[toán 10] BT

H

hatcat_sad

B

bosjeunhan

Câu 1: tính
a) A=cos 330*+ sin 315* +cos 250* + sin 160*
b) B= sin^2 1*+sin^2 2*+sin^2 3*+......+ sin^2 89*+sin^2 90*
( chú ý: *nghĩa là độ)

câu 2: cho tam giác ABC. chứng minh:
a) cos C= -cos(A+B+2C)
b) cos A= -cos(B+C)

Haizz, chém bừa chém bừa....

1) b
Ta có: sin^2 1* = cos^2 89*,....
Vậy nên B = (sin^89+cos^2 89*)+....+(sin^2 46* + cos^2 44*) +sin^2 45 * +sin^2 90 *
Mà sin^2 a + cos^2 a = 1
Đến đây chắc ok
 
J

jelouis

$A=cos 330^{\circ}+ sin 315^{\circ} +cos 250^{\circ} + sin 160^{\circ}$
$=cos(360^{\circ}-30^{\circ})+sin(360^{\circ}-45^{\circ})+cos(180^{\circ}-70^{\circ})+sin(180^{\circ}-20^{\circ})$
$=cos30^{\circ}-sin45^{\circ}-cos70^{\circ}+sin20^{\circ}=\frac{\sqrt{3}-\sqrt{2}}{2} - sin20^{\circ}+sin20^{\circ}=\frac{\sqrt{3}-\sqrt{2}}{2}$
$b/sin^289^{\circ}=cos^21^{\circ}$
$cos^21^{\circ}+sin^21^{\circ}=1$
$\Longrightarrow B=44+sin45^{\circ}+sin90^{\circ}=\frac{90+\sqrt{2}}{2}$
$c/ A+B+C=\pi$
$-cos(A+B+2C)=-cos(\pi+C)=cosC$
$d/-cos(B+C)=-cos(180-A)=cosA$
 
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