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

Hướng dẫn giải:

Đáp án đúng là: B

$\sin a\cdot \sin b=\frac{1}{2}[cos(a-b)-cos(a+b)]$.

Hướng dẫn giải:

Đáp án đúng là: A

$\cos 3x\cdot \cos 5x=\frac{1}{2}(cos8x+cos2x)$

$=\frac{1}{2}[cos(-2x)+cos8x]$ $=\frac{1}{2}(cos2x+cos8x)$.

Hướng dẫn giải:

Đáp án đúng là: B

$A = cos\frac{{2\pi }}{7} + cos\frac{{4\pi }}{7} + cos\frac{{6\pi }}{7}$$= \frac{{\sin \frac{\pi }{7}\left( {cos\frac{{2\pi }}{7} + cos\frac{{4\pi }}{7} + cos\frac{{6\pi }}{7}} \right)}}{{\sin \frac{\pi }{7}}}$

$$= \frac{{\sin \frac{\pi }{7}cos\frac{{2\pi }}{7} + \sin \frac{\pi }{7}cos\frac{{4\pi }}{7} + \sin \frac{\pi }{7}cos\frac{{6\pi }}{7}}}{{\sin \frac{\pi }{7}}}$$

Ta có $\sin \frac{\pi }{7}cos\frac{{2\pi }}{7}$$= \frac{1}{2}\left[ {sin\left( {\frac{\pi }{7} - \frac{{2\pi }}{7}} \right) + \sin \left( {\frac{\pi }{7} + \frac{{2\pi }}{7}} \right)} \right]$

$= \frac{1}{2}\left[ {sin\left( { - \frac{\pi }{7}} \right) + \sin \frac{{3\pi }}{7}} \right]$$= \frac{1}{2}\left( { - sin\frac{\pi }{7} + \sin \frac{{3\pi }}{7}} \right)$.

$\sin \frac{\pi }{7}cos\frac{{4\pi }}{7}$$= \frac{1}{2}\left[ {\sin \left( {\frac{\pi }{7} - \frac{{4\pi }}{7}} \right) + \sin \left( {\frac{\pi }{7} + \frac{{4\pi }}{7}} \right)} \right]$

$= \frac{1}{2}\left[ {\sin \left( { - \frac{{3\pi }}{7}} \right) + \sin \frac{{5\pi }}{7}} \right]$$= \frac{1}{2}\left( { - \sin \frac{{3\pi }}{7} + \sin \frac{{5\pi }}{7}} \right)$.

$\sin \frac{\pi }{7}cos\frac{{6\pi }}{7}$$= \frac{1}{2}\left[ {\sin \left( {\frac{\pi }{7} - \frac{{6\pi }}{7}} \right) + \sin \left( {\frac{\pi }{7} + \frac{{6\pi }}{7}} \right)} \right]$

$= \frac{1}{2}\left[ {\sin \left( { - \frac{{5\pi }}{7}} \right) + \sin \pi } \right]$$= \frac{1}{2}\left( { - \sin \frac{{5\pi }}{7} + \sin \pi } \right)$.

Do đó $\sin \frac{\pi }{7}cos\frac{{2\pi }}{7} + \sin \frac{\pi }{7}cos\frac{{4\pi }}{7} + \sin \frac{\pi }{7}cos\frac{{6\pi }}{7}$

$= \frac{1}{2}\left( { - sin\frac{\pi }{7} + \sin \frac{{3\pi }}{7}} \right)$$+ \frac{1}{2}\left( { - \sin \frac{{3\pi }}{7} + \sin \frac{{5\pi }}{7}} \right)$$+ \frac{1}{2}\left( { - \sin \frac{{5\pi }}{7} + \sin \pi } \right)$

$= \frac{1}{2}\left( { - sin\frac{\pi }{7} + \sin \pi } \right)$$= - \frac{1}{2}sin\frac{\pi }{7}$.

Do đó $A = \frac{{ - \frac{1}{2}sin\frac{\pi }{7}}}{{sin\frac{\pi }{7}}} = - \frac{1}{2}$.

Hướng dẫn giải:

Đáp án đúng là: A

$\tan \frac{\pi }{{24}} + \tan \frac{{7\pi }}{{24}}$$= \frac{{\sin \frac{\pi }{{24}}}}{{cos\frac{\pi }{{24}}}} + \frac{{\sin \frac{{7\pi }}{{24}}}}{{cos\frac{{7\pi }}{{24}}}}$$= \frac{{\sin \frac{\pi }{{24}}cos\frac{{7\pi }}{{24}} + cos\frac{\pi }{{24}}\sin \frac{{7\pi }}{{24}}}}{{cos\frac{\pi }{{24}}cos\frac{{7\pi }}{{24}}}}$

$= \frac{{\sin \left( {\frac{\pi }{{24}} + \frac{{7\pi }}{{24}}} \right)}}{{\frac{1}{2}\left[ {cos\left( {\frac{\pi }{{24}} - \frac{{7\pi }}{{24}}} \right) + cos\left( {\frac{\pi }{{24}} + \frac{{7\pi }}{{24}}} \right)} \right]}}$$= \frac{{\sin \frac{\pi }{3}}}{{\frac{1}{2}\left[ {cos\left( { - \frac{\pi }{4}} \right) + cos\frac{\pi }{3}} \right]}}$

$= \frac{{\frac{{\sqrt 3 }}{2}}}{{\frac{1}{2}\left[ {\frac{{\sqrt 2 }}{2} + \frac{1}{2}} \right]}}$$= \frac{{2\sqrt 3 }}{{\sqrt 2 + 1}}$$= 2\sqrt 3 \left( {\sqrt 2 - 1} \right)$$= 2\left( {\sqrt 6 - \sqrt 3 } \right)$.

Hướng dẫn giải:

Đáp án đúng là: D

$P = cos(a + b)cos(a - b)$$= \frac{1}{2}\left\{ {cos\left[ {(a + b) - (a - b)} \right] + cos\left[ {(a + b) + (a - b)} \right]} \right\}$

$= \frac{1}{2}\left[ {cos(2b) + cos(2a)} \right]$$= \frac{1}{2}\left[ {2{{\cos }^2}a - 1 + 2{{\cos }^2}b - 1} \right]$$= {\cos ^2}a + {\cos ^2}b - 1$

$= {\left( {\frac{1}{3}} \right)^2} + {\left( {\frac{1}{4}} \right)^2} - 1$$= - \frac{{119}}{{144}}$.

Hướng dẫn giải:

Đáp án đúng là: C

cos10°×cos30°×cos50°×cos70°

= $cos10^\circ \cdot cos30^\circ \cdot \frac{1}{2}\left[ {cos\left( {50^\circ - 70^\circ } \right) + cos\left( {50^\circ + 70^\circ } \right)} \right]$

$= cos10^\circ \cdot \frac{{\sqrt 3 }}{2} \cdot \frac{1}{2}\left( {cos20^\circ + cos120^\circ } \right)$

$= cos10^\circ \cdot \frac{{\sqrt 3 }}{2} \cdot \frac{1}{2}\left( {cos20^\circ - \frac{1}{2}} \right)$

$= \frac{{\sqrt 3 }}{4}cos10^\circ \left( {cos20^\circ - \frac{1}{2}} \right)$

$= \frac{{\sqrt 3 }}{4}cos10^\circ cos20^\circ - \frac{{\sqrt 3 }}{8}cos10^\circ$

$= \frac{{\sqrt 3 }}{4} \cdot \frac{1}{2}\left[ {cos\left( {10^\circ - 20^\circ } \right) + cos\left( {10^\circ + 20^\circ } \right)} \right] - \frac{{\sqrt 3 }}{8}cos10^\circ$$= \frac{{\sqrt 3 }}{4} \cdot \frac{1}{2}\left( {cos10^\circ + cos30^\circ } \right) - \frac{{\sqrt 3 }}{8}cos10^\circ$

$= \frac{3}{{16}}$.

Hướng dẫn giải:

Đáp án đúng là: A

$A = \frac{1}{{2\sin 10^\circ }} - 2\sin 70^\circ$$= \frac{{1 - 4\sin 70^\circ \sin 10^\circ }}{{2\sin 10^\circ }}$

$= \frac{{1 - 4 \cdot \frac{1}{2}\left[ {cos\left( {70^\circ - 10^\circ } \right) - cos\left( {70^\circ + 10^\circ } \right)} \right]}}{{2\sin 10^\circ }}$

$= \frac{{1 - 4 \cdot \frac{1}{2}\left( {cos60^\circ - cos80^\circ } \right)}}{{2\sin 10^\circ }}$

$= \frac{{1 - 4 \cdot \frac{1}{2}\left( {\frac{1}{2} - cos80^\circ } \right)}}{{2\sin 10^\circ }}$

$= \frac{{1 - \left( {1 - 2cos80^\circ } \right)}}{{2\sin 10^\circ }}$

$= \frac{{2cos80^\circ }}{{2\sin 10^\circ }}$

$= \frac{{2cos\left( {90^\circ - 10^\circ } \right)}}{{2\sin 10^\circ }}$

$= \frac{{2\sin 10^\circ }}{{2\sin 10^\circ }} = 1$.

Hướng dẫn giải:

Đáp án đúng là: D

$cos2x \cdot \sin 5x \cdot cos3x = cos2x \cdot \frac{1}{2} \cdot \left[ {\sin \left( {5x - 3x} \right) + \sin \left( {5x + 3x} \right)} \right]$

$= cos2x \cdot \frac{1}{2} \cdot \left( {\sin 2x + \sin 8x} \right)$

$= \frac{1}{2}cos2x\sin 2x + \frac{1}{2}cos2x\sin 8x$

$= \frac{1}{4}\sin 4x + \frac{1}{4}\left[ {\sin \left( {8x - 2x} \right) + \sin \left( {8x + 2x} \right)} \right]$

$= \frac{1}{4}\sin 4x + \frac{1}{4}\left( {\sin 6x + \sin 10x} \right)$

$= \frac{{\sin 10x + \sin 6x + \sin 4x}}{4}$.

Hướng dẫn giải:

Đáp án đúng là: B

$2\sin (\frac{π}{4}+α)\sin (\frac{π}{4}-α)$

$=2\cdot \frac{1}{2}\{cos[(\frac{π}{4}+α)-(\frac{π}{4}-α)]-cos[(\frac{π}{4}+α)+(\frac{π}{4}-α)]\}$

$=2\cdot \frac{1}{2}(cos2α-cos\frac{π}{2})$

$=cos2α$

Hướng dẫn giải:

Đáp án đúng là: C

$A=cos\frac{π}{12}\cdot cos\frac{5π}{12}=\frac{1}{2}[cos(\frac{π}{12}-\frac{5π}{12})+cos(\frac{π}{12}+\frac{5π}{12})]$

$=\frac{1}{2}[cos(-\frac{π}{3})+cos\frac{π}{2}]$

$=\frac{1}{2}(cos\frac{π}{3}+cos\frac{π}{2})$

$=\frac{1}{4}$