Cos2x formula cos2x=1-2sinx^2 cos2x=2cosx^2-1 How is the formula derived?

Cos2x formula cos2x=1-2sinx^2 cos2x=2cosx^2-1 How is the formula derived?

cos2x=cos^2 x-sin^2 x=cos^2 x-sin^2 x+1-1
=cos^2 x-sin^2 x+cos^2 x+sin^2 x -1
=1-2sinx^2=2cosx^2-1

Three formulas of cos2x = in high school mathematics?

cos2x=cos²x-sin²x
cos2x=1-2sin²x
cos2x=2cos²x-1
Cos (a + b) = cosacosb sinasinb replace B in this formula with a to obtain the formula of double angle

Simplified formula of SiNx ^ 2 cos2x? Don't you know how to solve 2Sin {X / 2) ^ 2 = 1-cosx cos2x = cosx ^ 2-sinx ^ 2?

In fact, these two are the same,
cos2x=cosx^2-sinx^2
cos2x=1-2sinx^2
cos2x=2cosx^2-1
These are the three basic formulas of the angle double of trigonometric function
In addition, as long as SiNx ^ 2 + cosx ^ 2 = 1, these three formulas can also be converted to each other

How to simplify f (x) = (cos2x-1) / sin2x

f(x)=(1-2sin²x-1)/2sinxcosx
=-2sin²x/2sinxcosx
=-sinx/cosx
=-tanx

How to simplify (6cos ^ 4x-sin ^ 2x-1) / (2 (2cos ^ 2x - 1))

(6cos^4x-sin^2x-1)/[2(2cos^2x -1)]=(6cos^4 x + cos²x -2) /2(2cos^2x -1)]=[(3cos²x+2)(2cos²x-1)]/2(2cos^2x -1)]=(3cos²x+2)/2=(3/2)*cos²x +1

If cos (x - π / 2) = 3 / 5, then cos 2x=

sinx=cos(x-π/2)=3/5
cos2x=1-2sin^2(x)
=1-18/25
=7/25

If x ∈ (− 3 π) 4，π 4) And COS (π 4−x)＝−3 5 then the value of cos2x is () A. −7 Twenty-five B. −24 Twenty-five C. 24 Twenty-five D. 7 Twenty-five

0

0

prove:
sin(π/4+x)/sin(π/4-x)+cos(π/4+x)/(π/4-x) ?
Finally, a trigonometric function symbol is missing. Please check the title and ask

If x ∈ (− 3 π) 4，π 4) And COS (π 4−x)＝−3 5 then the value of cos2x is () A. −7 Twenty-five B. −24 Twenty-five C. 24 Twenty-five D. 7 Twenty-five

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0

∵x∈(−3π
4，π
4)
∴π
4-x∈（0，π）
∴sin（π
4−x）=
1−(−3
5)2=4
Five
sin（π
2-2x）=sin[2（π
4−x）]=2sin（π
4−x）cos（π
4−x）=2×4
5×(−3
5)=-24
Twenty-five
cos2x=-24
Twenty-five
Therefore, B