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Cos (- 5 π / 6) and COS (- 3 π / 4) are compared. Thank you
cos(-5π/6)=cos(5π/6)=cos(π-π/6)=-cosπ/6
cos(-3π/4)=cos(3π/4)=cos(π-π/4)=-cosπ/4
cosπ/6=1/2
cosπ/4=√2/2
cosπ/6-cosπ/4
That is cos (- 5 π / 6) > cos (- 3 π / 4)
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If cos (α + π / 6) = 4 / 5, then sin (2 α + π / 3) + cos (2 α + π / 3) If cos (α + π / 6) = 4 / 5, then sin (2 α + π / 3)+ cos（2α+π/3）=
The answer is 31 in 25
If cos (π / 6 + α) = √ 2 / 4, the value of COS (5 π / 6 - α) is ()
cos(5π/6-α)
=cos[π-(π/6+α)]
=-cos(π/6+α)=√2/4
How to calculate cos (3 π / 2 - α)?
The original formula = cos [π + (π / 2 - α)] = - cos (π / 2 - α) = - sin α
(9) calculation: cos (2 Π / 7) + cos (4 Π / 7) + cos (6 Π / 7) The answer is: - 1 / 2 How to do without a calculator? Please write down the detailed process and ideas Thank you~~
The difference is: cos (2 π / 7) + cos (4 π / 7) + cos (4 π / 7) + cos (4 π / 7) + cos (6 π / 7) = 2Sin (2 π / 7) [cos (2 π / 7) + cos (4 π / 7) + cos (4 π / 7) + cos (6 π / 7) + cos (6 π / 7)] / 2Sin (2 π / 7) = [sin (4 π / 7) + sin (6 π / 7) + sin (6 π / 7) + sin (- 2 π / 7) + s (- 2 π / 7) + s (2 π / 7) + s (2 π / 7) + s (7) + s (7) + s (7) + s (7) + s (7) + sin (8 π / 7) +
How to calculate cos (2 π a), a is householder matrix For example, give the calculation process or solution
A can be diagonalized and its eigenvalues are 1 and - 1, so cos (2 π a) can also be diagonalized with eigenvalues of 1, that is, cos (2 π a) = I
How to operate cos [(a + b) - A]
cos（a+b）*cos（a）+sin（a+b）*sin（a）
Help us calculate ∫ (π / 6, π / 2) (COS ^ 2x) DX and ∫ - 2, - 1) DX / (11 + 5x) ^ 3 ditto
∫（π/6,π/2)(cos^2x)dx=(1/2)[∫（π/6,π/2)(2cos^2x - 1)dx ]+(1/2)∫（π/6,π/2)dx=(1/2)[∫（π/6,π/2)(cos2x)dx ]+(1/2)（π/2-π/6)=(1/4)[∫（π/6,π/2)(cos2x)d2x ]+(1/2)*(π/3)=(1/4)*sin2x |（π/6...
If cos (PAI + a) = - 1 / 2, 3 Pai / 2
Because cos (π + a) = - 1 / 2
Then cosa = 1 / 2
Then sin (5 π / 2 + a) = sin (π / 2 + a) = cosa = 1 / 2
cos^2(π/3-a)+cos^2(π/6+a) Please write down the detailed process
cos^2(π/3-a)+cos^2(π/6+a)
=cos^2(π/3-a)+sin^2(π/3-a)
=1
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