§1. Giá trị lượng giác của một góc bất kỳ từ 0 (độ) đến 180 (độ)

a: Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(\widehat{A}=180^0-\widehat{B}-\widehat{C}\)

=>\(sinA=sin\left(180-B-C\right)\)

=>\(sinA=sin\left(B+C\right)\)

b: Ta có: \(\widehat{A}=180^0-\widehat{B}-\widehat{C}\)

=>\(cosA=cos\left(180-\left(B+C\right)\right)\)

=>\(cosA=-cos\left(B+C\right)\)

cosx(1 + 1/cosx + tanx)(1 - 1/cosx + tanx)

= cosx[(1 + tanx)² - 1/cos²x]

= cosx(1 + tan²x + 2tanx - 1 - tan²x)

= 2tanxcosx

= 2sinxcosx/cosx

= 2sinx

Chọn D

\(=cosx\left(1+\dfrac{1}{cosx}+\dfrac{sinx}{cosx}\right)\left(1-\dfrac{1}{cosx}+\dfrac{sinx}{cosx}\right)\)

\(=cosx\left(\dfrac{sinx+1+cosx}{cosx}\right)\left(\dfrac{cosx+sinx-1}{cosx}\right)\)

\(=\left(cosx+sinx\right)^2\)\(-1\)

\(=cos^2x+2cosxsinx+sin^2x-1\)

\(=cos^2x+sin^2x+2cosxsinx-1\)

\(=1+2cosxsinx-1\)

\(=2cosxsinx\)

Áp dụng đẳng thức góc nhân đôi cho sin

\(=2sinx\)

Chứng minh các biểu thức khồn phụ thuộc vào x

\(E=\left(cos^6x+sin^6x\right)+sin^4x\left(sin^2x+cos^2x\right)+4sin^2x.cos^2x-sin^2x\)

\(=\left(cos^2x+sin^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+sin^4x+4sin^2x.cos^2x-sin^2x\)

\(=1+sin^2x.cos^2x+sin^4x-sin^2x\)

\(=cos^2x+sin^2x\left(sin^2x+cos^2x\right)\)

\(=cos^2x+sin^2x=1\)

\(=cos0^0+cos20^0+cos40^0+...+cos\left(180^0-40^0\right)+cos\left(180^0-20^0\right)+cos\left(180^0-0^0\right)\)

\(=cos0^0+cos20^0+cos40^0+...+\left(-cos40^0\right)+\left(-cos20^0\right)+\left(-cos0^0\right)\)

\(=0\)

\(=\left(\dfrac{\sqrt{2}}{2}\right)^2-2sin^2\left(90^0-40^0\right)+3\left(\dfrac{\sqrt{2}}{2}\right)^2-2sin^240^0+4tan55^0.tan\left(90^0-55^0\right)\)

\(=\dfrac{1}{2}+\dfrac{3}{2}-2cos^240^0-2sin^240^0+4tan55^0.cot55^0\)

\(=2-2\left(sin^240^0+cos^240^0\right)+4\)

\(=2-2+4=4\)

A=cot20+cot160+cot40+cot140+...+cot80+cot100

=0+0+0+0

=0

\(\text{A=cot20+cot160+cot40+cot140+...+cot80+cot100}\)

\(=0+0+0+0\)

\(=0\)

\(sinacosa=\dfrac{\left(sina+cosa\right)^2-1}{2}=\dfrac{1-1}{2}=0\)

\(sin^3a+cos^3a=\left(sina+cosa\right)^3-3cosa\cdot sina\left(sina+cosa\right)=1\)

\(sinacosa=\dfrac{\left(sina+cosa\right)^2-1}{2}=\dfrac{1-1}{2}=0\)

\(sin^3a+cos^3a=\left(sina+cosa\right)^3-3cosa.sina\)\(\left(sina+sosa\right)=1\)

`@1+tan^2 a=1/[cos^2 a]=>1+3^2=1/[cos^2 a]=>cos^2 a=1/10`

`@tan a=3=>[sin a]/[cos a]=3=>sin a=3cos a`

Thay vào `A` có: `A=[2.3cos a-cos a]/[(3cos a)^3+2cos^3 a]`

                          `A=[5cos a]/[29cos^3 a]=5/[29 cos^2 a]=5/[29. 1/10]=50/29`

tan a=3 nên sin a/cosa=3

=>sin a=3 cosa

\(A=\dfrac{2\cdot3cosa-cosa}{27\cdot cos^3a+2cos^3a}=\dfrac{5cosa}{29cos^3a}=\dfrac{5}{29cos^2a}\)