Giải bài tập 3.1 trang 32 SBT toán 10 tập 1 kết nối

Trả lời:

a) A = sin$45^{o}$ + 2sin$60^{o}$ + tan$120^{o}$ + cos$135^{o}$

A = $\frac{1}{\sqrt{2}}$ + 2.$\frac{\sqrt{3}}{2}$ + $-\sqrt{3}$ - $\frac{1}{\sqrt{2}}$

A = 0

b) B = tan$45^{o}$ . cot$135^{o}$ - sin$30^{o}$ . cos$120^{o}$ - sin$60^{o}$ . cos$150^{o}$

B = 1 . (-1) - $\frac{1}{2}$ . $\frac{1}{2}$ - $\frac{\sqrt{3}}{2}$ . (-$\frac{\sqrt{3}}{2}$)

B = 0

c) C = $cos^{2}$$5^{o}$ + $cos^{2}$$25^{o}$ + $cos^{2}$$45^{o}$ + $cos^{2}$$65^{o}$ + $cos^{2}$$85^{o}$

Do $5^{o}$ = $90^{o}$ - $85^{o}$, $25^{o}$ = $90^{o}$ - $65^{o}$

Như vậy, ta có: $cos5^{o}$ = $sin45^{o}$, $cos^{2}$$25^{o}$ = $sin65^{o}$

C = ($sin^ {2}$$85^{o}$ + $cos^{2}$$85^{o}$) + ($sin^ {2}$$65^{o}$ +  $cos^{2}$$65^{o}$) + $cos^{2}$$45^{o}$

C = 1 + 1 + $\frac{1}{2}$ 

C = $\frac{5}{2}$

d) D = $\frac{1}{1 + tan^{2}73^{o}}$ - 4tan$75^{o}$ . cot$105^{o}$ + $12sin^{2}$$107^{o}$ - 2tan$40^{o}$ . cos$60^{o}$ . tan$50^{o}$;

Do $73^{o}$ + $107^{o}$ = $75^{o}$ + $105^{o}$ = $100^{o}$

Như vậy, ta có:

(1) $\frac{1}{1 + tan^{2}73^{o}}$ + $12sin^{2}$$107^{o}$ = 12($cos^{2}$$73^{o}$ + $sin^{2}$$73^{o}$ = 12

(2) tan$75^{o}$ . cot$105^{o}$ = tan$75^{o}$ . (-cot$75^{o}$) = -1

Do $40^{o}$ + $50^{o}$ = $90^{o}$

Như vậy, ta có:

(3) tan$40^{o}$ . tan$50^{o}$ = tan$40^{o}$ . cot$40^{o}$

Từ (1), (2) và (3) suy ra D = 12 - 4 . (-1) - 2 . 1 . $\frac{1}{2}$ = 15

e) E = 4tan$32^{o}$ . cos$60^{o}$ . cot$148^{o}$ + $\frac{5cot^{2}108^{o}}{1 + tan^{2}18^{o}}$ + 5$sin^{2}$$72^{o}$

Do $148^{o}$ + $32^{o}$ = $108^{o}$ + $72^{o}$ = $180^{o}$

Và $72^{o}$ + $18^{o}$ = $90^{o}$

Nên E = 4tan$32^{o}$ . (-cot$32^{o}$) . cos$60^{o}$ + 5(-cot$72^{o})^{2}$ .  $cos^{2}18^{o}$ + 5$cos^{2}18^{o}$

E = 4 . (-1) . $\frac{1}{2}$ + 5 . 1 = 3