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If TaNx = - 1 / 2, find 2Sin ^ 2x + SiNx times cosx-3cos ^ X
Standard Practice
2Sin ^ 2x + SiNx times cosx-3cos ^ x
=(2Sin ^ 2x + SiNx times cosx-3cos ^ x) / (SIN) ² x+cos ² x)
=(2tan ² x+tanx-3)/(tan ² x+1)
=(1/2-1/2-3)/(1/4+1)
=-12/5
It is known that cosx SiNx = 3 times the root sign 2 / 5,17 π / 12
Cosx SiNx = 3 times the root sign 2 / 5, the square on both sides of the formula can find 2cosx * SiNx = 7 / 25, then add 1 on both sides to get cosx + SiNx = 4 times the root sign 2 / 5, add and subtract to get the results of SiNx and cosx, and then get TaNx = 1 / 7
Sin2x = 2cosx * SiNx, sin ^ 2x = (2cosx * SiNx) has obtained these values, which can be carried in
Given that x is greater than negative row and less than zero, SiNx + cosx = 5 points 1, find the value of SiNx cosx (2) find the value of 1-tanx points sin2x + 2Sin square X
The first question: SiNx + cosx = 5 points, 1 and the normal square of cosx + the square of SiNx = 1. We can calculate the values of cosx and SiNx respectively, and then substitute it to calculate the result of the first question. The second question: first reduce the specialization with TaNx to the evaluation with SiNx and con
Given TaNx = - 3 / 4, find the values of cosx and SiNx
sinx/cosx=tanx=-3/4
sinx=-3cosx/4
sin ² x+cos ² x=25cos ² x/16=1
cos ² x=16/25
cosx=±4/5,sinx=±3/5
Because cosx and SiNx have different signs, so
SiNx = 3 / 5, cosx = - 4 / 5 or SiNx = - 3 / 5, cosx = 4 / 5
Given SiNx = - 1 / 2, find the values of cosx and TaNx. Find the maximum value and minimum value of the function y = 1-1 / 2cos (2x + Pie / 4) Given SiNx = - 1 / 2, find the values of cosx and TaNx Find the function y = 1-1 / 2cos (2x + Pie / 4) to obtain the maximum value. The minimum value makes the set of independent variables X, and write the maximum and minimum values respectively Sin25 / 6 faction + cos25 / 3 faction + Tan (- 25 / 4 faction) + sin (26 / 3 faction)
Cosx = ±√ (1-sinx ^ 2) = ±√ (1-1 / 4) = ±√ 3 / 2tanx = SiNx / cosx = ±√ 3 / 3Y = 1-1 / 2cos (2x + Pie / 4) to get the maximum value, it is necessary to make cos (2x + π / 4) get the minimum value - 1, that is, 2x + π / 4 = 2K π + π x + π / 8 = k π + π / 2x = k π + 3 π / 8, and the maximum value of K ∈ ZY is the set of independent variables X of 1 + 1 / 2 = 3 / 2
Known 0 < x < π 2, then LG (cosx • TaNx + 1 − 2sin2x 2)+lg[ 2cos(x−π 4)]−lg(1+sin2x)=______．
The original formula = LG (cosx • sinxcosx + cosx) + LG 2 (cosx • 22 + SiNx • 22) - LG (sin2x + cos2x + 2sinxcosx) = LG (SiNx + cosx) + LG (cosx + SiNx) - LG (SiNx + cosx) 2 = LG (SiNx + cosx) 2 (SiNx + cosx) 2 = LG1 = 0, so the answer is 0
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