Bài tập về tính giá trị lượng giác

Hôm nay các em học sinh hãy ứng dụng các công thức đã cho để làm bài tập về tính giá trị lượng giác của các góc.

Bài 1: $\displaystyle \sin \alpha =-\frac{3}{5}\left( {\pi <\alpha <\frac{{3\pi }}{2}} \right).T\text{ }\!\!\acute{\mathrm{i}}\!\!\text{ nh cos}\alpha \text{,tan}\alpha \text{,cot}\alpha \text{.}$

Bài 2: Cho 5cosa + 4 = 0 Với $\displaystyle \left( {{{{180}}^{o}}<a<{{{270}}^{o}}} \right)$
 

Tính sina , tana, cota.

Bài 3: Cho $\displaystyle \tan {{15}^{o}}=2-\sqrt{3}.\,\,\,\,\,T\text{ }\!\!\acute{\mathrm{i}}\!\!\text{ nh}\,\,\text{sin1}{{\text{5}}^{\text{o}}},\cos {{15}^{o}},\cot {{15}^{o}}.$

Bài 4: Tính $\displaystyle A=\frac{{\tan x+\cot x}}{{\tan x-\cot x}}$ biết $\displaystyle \text{sinx = }\frac{\text{1}}{\text{3}}$

Tính $\displaystyle B=\frac{{2\sin x+3\cos x}}{{3\sin x-2\cos x}}$ biết tanx = -2

Tính $\displaystyle C=\frac{{{{{\sin }}^{2}}x+3\sin x\cos x-2{{{\cos }}^{2}}x}}{{1+4{{{\sin }}^{2}}x}}$ biết cotx = -3

Bài 5: Chứng minh:  

a, $\displaystyle \text{si}{{\text{n}}^{\text{4}}}\text{x+co}{{\text{s}}^{\text{4}}}\text{x=1-2si}{{\text{n}}^{\text{2}}}\text{xco}{{\text{s}}^{\text{2}}}\text{x}$
 

b, $\displaystyle \text{si}{{\text{n}}^{\text{6}}}\text{x+co}{{\text{s}}^{\text{6}}}\text{x=1-3si}{{\text{n}}^{\text{2}}}\text{xco}{{\text{s}}^{\text{2}}}\text{x}$

c, $\displaystyle \text{ta}{{\text{n}}^{\text{2}}}\text{x = si}{{\text{n}}^{\text{2}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{x}\text{.ta}{{\text{n}}^{\text{2}}}\text{x}$

d, $\displaystyle \text{si}{{\text{n}}^{\text{2}}}\text{x}\text{.tanx + co}{{\text{s}}^{\text{2}}}\text{x}\text{.cotx + 2sinx}\text{.cosx = tanx + cotx}$

Bài 6: Chứng minh các đẳng thức dưới đây:  

a, $\displaystyle \frac{{\text{1-2co}{{\text{s}}^{\text{2}}}\text{x}}}{{\text{si}{{\text{n}}^{\text{2}}}\text{x}\text{.co}{{\text{s}}^{\text{2}}}\text{x}}}\text{ = ta}{{\text{n}}^{\text{2}}}\text{x-co}{{\text{t}}^{\text{2}}}\text{x}$

b, $\displaystyle \frac{{\text{1+si}{{\text{n}}^{\text{2}}}\text{x}}}{{\text{1-si}{{\text{n}}^{\text{2}}}\text{x}}}\text{ = 1+2ta}{{\text{n}}^{\text{2}}}\text{x}$

c, $\displaystyle \frac{{\text{cosx}}}{{\text{1+sinx}}}\text{+tanx = }\frac{\text{1}}{{\text{cosx}}}$

d, $\displaystyle \frac{{\text{sinx}}}{{\text{1+cosx}}}\text{+}\frac{{\text{1+cosx}}}{{\text{sinx}}}\text{ = }\frac{\text{2}}{{\text{sinx}}}$

e, $\displaystyle \frac{{\text{1-sinx}}}{{\text{cosx}}}\text{ = }\frac{{\text{cosx}}}{{\text{1+sinx}}}$

f, $\displaystyle \frac{{\text{sinx+cosx-1}}}{{\text{sinx-cosx+1}}}\text{ = }\frac{{\text{cosx}}}{{\text{1+sinx}}}$

Bài 7: Chứng minh các biểu thức sau đây độc lập đối với x:

$\displaystyle \begin{array}{l}\text{A=2}\left( {\text{si}{{\text{n}}^{\text{6}}}\text{x+co}{{\text{s}}^{\text{6}}}\text{x}} \right)\text{-3}\left( {\text{si}{{\text{n}}^{\text{4}}}\text{x+co}{{\text{s}}^{\text{4}}}\text{x}} \right)\text{;   B=co}{{\text{s}}^{\text{4}}}\text{x}\left( {\text{2co}{{\text{s}}^{\text{2}}}\text{x-3}} \right)\text{+si}{{\text{n}}^{\text{4}}}\text{x}\left( {\text{2si}{{\text{n}}^{\text{2}}}\text{x-3}} \right)\\\text{C=2}{{\left( {\text{si}{{\text{n}}^{\text{4}}}\text{x+co}{{\text{s}}^{\text{4}}}\text{x+si}{{\text{n}}^{\text{2}}}\text{xco}{{\text{s}}^{\text{2}}}\text{x}} \right)}^{\text{2}}}\text{-}\left( {\text{si}{{\text{n}}^{\text{8}}}\text{x+co}{{\text{s}}^{\text{8}}}\text{x}} \right)\text{;   D=3}\left( {\text{si}{{\text{n}}^{\text{8}}}\text{x-co}{{\text{s}}^{\text{8}}}\text{x}} \right)\text{+4}\left( {\text{co}{{\text{s}}^{\text{6}}}\text{x-2si}{{\text{n}}^{\text{6}}}\text{x}} \right)\text{+6si}{{\text{n}}^{\text{4}}}\text{x}\\\text{E=}\sqrt{{\text{si}{{\text{n}}^{\text{4}}}\text{x+4co}{{\text{s}}^{\text{2}}}\text{x}}}\text{+}\sqrt{{\text{co}{{\text{s}}^{\text{4}}}\text{x+4si}{{\text{n}}^{\text{2}}}\text{x}}}\text{;   F=}\frac{{\text{si}{{\text{n}}^{\text{6}}}\text{x+co}{{\text{s}}^{\text{6}}}\text{x-1}}}{{\text{si}{{\text{n}}^{\text{4}}}\text{x+co}{{\text{s}}^{\text{4}}}\text{x-1}}}\text{;    G=}\frac{{\text{si}{{\text{n}}^{\text{4}}}\text{x+3co}{{\text{s}}^{\text{4}}}\text{x-1}}}{{\text{si}{{\text{n}}^{\text{6}}}\text{x+co}{{\text{s}}^{\text{6}}}\text{x+3co}{{\text{s}}^{\text{4}}}\text{x-1}}}\\\text{H=cosx}\sqrt{{\text{1-sinx}\sqrt{{\text{1-cosx}\sqrt{{\text{1-si}{{\text{n}}^{\text{2}}}\text{x}}}}}}}\text{+sinx}\sqrt{{\text{1-cosx}\sqrt{{\text{1-sinx}\sqrt{{\text{1-co}{{\text{s}}^{\text{2}}}\text{x}}}}}}};(x\in \left[ {0;\frac{\pi }{2}} \right])\end{array}$