a: \(A=sin^210^0+sin^280^0+cos^220^0+sin^270^0\)

\(=sin^210^0+cos^210^0+sin^270^0+sin^270^0\)

\(=2\cdot sin^270^0+1\)

b: \(=sin^215^0+sin^275^0+sin^235^0+sin^255^0\)

\(=sin^215^0+cos^215^0+sin^235^0+cos^235^0\)

=1+1

=2

\(A=sin^210^0+sin^280^0+cos^220^0+sin^270^0\)

\(=sin^210^0+cos^210^0+sin^270^0+sin^270^0\)

\(=2sin^270^0+1\)

\(B=sin^215^0+sin^275^0+sin^235^0+sin^255^0\)

\(=sin^215^0+cos^215^0+sin^235^0+cos^235^0\)

=1+1

=2

\(A=sin^215^o+sin^225^o+sin^235^o+sin^245^o+sin^255^o+sin^265^o+sin^275^o\)

\(A=cos^275^o+cos^265^o+cos^255^o+cos^245^o+sin^255^o+sin^265^o+sin^275^o\)

\(A=\left(cos^275^o+sin^275^o\right)+\left(cos^265^o+sin^265^0\right)+\left(cos^255^o+sin^255^o\right)+cos^245^o\)

\(A=1+1+1+0,5\)

\(A=3,5\)

a: 

b: \(B=3-sin^290^0+2\cdot cos^260^0-3\cdot tan^245^0\)

\(=3-1+2\cdot\left(\dfrac{1}{2}\right)^2-3\cdot1^2\)

\(=2-3+2\cdot\dfrac{1}{4}=-1+\dfrac{1}{2}=-\dfrac{1}{2}\)

c: \(C=sin^245^0-2\cdot sin^250^0+3\cdot cos^245^0-2\cdot sin^240^0+4\cdot tan55\cdot tan35\)

\(=\left(\dfrac{\sqrt{2}}{2}\right)^2+3\cdot\left(\dfrac{\sqrt{2}}{2}\right)^2-2\cdot\left(sin^250^0+sin^240^0\right)+4\)

\(=\dfrac{1}{2}+3\cdot\dfrac{1}{2}-2+4\)

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

A = ( sin² 10^o + sin² 80^o) + (sin² 20^o  + sin² 70^o) + ...+ (sin²40^o + sin² 50^o)

A =  ( sin² 10^o + cos² 10^o) + (sin² 20^o  + cos² 20^o) + ...+ (sin²40^o + cos² 40^o)

A = 1 + 1 + 1 + 1 = 4 ( Vì  ( sin² a + cos² a =  1 với mọi a)

Bài làm

A = ( sin² 10^o + sin² 80^o) + (sin² 20^o  + sin² 70^o) + ...+ (sin²40^o + sin² 50^o)

A =  ( sin² 10^o + cos² 10^o) + (sin² 20^o  + cos² 20^o) + ...+ (sin²40^o + cos² 40^o)

A = 1 + 1 + 1 + 1 = 4 

hok tốt

\(A=sin^210^o+cos^220^o+sin^280^o+cos^270^o\)

\(A=\left(sin^210^o+sin^280^o\right)+\left(cos^220^o+cos^270^o\right)\)

\(A=0+0\)

\(A=0\)

- Nhập \(sin^2\left(20^o\right)+sin^2\left(30^o\right)+sin^2\left(40^o\right)+sin^2\left(50^o\right)+sin^2\left(60^o\right)+sin^2\left(70^o\right)\)

vào màn hình bấm \(=3\)

- Nhập \(sin^2\left(36^o\right)+sin^2\left(54^o\right)-2tan\left(25^o\right).tan\left(65^0\right)\)vào màn hình bấm \(=-0,6031977533\)

Để mình giúp@Trịnh Ngọc Hân cho :

\(\sin\left(20^0\right)^2+\sin\left(30^0\right)^2+\sin\left(40^0\right)^2+\sin\left(50^0\right)^2+\sin\left(60^0\right)^2+\sin\left(70^0\right)^2+\sin\left(36^0\right)^2+\sin\left(54^0\right)^2-2\tan25^0.\tan65^0\)

\(=\left(\sin\left(20^0\right)^2+\sin\left(70^0\right)^2\right)+\left(\sin\left(30^0\right)^2+\sin\left(60^0\right)^2\right)+\left(\sin\left(40^0\right)^2+\sin\left(50^0\right)^2\right)+\left(\sin\left(36^0\right)^2+\sin\left(54^0\right)^2\right)-2\tan25^0.\tan65^0\)

\(=1+1+1+1-2.1=2\)

Học tốt !

Lời giải:

Ta có tính chất:

$\sin x=\cos (90-x)$ và $\sin ^2x+\cos ^2x=1$. Áp dụng vào bài ta có:

$\sin ^227+\sin ^241+\sin ^263+\sin ^245+\sin ^249$

$=(\sin ^227+\sin ^263)+(\sin ^241+\sin ^249)+\sin ^245$

$=(\sin ^227+\cos ^227)+(\sin ^241+\cos ^241)+\sin ^245$

$=1+1+(\frac{\sqrt{2}}{2})^2=\frac{5}{2}$