Bài tập Áp dụng công thức biến đổi tích thành tổng lớp 11 (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}}$.