A is the inner angle of the triangle and sin2A3 5, then cos A4

cos(a+π/4)=cos a cosπ/4-sin a sinπ/4=√2/2(cosa-sina)

Because sin2a=2sinacosa=-3 5sin 2a-2sinacosa+cos 2a=1+3 5=8 5cosa-sina) 2=8 5

cosa-sina=2 10 5 or -2 10 5 so. cos(a+4)= 2 2(2 10 5)=2 5 5 or -2 5 5

Because sin2a = -3 5, 2a >

a>π/2

a+π/4>π/2

Rule. cos (a+π/4)<0

cos(a+π/4)=-2√5/5

sin2a 2sinacosa 3 5, and a is the inner angle of the triangle.

a is an obtuse angle, i.e., 2 a, 3 4 a 4 5 4cos(a+ 4) 0

cos (a+ 4) [1 cos 2a 2 ] 2 1 sin2a 2

cos﹙a+π/4﹚＝－2√5／5

From a being the inner angle of the triangle and sin2a=-3 5, we get cos2a=-4 5cos(a+ 4)= 2 2*cosa- 2 2*sina=)= 2 2* (cosa-sina).

cos2a=2cos 2a-1=-4 5 The solution gives coaa= 10 10cos2a=1-2sin 2a=-4 5 The solution gives sina=3 10 10

cos(a+π/4)=√2/2*cosa-√2/2*sina=)=√2/2* (cosa-sina)=-√5/5

sin2a 2sinacosa 3 5, and a is the inner angle of the triangle.

a is an obtuse angle, i.e. 2 a, 3 4 a 4 5 4

cos(a+π/4)＜0

cos (a+ 4) [1 cos 2a 2] 2

1－sin2a﹚／2

cos a+ 4 2 5 5,4, which is solved by a as the inner angle of the triangle and sin2a=-3 5 gives cos2a=-4 5

cos(a+π/4)=√2/2*cosa-√2/2*sina=)=2/2* (cosa-sina)

cos2a=2cos2a-1=-4 5 The solution yields coaa= 10 10

cos2a=1-2sin 2a=-4 5 solution gives sina=3 10 10

cos(a+π/4)=√2/2*cosa-√2/2*sina=)=2/2* (cosa-sina)=-5/5,2,cos(a+π/4)=cos a cosπ/4-sin a sinπ/4=√2/2(cosa-sina)

Because sin2a=2sinacosa=-3 5

sin^2a-2sinacosa+cos^2a=1+3/5=8/5

cosa-sina)^2=8/5

cosa-sina=2 10 5 or -2 10 5

So cos(a+ 4)= 2 2(..2. There's an answer upstairs. ,2,

Summary. Knowing that a is the inner angle of the triangle and cosa=5, the value of sina is 2 root number 6 5. The specific process will be sent by the teacher later**.

317.It is known that a is the inner angle of the triangle, and cosa=5(1) finds the value of sina.

317.It is known that a is the inner angle of the triangle, and cosa=5(1) finds the value of sina.

Knowing that a is the inner angle of the triangle and cosa=5, the value of sina is 2 root number 6 5. The specific process will be sent by the teacher later**.

First of all, the judgment question type is trigonometric function. Secondly, a is the inner corner of the trapped dust, then 0 a. Then, according to the triangular Pythagorean theorem, the other side can be obtained. Finally, the answer can be obtained by simplifying the four algorithms.

Expand 1Sine (sin) In a right triangle, the ratio of the opposite side of any acute angle a to the hypotenuse is called the sine of a. Sparrow void sin30 ° = 1 2 sin45 ° = 2 2 sin60 ° = 3 22

Cosine (cos) In a right-angled triangle, the ratio of the critical edge to the hypotenuse of any acute angle a is called the cosine of a, which is denoted as cosa, that is, cosa = the critical edge of a hypotenuse. cos30°=√3/2 cos45°=√2/2 cos60°=1/23.Tangent (tan) In a triangle of straight holes, the ratio of the opposite side of any acute angle a to the critical edge is called the tangent of a, which is denoted as tana, that is, tana = the opposite side of a is the critical edge.

tan30°=√3/3 tan45°=1 tan60°=√3

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A quick hint of this elliptical question.

The first question is e c a, and the key is to find the value of c.

The second question is that the product of the slope of the 2 straight line is -1, and then the point slope is in the same formula.

The third question is to set up the analytic formula of the parabola, and then solve the unknowns with a dot.

sina + cosa = -7 yelling 13 squares to get 1 + 2 sinacosa = 49 169

i.e. sinacosa= -60 169

Combined with sina + cosa = -7 13 and a should be rolled the bench is obtuse because they add up to be negative.

The solution is sina=5 13 cosa= -12 13, then tana= -5 12

tan( 4+a)= 1+tana) (1-tana)=7 17 Method 2: Auxiliary Angle Formula.

sina + cosa = root number 2 * sin(a + brigade 4) = 7 17

From this we calculate sin(a+ 4).

Then calculate cos(a+4) and divide the two to get tan(4+a).

sina 2+cosa 2+2sina*cosa=sin2a=2sina*cosa=1 4-1=-3 4a is within 0 to 180, then 2a is 180 to 360

cos2a = (1-sin2a 2) (1 2) = plus or minus (1-9 16) 1 2

Plus or minus 7 (1 2) 4

sinacosπ/4+cosasinπ/4=3/5(√2/2)(sina+cosa)=3/5sina+cosa=3√2/5

square sin a + cos a + 2 sinacosa = 18 251 + 2 sinacosa = 18 25

sinacosa=-7/50

sina+cosa=3√2/5

By the Vedic theorem.

Sina and Cosa are the root of x=7 2x 5-7 50=0 x=7 2 10, x=- 2 10

Because sina >0

So cosa = - 2 10

5 root number 2 = root number 50

2*3=6=root number 36

So 5 root number 2>3*2

Root number 2 2 > 3 5

So sin( 4) = sin(3 4)>sin(a+ 4)a (0, ).

So 2, so cos(a+3 4)=-4 5

cosa=cos（a+π/4-π/4）=cos(a+π/4)cosπ/4+sin(a+π/4)sinπ/4

4 5 * root number 2 2 + 3 5 root number 2 2 = - root number 2 10

Solution: sin2a = 3 5

2la＞180°→180°＞la＞90°

2tanasin2a= —

1+tan²a

2tana1+tan²a

3tan²a+10tana+3=→（1+3tana）（tana+3）=0

tana = -1 3 or ·tana = -3

If tana = -1 3

By 1+tan a=sec a has: seca = - root number (1 + 1 9) = root number 10 3

cosa=1 seca=-(3 roots, number 10) 10

sina = tanacosa = (1 3) [3 root number 10) 10] = root number 10 10

cos(a+6)=cosacos 6-sinasin 6=[-3 root number 10) 10] root number 3 2 - (root number 10 10) 1 2

Root number 10 2

2] If tana=-3

seca = - root number (1+9) = - root number 10

cosa=1 seca= - root number 10 10

sina = tanacosa = (3) (root number 10 10) = 3 root number 10 10

cos(a+ 6)=cosacos 6-sinasin 6==[root number 10) 10] root number 3 2 - (3 root number 10 10) 1 2

3 root No. 10 10

Knowing sina 2+cosa 2=1 and sina + cosa = 7 13, we can get sina*cosa = [(sina + cosa) 2-(sina 2+cosa 2) ] 2=-60 169 is less than 0, and because a is an inner angle of a triangle, a is an obtuse angle (if it is an acute angle, sina*cosa is greater than 0).

It can be solved with sina = 12 13 and cosa = -5 13

(sina+cosa)*(sina+cosa)=49/169sina*sina+cos*cosa=1

sina*sina+cos*cosa+2sina*cosa=49/169

Then 2sina*cosa = -120 169

cosa-sina)*(cosa-sina)=cosa*cosa+sina*sina-2cosa*sina=1-(-120/169)=289/169

Then the square root of cosa-sina = 289 169, you can ask for it yourself, and the negative is worth it, because a must be greater than 90 degrees.