Given Sina + cosa = 1 / 2, find Sina ^ 3 + cosa ^ 3 and sin ^ 4 + cosa ^ 4 Notice that the third and fourth power are the third and fourth power of the angle

sina+cosa=1/2
Two sides square
sin²a+2sinacosa+cos²a=1/4
2sinacosa=-3/4
sinacosa=-3/8
sin³a+cos³a=(sina+cosa)(sin²a-sinacos+cos²a)
=1/2(1+3/8)=11/16
sin^4 a+cos^4 a=(sin²a+cos²a)²-2sin²acos²a=1-2×9/64=23/32

Prove that the sum of inner angles of n-polygon is equal to (n-2) * 180

Utilization
prove
1. When n = 3, the sum of internal angles is 180, which is equal to (3-2) x180
2. If n = k, then
(k-2)X180+180=((k+1)-2)X180
That is, when n = k is true, n = K + 1 is also true
3. Get proof
If you don't understand, you are welcome to ask,

It is proved that the sum of internal angles of N deformation is equal to (n-2) * 180 degree
Prove that chapter with volume two
Prove that the sum of inner angles of n-sided shape is equal to (n-2) = 180 degree

Take any point O in the n-polygon and connect it with each vertex. There are n triangles. The sum of the internal angles of the triangles is 180 degrees, n * 180 degrees in total. The calculated angle is exactly a circumference of 360 degrees, so - 360 degrees is the sum of the internal angles of the n-polygon = n * 180 degrees - 360 degrees = n * 180 degrees - 2 * 180 degrees = (n-2) * 180 degrees

How to prove that the sum of the exterior angles of a triangle is equal to 360 degrees?

Let the three vertices of the triangle be a, B and C. let AB lead to the ray ad, BC lead to the ray be, CA lead to the ray CF ∵ ABC + ∵ BAC + ∵ BCA = 180 ° and ∵ ABC + ∵ DBC = 180, ∵ BCA + ∵ ECA = 180, ∵ bac + ∵ Fab = 180

Given Sina + cosa = 1 / 2, find Sina ^ 3 + cosa ^ 3 and sin ^ 4 + cosa ^ 4 Notice that the third and fourth power are the third and fourth power of the angle