What is 1 - & frac12; sin & sup2; 15 & ordm; equal to/

What is 1 - & frac12; sin & sup2; 15 & ordm; equal to/

Sin ^ 2 (α / 2) = (1-cos α) / 2, replace this formula, 15 & ordm; = 30 & ordm. / 2
zero point nine six six five
If a and B are opposite to each other, C and D are reciprocal to each other, and the absolute value of X is equal to 0, then 2 (a + b) + (CD) & sup2; & ordm
a. B is opposite to each other
a+b=0
c. D is reciprocal to each other
cd=1
X=1
2（a+b）+（cd）²º
=0+1
=1
One
a. B is opposite to each other, C and D are reciprocal to each other, and the absolute value of X is equal to 0
Then a + B = 0, CD = 1. X = 0
2（a+b）+（cd）²º
=2*0+1²º
=0+1
=1
∵ A and B are opposite to each other
∴a+b=0
∵ C and D are reciprocal
∴cd=1
2（a+b）+（cd）²º
=2*0+1^20
=1
The answer is 0
In △ ABC, ∠ a - ∠ B = ∠ B - ∠ C = 15 ° calculate the degree of ∠ a, ∠ B, ∠ C
If the ratio of the degrees of the three outer angles of a triangle is 2:3, the degrees of the inner angles of the triangle are
If the circumference of isosceles △ ABC is 18cm, and ab: BC = 1: then the three sides of triangle are
A-B=15
B-C=15
A+B+C=180
A=75 B=60 C=45
A=75 B=60 C=45
A 75du
b 60du
c 45du
∠A+∠C=2∠B,
∠A+∠B+∠C=180
The results show that ∠ B = 60, a = 75, C = 45
Angle A75 ° angle B60 ° angle B45 °
1),∠A-∠B=∠B-∠C=15°，
,∠A+∠B+∠C=180°，,∠A=∠B+15，∠C=∠B-15°，
∠B+15+∠B+∠B-15°=180°，∠B=60°，∠A=75°，∠B=45°
2)180*2/(2+3+4)=40°，
180*3/(2+3+4)=60°
180*4/(2+3+4)=80°
3) If the circumference of isosceles △ ABC is 18cm, and ab... Unfolds
1),∠A-∠B=∠B-∠C=15°，
,∠A+∠B+∠C=180°，,∠A=∠B+15，∠C=∠B-15°，
∠B+15+∠B+∠B-15°=180°，∠B=60°，∠A=75°，∠B=45°
2)180*2/(2+3+4)=40°，
180*3/(2+3+4)=60°
180*4/(2+3+4)=80°
3) If the circumference of isosceles △ ABC is 18cm, and ab: BC = 1:2, then the three sides of triangle are
AB = 18 * 1 / (1 + 2 * 2) = 18 / 5, AC = BC = 36 / 5
BC as the base, BC = 18 * 2 / (2 + 1 * 2) = 9, ab = AC = 4.5
∵∠A-∠B=15°
∴∠A=∠B+15°
∵∠B-∠C=15°
∴∠C=∠B-15°
∵∠A+∠B+∠C=180°
And ∵ ∠ a = ∠ B + 15 °
∠C=∠B-15°
∴∠B+15°+∠B+∠B-15°=180°
∴∠B=60°
∴∠A=75°
∠C=45°
2sin20°+cos10°+tan20°sin10°=______ .
2sin20 ° + cos10 ° + tan20 ° sin10 ° = 2sin20 ° + cos10 ° + sin20 ° sin10 ° cos20 ° = 2sin20 ° cos20 ° + cos10 ° cos20 ° = sin40 ° + cos10 ° cos20 ° = cos50 ° + cos10 ° cos20 ° = 2cos30 ° · cos20 ° cos20 ° = 2cos30 ° = 3
Dissolve 10 grams of sugar in 100 grams of water. How much sugar accounts for
emergency
1/11
sin80°cos35°-cos80°sin35°=?
sin80°cos35°-cos80°sin35°
=sin(80-35)
=sin45
=√2/2
Original formula = sin (80-35)
=sin45
=Question: cos α = - 5 / 13, and α is the second quadrant angle, then sin α =?
Dissolve 10 grams of sugar in 90 grams of water. Sugar accounts for 90 percent of the sugar______ %．
1010 + 90 × 100% = 10% a: the salt content of the brine is 10%
Cos80 ° cos35 ° + sin80 ° sin35 ° = cos (80 ° - 35 °) = cos45 ° why = two-thirds root sign
Because the 45 degree cosine is the root of two
Cosacosb + sinasinb = cos (a-b) is the formula of cosine operation
Formula of sum of two angles
　　cos(α+β)=cosαcosβ-sinαsinβ 　　cos(α-β)=cosαcosβ+sinαsinβ 　　sin(α+β)=sinαcosβ+cosαsinβ 　　sin(α-β)=sinαcosβ -cosαsinβ
Others ask why cos45 ° is equal to two-thirds of the root sign.
Motherland!! I am worried about your future!!!!
…………………………
Isn't cos 45 ° equal to the root of two?
When 10 grams of salt is dissolved in 100 grams of water, salt accounts for ()
A. 12B. 18C. 1/11D. 14
A: salt accounts for 111% of salt water
sin10°cos35°+sin80°sin35°=?
sin10°cos35°+sin80°sin35°
=sin10°cos35°+cos10°sin35°
=sin(10+35)°
=√2/2
Cos80cos35 + cos10cos55 = sin10cos35 + cos10sin35 = sin45 = root 2 / 2 cos80cos35 + cos10cos55 = cos80cos35 + sin80sin35 = cos (80-35) = cos45 = √ 2/