Known sinA 1 cosA 1 2 find cosA sinA 1

In the equation on the left, (sina+cosa) is regarded as a whole, and in the result of the squared calculation, there is also the square of (sina + cosa), and in the pants, it is enough to calculate the slippery pure world. Finally, the equation on the left is calculated to see if it is equal. This proves the letter limb.

This type of problem is a trigonometric equation. The square difference formula a -b = (a + b) (a-b) and the basic trigonometric formula sin +cos = 1 are used to simplify, and then the unary quadratic equation is solved, and then the inverse trigonometric function is obtained

Solution: (sin +cos) (sin -cos) sin -cos

2sin²α-1

2sin²α-1=1/2

sin²α=3/4

sinα=√3/4)=√3)/2

arcsin(√(3)/2)=60°

For (1+sina) cosa=-1 2, the numerator denominator Zheng Zelimb shouts that the world is multiplied by 1-sina at the same time, and it is obtained.

1*1-sina*sina) cosa(1-sina)=-1 2 simplification to cosa*cosa staring cosa(1-sina)=cosa(1-sina)=-1 2

So cosa (sina-1) = 1 2

sina/cosa=-3/2

2sina=-3cosa

2/3=cosa/sina

5/3=cosa/sina-1

It's been a long time since I've seen the curved buried number of buried ants, but these are still very simple orange lists, hehe

There are two key points in this problem: one is the cube sum (difference) formula, and the other is to narrow the range of a to determine the positive or negative of sina-cosa

I hope it can help you and strengthen your learning.

(1+sina)(1+cosa)=1, which can be reduced to:

1+(sina+cosa)+sinacosa=1sina+cosa= - sinacosa ,, both sides squared;

1+2sinacosa=(snacosa) (snacosa) -2(snacosa)-1=0sinacosa=[2 2 2] 2=1 2 and sinacosa=(1 2)sin2a 1 2, so sinacosa=1- 2

1-sina)(1-cosa)=1-(sina+cosa)+sinacosa

1+2sinacosa=1+2(1-2)=3-2 2 Therefore: (1-sina)(1-cosa)=3-2 2

Because sin a+cos a=1

cos a=1-sin a=(1+sina)(1-sina) so(1+sina) cosa=cosa (1-sina)=-1 2, so primitive=1 2