Knowing that angle 1 is the first quadrant angle, try the trigonometric definition proof. 1 sin cos

The proof to the right of the inequality is equivalent to (sin +cos) 2 2, i.e., sin 2 +cos 2 +2sin ·cos 2 --equation (1).

From the trigonometric identity, we know that 2=2 (sin 2 +cos 2), that is, 2 = 2sin 2 +2cos 2 - Eq. (2) Replace the number 2 on the right side of Eq. (1) with 2sin 2 +2cos 2.

sin^2α +cos^2α +2sinα·cosα≤2sin^2α +2cos^2α

Simplification yields sin 2 +cos 2 +2sin ·cos 0, i.e. (cos -sin ) 2 0

The angle is the first quadrant angle, i.e. (0°, 90°).

sinα>0,cosα>0

sinα+cosα>0

In addition2 (0°, 180°), sin2 = 2sin *cos (0,1].

sinα+cosα)²=sin²α+cos²α+2sinα*cosα=1+2sinα*cosα=1+sin2α∈（1,2]

1< sin + cos 2, (minus) < p>

It seems that you don't understand the meaning of this analysis, I'll help you. "Substitution simplification" means replacing 2 substitutions with the formula "2(sin 2 +cos 2 )", you see: (sin +cos) 2 2 = 2(sin 2 +cos 2 ), that is, (sin +cos) 2 2(sin 2 +cos 2 ), and by shifting and simplifying this formula, we get (cos -sin) 2 0.

Can you read it?

Draw a unit circle, take a point A on the circle in the first quadrant to make the perpendicular line of the x-axis and intersect at point B, then oa=1, ab=sin, ob=cos, the sum of the two sides of the triangle is greater than the third side of the dust, so the hole sin + cos >1

Using the unit circle, you can draw a circle with a radius of one, establish a Cartesian coordinate system with the midpoint of the circle, draw a triangle image in the first elephant family, and draw a triangle image in the first elephant family, where bc sin ac cos, then ac bc sin +cos ac 1 sin +cos >1 is proven.

The landlord refers to not sin, cos(2),tan( lujube 2) ah. If you believe that it is cos(2), please ask again. This value is rarely present.

When is the first quadrant angle, 2k , 2+2k ), k z; α/2∈(kπ,π4+kπ),k∈z

The sin must be positive, huh. The reason is that the sin value, cos value, and tan value of the first quadrant angle are all positive] (this is how it is defined, cos(2) is not necessarily positive, because although (2k, 2+2k), k z, but 2 (k, 4+k), k z, may be in the first quadrant, or it may be in the third quadrant (for example, 380 degrees is the first quadrant angle, but 190 degrees is in the third quadrant), [the cos value of the third quadrant angle is negative], so cos(2) may be positive, Possibility of negative;

tan( 2) must be positive, as mentioned above, 2 may be in the first or third quadrant, but [the tan value of the third quadrant angle is positive], so tan( 2) must be positive.

In summary, sin and tan(2) must be positive.

2k1 2k1 ,2k2 2k2 potato honorifics,k1,manuscript state k2 z,2(k1 k2) 2(k1 k2) 2Let k1 k2 n z, 2n number search 2n 2 ,n z

tanα=-3

tan''α9

Bright stove digging'refers to the square].

1+tan''α10=sin''αcos''αcos''α1/cos''α

then cos''Respect nucleus =1 10

sin''α1-cos''α9/10

is the angle of the second quadrant, then cos is negative, cos = 10 10

sin is positive and is equal to 3 10 10

sin = 3 times the root number 10, cos = 10 times the root number 10

Solution tan = sin cos = 3

sin∝=3cos∝

Angle In the first quadrant, sin 0, cos 0 sin +cos = 1

Solve the system of equations.

sin∝=3/✔10

cos∝=1/✔10

tan = 3 is known

Then sin cos = 3

sinα=3cosα

Because sin +cos = 1

So (3cos) cos = 1

i.e. cos = 1 10

Because is the first quadrant angle.

So cos = (1 10) = 10 10 so sin = 3 cos = 3 10 10

If you don't understand, please ask, and I wish you a happy study!