=
-5/ 13，cos(α+π/4)= 1
√2/2cosα
-
√2/2sinα
=
17√2/26
2, the original type
=(sinα)^2+(sinβ)^2-2sinαsinβ
+(cosα)^2+(cosβ)^2+2cosαcosβ
=
2
+
2cosαcosβ
-
2 sine α sine β
=
2
+
2co(α+β)
=
8/3
3, α and β are acute angles, cosα.
=
√( 1-(sinα)^2)
=
(2√5)/5
,
Similarly, cosβ=
(3√ 10)/ 10,
cos(α+β)= 1
Coase α Coase β
-
Octagonal α-Octagonal β
Alternative calculation, cos(α+β)= 1
√2/2
Because the values of sinα and sinβ are less than √2/2, it can be seen that both angles are less than π/4.
So the value of α+β is π/4.
4. Similar to the third question, cosα
=(2√5)/5，sinβ
=
(3√ 10)/ 10
cos(α+β)= 1
Coase α Coase β
-
Octagonal α-Octagonal β
=
-(√2)/ 10
Because α and β are acute angles, 0
=
arccos[-(√2)/ 10]