Tan (a + π / 4) = 2 a ∈ (0, π / 2), find Tana What about the value of COS (π / 2-A) + cos (3 π + a)?

Tan (a + π / 4) = 2 a ∈ (0, π / 2), find Tana What about the value of COS (π / 2-A) + cos (3 π + a)?

Tan (a + π / 4) = 2 is positive, plus a ∈ (0, π / 2), then a ∈ (0, π / 4)
tan(a+π/4)=2
(tana+1)/(1-tana)=2
tana=1/3
So Sina = 1 / radical 10 cosa = 3 / radical 10
cos(π/2-a)+cos(3π+a）
=sina-cosa
It's OK to count on behalf of others

12¹³/（3¹º*4¹¹）

12^13 / (3^10*4^11)
=(3^13/3^10)*(4^13/4^11)
=27*16
=432

Mathematics of grade two in junior high school 12 & sup1; & sup3; △ 3 & sup1; & ordm; × 4 & sup1; & sup1;)
12 & sup1; & sup3;? (3 & sup1; & ordm; × 4 & sup1; & sup1;) solution

12¹³÷（3¹º×4¹¹）
=（3*4）^13/(3^10*4^10*4)
=3^13*4^13/(3^10*4^10*4)
=(1/4)*3^3*4^3
=(1/4)*27*64
=432

Why is it known that Q & sup1; & ordm; = & frac14;, we can get the quintic power of q = ± & frac12;? How to calculate? How to find the algorithm

You think of Q to the fifth power as X
So the known condition is actually that the square of x = a quarter
Square both sides to get x = ± half

(-2)²º¹º×(-½)²º¹¹=

(-2)²º¹º×(-½)²º¹¹=
=(-2*(-1/2))²º¹º ×(-½)
=-1/2

a. If B is opposite to each other, C and D are reciprocal to each other, and the absolute value of X is equal to twice of its opposite number, then the quadratic power of X + ABCD + (a + b) CD = -————

a. B is opposite to each other, C and D are reciprocal to each other, and the absolute value of X is equal to twice of its opposite number,
So a + B = 0, CD = 1, x = 0
The second power of X + ABCD + (a + b) CD
=0+cd（ab+a+b）
=1

It is known that in △ ABC, angle B is 20 greater than angle A and angle B is 20 smaller than angle C. The degree of the sum of the three inner angles of △ ABC is obtained

Let the angle B be x degrees
（x-20)+x+(x+20)=180
The solution is x = 60
Angle A: 60-20 = 40
Angle B: 60
Angle c: 60 + 20 = 80
PS: remember to give points~

In △ ABC, it is known that ∠ a = 2 ∠ B, the degree of ∠ C is 20 ° less than that of ∠ a, and the degree of three internal angles of △ ABC is calculated

∠A=（180+20）÷（1+1+1/2）=200÷2.5=80°
∠B=80×1/2=40°
∠C=80-20=60°
Have a good time

In △ ABC, ∠ A is twice as much as ∠ B, and ∠ C is 15 ° larger than ∠ a?

∠A+∠B+∠C=180°
According to the meaning of the title: 2 ∠ B + ∠ B + 2 ∠ B + 15 ° = 180 °
∠B = 33°
Then ∠ a = 66 °∠ C = 81 °

In the triangle ABC, the angle a is greater than the angle B = 3:2, the angle a-angle C = 60, and the degree of each angle is calculated

If angle A: angle B = 3:2, then angle B = 2, angle B = 2 / 3 * angle a-angle C = 60 & # 186; angle c = angle A-60 & # 186; because angle a + angle B + angle c = 180 & # 186; so angle a + 2 / 3 * angle a + angle A-60 & # 186; = 180 & # 186; (1 + 2 / 3 + 1) * angle a = 240 & # 186; angle a = 90 & # 186; angle B = 2 / 3 * 90 = 60 & # 186; angle c = 30