Cos (- Pai / 6) + cos (Pai Pai / 6) cos (2 Pai / 6) + cos (3 Pai / 6) + +Cos (2010 PAI - Pai / 6) Cos (- Pai / 6) + cos (Pai Pai / 6) cos (2 Pai / 6) + cos (3 Pai / 6) + +Cos (2010 Pai Pai / 6)

Cos (- Pai / 6) + cos (Pai Pai / 6) cos (2 Pai / 6) + cos (3 Pai / 6) + +Cos (2010 PAI - Pai / 6) Cos (- Pai / 6) + cos (Pai Pai / 6) cos (2 Pai / 6) + cos (3 Pai / 6) + +Cos (2010 Pai Pai / 6)

=cos(-30)+cos150+cos(330)+cos510.+cos(2010π-30)
=cos30-cos(30)+cos30-cos30.+cos30
=cos30+0+0+0...+0
=cos30
=√3/2

5cos^2(a/2)+6sin(a/2)*cos(a/2)-3sin^2(a/2)-1 (√3tan70+1)/【（4cos²70-2）sin70】

The original formula = 5 (1 + COSA) / 2 + 3sina-3 (1-cosa) / 2-1
=3sina+4cosa
=5sin(a+b)
Where tanb = 4 / 3

5cos^2α=1 cosα=?

5cos^2α=1
cos^2α=1/5
Cosa = positive root (1 / 5) or cosa = negative root (1 / 5)

If cos (a + b) = 1.5cos (a-b), what is tanatanb Quick solution! To answer the process, the answer is - 1 / 5

cos(a+b)=1.5cos(a-b)
cosacosb-sinasinb=1.5cosacosb+1.5sinasinb
-2.5sinasinb=0.5cosacosb
tanatanb=-1/5

If cos (α + β) = 3 / 5cos (α - β) = 12 / 130 If cos (α + β) = 3 / 5, cos (α - β) = 12 / 130

3 / 5 cos (α - β) = 12 / 13, then cos (α - β) is not greater than 1? Maybe the title is not written correctly
Here's your idea. Calculate sin (α + β), sin (α - β), and then use the sum angle formula to calculate (α - β) + (α + β) = 2 α

Is the square of COS (a-b) / 2 equal to 1 + cos (a-b) / 2

The square of COS (a-b) / 2 is equal to 1 + cos (a-b) / 2
Power reduction formula: (COSA / 2) ^ 2 = (1 + COSA) / 2

Evaluate [1 / cos ^ 2 (80 °) - 3 / cos ^ 2 (10 °)] 1 / cos20 °

[1/(cos80°)^2-3/(cos10°)^2] 1/cos20°
=[(1/cos80 + √3/cos10) * (1/cos80 - √3/cos10)] *[1/cos20]
=[(1/sin10 + √3/cos10) * (1/sin10 - √3/cos10)] *[1/cos20]
=[(cos10+√3sin10)/sin10cos10 * (cos10-√3sin10)/sin10cos10] *[1/cos20]
=[4sin40/sin20 * 4cos70/sin20] *[1/cos20]
=[16sin40/sin20] *[1/cos20]
=[32cos20] *[1/cos20]
=32

Evaluation: 3 − sin70 degrees 2−cos210°．

The original formula = 3 − sin (90 ° - 20 °)
2−1+cos20°
2=3−cos20°
2−1+cos20°
2=6−2cos20°
4−1−cos20°=2(3−cos20°)
3−cos20°=2．

Cos (- 8 π / 3) evaluation

cos（-8π/3）
=cos（-8π/3+2π）
=cos（-2π/3）
=-cos（-2π/3+π）
=-cos（π/3）
=-1/2

Evaluation: cos (arcsin1 / 3)=

Let arcsin (1 / 3) = a
Then Sina = 1 / 3
According to the property of inverse function 0