c) Đúng. d) Sai.

a) Đúng. cosa + cos2a + cos3a + cos4a = (cosa + cos3a) + (cos2a + cos4a)

= 2cos2acosa + 2cos3acosa = 2(cos2a + cos3a)cosa = $4\cos \frac{{5a}}{2} \cdot \cos \frac{a}{2} \cdot \cos a$.

b) Đúng. 1 + sina + cos3a – (cosa + sin2a + cos2a)

= (1 − cos2a) + (cos3a – cosa) + (sina – sin2a)

= 2sin$^{2}$a – 2sin2asina + (sina – sin2a)

= 2sina(sina – sin2a) + (sina – sin2a) = (sina – sin2a)(2sina +1)

= $2\cos \frac{{3a}}{2}\sin \left( {\frac{{ - a}}{2}} \right) \cdot \left( {2\sin a + 1} \right)$.

c) Đúng. cosa + cosb + cosc + cos(a + b + c)

= (cosa + cosb) + [cosc + cos(a + b +c )]

$= 2\cos \frac{{a + b}}{2} \cdot \cos \frac{{a - b}}{2} + 2\cos \frac{{c + a + b + c}}{2} \cdot \cos \frac{{c + a + b - c}}{2}$

$= 2\cos \frac{{a + b}}{2} \cdot \cos \frac{{a - b}}{2} + 2\cos \frac{{a + b + 2c}}{2} \cdot \cos \frac{{a + b}}{2}$

$= 2\cos \frac{{a + b}}{2} \cdot \left( {\cos \frac{{a - b}}{2} + \cos \frac{{a + b + 2c}}{2}} \right) = 2\cos \frac{{a + b}}{2} \cdot 2\cos \frac{{b + c}}{2} \cdot \cos \frac{{c + a}}{2}$

$= 4\cos \frac{{a + b}}{2} \cdot \cos \frac{{b + c}}{2} \cdot \cos \frac{{c + a}}{2}$.

d) Sai. sin x + sin(x + a) + sin(x + 2a)

= [sin(x + 2a) + sinx] + sin(x + a) = $2\sin \frac{{x + 2a + x}}{2} \cdot \cos \frac{{x + 2a - x}}{2} + \sin (x + a)$

$= 2\sin \frac{{2x + 2a}}{2} \cdot \cos \frac{{x + 2a - x}}{2} + \sin (x + a) = 2\sin (x + a)\cos a + \sin (x + a)$

= (2cosa + 1)sin(x + a).

Đáp án đúng là: 3

$P = \frac{{\cos \alpha + 2\cos 3\alpha + \cos 5\alpha }}{{\sin \alpha + 2\sin 3\alpha + \sin 5\alpha }} = \frac{{\left( {\cos \alpha + \cos 5\alpha } \right) + 2\cos 3\alpha }}{{\left( {\sin \alpha + \sin 5\alpha } \right) + 2\sin 3\alpha }} = \frac{{2\cos 3\alpha \cdot \cos 2\alpha + 2\cos 3\alpha }}{{2\sin 3\alpha \cdot \cos 2\alpha + 2\sin 3\alpha }}$

$= \frac{{2\cos 3\alpha \cdot \left( {\cos 2\alpha + 1} \right)}}{{2\sin 3\alpha \cdot \left( {\cos 2\alpha + 1} \right)}} = \cot 3\alpha$.

Vậy a = 3.

Đáp án: 2

$P = \frac{{\sin \alpha + \sin 2\alpha + \sin 3\alpha }}{{\cos \alpha + \cos 2\alpha + \cos 3\alpha }} = \frac{{\left( {\sin \alpha + \sin 3\alpha } \right) + \sin 2\alpha }}{{\left( {\cos 2\alpha + \cos 3\alpha } \right) + \cos 2\alpha }} = \frac{{2\sin 2\alpha \cdot \cos \alpha + \sin 2\alpha }}{{2\cos 2\alpha \cdot \cos \alpha + \cos 2\alpha }}$

$= \frac{{\sin 2\alpha \cdot \left( {2\cos \alpha + 1} \right)}}{{\cos 2\alpha \cdot \left( {2\cos \alpha + 1} \right)}} = \tan 2\alpha$.

Vậy k = 2.

Đáp án: 6

B = sina + sin2a + sin3a + sin4a = (sina + sin3a) + (sin2a + sin4a)

$= 2\sin \frac{{a + 3a}}{2} \cdot \cos \frac{{a - 3a}}{2} + 2\sin \frac{{2a + 4a}}{2} \cdot \cos \frac{{2a - 4a}}{2} = 2\sin 2a\cos a + 2\sin 3a\cos a$

$= 2\cos a\left( {\sin 2a + \sin 3a} \right) = 2\cos a \cdot 2\sin \frac{{5a}}{2} \cdot \cos \frac{a}{2} = 4\cos a \cdot \sin \frac{{5a}}{2} \cdot \cos \frac{a}{2}$.

Vậy m + n = 6.

Đáp án: 6

$C = {\mathop{\rm s}\nolimits} {\rm{inx}} + \cos x - \sin \left( {x - \frac{\pi }{6}} \right) + \cos \left( {x - \frac{\pi }{6}} \right)$

$= \sqrt 2 \cos \left( {x - \frac{\pi }{4}} \right) + \cos \left( {x + \frac{\pi }{3}} \right) + \cos \left( {x - \frac{\pi }{6}} \right)$

$= \sqrt 2 \cos \left( {x - \frac{\pi }{4}} \right) + 2\cos \left( {x + \frac{\pi }{{12}}} \right) \cdot \cos \frac{\pi }{4}$

$= \sqrt 2 \left[ {\cos \left( {x - \frac{\pi }{4}} \right) + \cos \left( {x + \frac{\pi }{{12}}} \right)} \right] = 2\sqrt 2 \cos \left( {x - \frac{\pi }{{12}}} \right) \cdot \cos \frac{\pi }{6}$

$= \sqrt 6 \cos \left( {x - \frac{\pi }{{12}}} \right)$.

Vậy a = 6.

Đáp án: 2

$M = \cos \left( {x + \frac{\pi }{4}} \right) - \cos \left( {x - \frac{\pi }{4}} \right) = - 2\sin \left( {\frac{{x + \frac{\pi }{4} + x - \frac{\pi }{4}}}{2}} \right) \cdot \sin \left( {\frac{{x + \frac{\pi }{4} - x + \frac{\pi }{4}}}{2}} \right)$

$= - 2\sin x \cdot \sin \frac{\pi }{4} = - \sqrt 2 {\mathop{\rm s}\nolimits} {\rm{inx}}$.

Vậy a = 2.