Evaluation: cos40 ° (1+ 3tan10°)=______ .

Evaluation: cos40 ° (1+ 3tan10°)=______ .

cos40°(1+
3tan10°)=sin50°(1+
3tan10°)=sin50°(cos10°+
3sin10°)
cos10°=2sin50°sin(30°+10°)
cos10°=2cos40° sin40°
cos10°=sin80°
cos10°=1
So the answer is: 1

Evaluate sin50 degrees (1 + Radix 3 times tan10 degrees) Urgent! Online wait, seek process!

Sin50 (1 + root 3 * sin10 / cos10) = sin50 * (cos10 + root 3 * sin10) / cos10 (cos10 + root 3 * sin10) = 2 (0.5 cos10 + 0.5 root 3 * sin10) = 2 (cos60cos10 + sin60sin10) = 2cos (60-10) = 2cos50, so the original formula = sin50 * 2cos50 / cos10 = sin100 / cos10 = cos10 / cos10

What is the root of 18.75 * 2.5

Thank you for your adoption

If Tana = 3 and a is the second quadrant angle, then cosa= What kind of knowledge is used,

tana=3
sina/cosa=3
sina^2+cosa^2=1
A is the second quadrant angle
cosa<0
Solve the above equations
Cosa = - 1 / root 10

It is known that Tan (α + π / 4) = - 1 / 2, π / 2

tan(α+π/4）=-1/2
(tanα+tanπ/4)/(1-tanαtanπ/4) = -1/2
(tanα+1)/(1-tanα) = -1/2
2tanα+2 = -1+tanα
tanα = -3
π/2

If a = [cot (4 π + a) cos (a + π) sin ^ 2 (3 π + a)] / [Tan (π + a) cos ^ 3 (- A - π)], the value of a ^ 2 + A + 1 is equal to I used to calculate 1, which is wrong. I recalculated and the correct answer was 3

a=[cot(4π+a)cos(a+π)sin^2(3π+a)]/[tan(π+a)cos^3(-a-π)]
=[cota*(-cosa)*sin^2a]/[(-tana)*(-cos^3a)]
=[-cota*tan^2a/tana]
=-1.
Then a ^ 2 + A + 1 = (- 1) ^ 2 + (- 1) + 1 = 1

Given that a is the second quadrant angle and COS a = - 4 / 5, then sin a =?

(sina)^2=1-(cosa)2=1-16/25=9/25
Moreover, a is the second quadrant angle, ﹥ Sina = 3 / 5

If the point P (sin θ, cos θ, 2cos θ) is located in the third quadrant, then the quadrant where the angle θ is located is () A. First quadrant B. Second quadrant C. The third quadrant D. Fourth quadrant

The point P (sin θ, cos θ, 2cos θ) is located in the third quadrant,
∴sinθcosθ＜0
2cosθ＜0，
∴sinθ＞0，
cosθ＜0
θ is the angle of the second quadrant
Therefore, B is selected

Given Tan α =2, find the value of 3sin (π + α) +cos (- α) /4sin (- α) -cos (9 π + α)

Firstly, the induction formula is used to simplify
[3sin(π+α)+cos(-α)]/[4sin(-α)-cos(9π+α)]
=(- 3sin α + cos α) / (- 4sin α + cos α) and denominator divided by cos α at the same time
=(-3tanα+1)/(-4tanα+1)
=5/7.

Given Tan ^ 2 α = 2tan ^ β + 1, it is proved that sin ^ 2 β = asin ^ 2 α - 1 fast

It is proved that: sin ^ 2 β = 2Sin ^ 2 α - 1 Tan ^ 2 α = 2tan ^ 2 β + 1sin ^ 2 α / cos ^ 2 α = 2Sin ^ 2 β + 1sin ^ 2 α / cos ^ 2 α = (2Sin ^ 2 β + cos ^ 2 β) / cos ^ 2 β sin ^ 2 α / cos ^ 2 α = (sin ^ 2 β + 1) / cos ^ 2 β sin ^ 2 α / (1-sin