Try to prove 1 2sin cos cos square sin square tan 4 .

Certainly not true.

As a counter-example: let =30° (i.e. = 6).

Left = 1 + (2 3 2) (3 4 1 4) = 1 3 Right = tan15°, check the table, right = 2 3 Obviously, the left and right are not equal, so the equation is not true.

Error cause: Left = 1+2tan (1 tan ) The numerator and denominator are divided by cos

1 tan +2tan ) (1 tan ) is divided.

Right = (tan 4 tan ) (1 + tan 4 tan ) - difference angle formula.

1 tan ) (1+tan ) substitutes tan 4=1.

1 2tan + tan ) (1 tan ) numerator denominator multiplication (1 tan ).

1+2tan tan ≠1 2tan +tan left ≠ right (the equation only holds when tan takes a very specific value such as 0).

Because the socks are definitely He Yin Tan ( +2, so Tan = -2 is the posture sin = -2cos

And because sin squared + cos squared = 1

So cos squared = 1 5, sin squared = 4 5 original = 1-3sin +4cos = 1-2 5 5 or 1+2 5 5

The square of the three-thirds root number is two-by-two-thirds of the root number three, and one-third plus three roots are three.

The root number of the second half of the number is one - the square of the root number of the third of the root number - the third of the root number is one-third of the root number and the fourth of the root number is two-third of the root number three.

The mobile phone party can't afford to hurt ah.

Because digging (sin -cos) 2=(sin) 2-2sin *cos + (cos) 2=1-2sin *cos

According to the first question, (sin -cos) 2=1-2sin *cos =1-2*6 13=1 13

Square the two sides of the above equation, i.e., the judgment code (sin -cos) 2= sin -coa =1 13) = 13 13

i.e. sin -coa =13 13

The root of 13 is No. 13

It's just to let the book brigade be like this.

Left and right bungalows. sin^2+cos^2+2sincos2sincos

sincos

1/5-cos)cos

The cos of the solution is sin, cos

Answer: sin cos=-3 4 or -4 3