What are sin5 cos5 SIN3 cos3 respectively? (not degrees, should be radians) fast~

What are sin5 cos5 SIN3 cos3 respectively? (not degrees, should be radians) fast~

Sin (- 5), sin5 compare size

Because,
5 ≈ 286 & # 186;, this angle is in the third limit
-5 ≈ - 286, this angle is in the second limit
sin(-5) > 0
sin5 < 0
So, sin (- 5) > sin5

Evaluation problem and judgment size
It is known that f (x) = log2 ^ (x + m) [log x + m logarithm with base 2] m ∈ R
(1) If f (1), f (3) and f (6) form an arithmetic sequence, find the value of M
(2) Let a, B, C be positive numbers which are not equal to each other, and a, B, C form an equal ratio sequence, M > 0, judge the size of F (a) + F (c) and 2F (b)

(1)2f(3)=f(1)+f(6)
That is 2log2 ^ (3 + m) = log2 ^ (1 + m) + log2 ^ (6 + m)
(m+3)^2=(m+1)(m+6)
m=3
(2)b^2=ac,a>0,b>0,c>0
(a+c)^2>4ac=4b^2 ,a+c>=2b
2f(b)=(3+b)^2
f(a)+f(c)=(3+a)(3+c)
2f(b)-[f(a)+f(c)]
=6b-3(a+c)

The sizes of sin194 ° and cos166 ° were compared

sin194°=sin(180+14)=-sin14°
cos166°=-cos(180-166)=-cos14°=-sin76°
sin76°>sin14°
-sin76° cos166°

Evaluate the answer of sin160 * cos160 * (tan340 * + cot340 *)

sin160*cos160*(tan340+cot340)=sin(180-20)*cos(180-20)*[tan(360-20)+cot(360-20)=sin20*(-cos20)*(-tan20-cot20)=sin20*cos20*(tan20+cot20)=sin20*cos20*(sin20/cos20+cos20/sin20)=sin^2 20+cos^2 20=1

Compare the process of (1) sin194, cos160 (2) sin7 / 4, cos5 / 3 (3) cos (- 47,10), cos (- 44 / 9)

(1)sin194=-sin14=-cos76
cos160=-cos20cos160
(2)sin7/4>0
cosπ>cos5/3>cos(π/2)=>cos5/3cos5/3
(3)cos(-47π/10)=cos(-47π/10+4π)=cos(-7π/10)=cos(7π/10)=cos(63π/90)
cos(-44π/9)=cos(-44π/9+4π)=cos(-8π/9)=cos(8π/9)=cos(80π/90)cos(-44π/9)

It is known that the right arrow on OA = (SIN3 / x, root 3cos3 / x), the right arrow on ob = (cos3 / x, cos3 / x) (x belongs to R), f (x) = on OA
It is known that the right arrow on OA = (SIN3 / x, root 3cos3 / x), the right arrow on ob = (cos3 / x, cos3 / x) (x belongs to R), f (x) = the right arrow on OA multiplied by OB right arrow, and the symmetry center and abscissa of F (x) image are obtained

Analysis: F (x) = vector OA? Vector ob = 1 / 2 * sin (2x / 3) + 3 (COS (x / 3)) ^ 2 = 1 / 2 * sin (2x / 3) + 3 / 2 * cos (2x / 3) + 3 / 2
F(x)=sin(2x/3+π/3)+√3/2
One axis of symmetry of F (x) is (2x / 3 + π / 3) = π / 2 = = > 2x / 3 = π / 6 = = > x = π / 4

If Tan (π - a) = 2 and Sina > 0, then cosa =? For detailed explanation

tan(π-a)=-tana=2
∴tana=-2
∴sina/cosa=-2
Sina = - 2cosa substituting Sin & # 178; a + cos & # 178; a = 1
The results show that 5 cos & # 178; a = 1, cos & # 178; a = 1 / 5
∵sina=-2cosa>0,cosa

Given that α is the third quadrant angle, Cos2 α = - (3 / 5), then what is tan2 α equal to
No one will?

Because a is the third quadrant angle, 2a is in the range of [2 π, 3 π]
So we can judge that sin2a > 0
According to sin2a ^ 2 + cos2a ^ 2 = 1
The solution is sin2a = 4 / 5
tan2a=sin2a/cos2a=-4/3

It is known that Tan α = - 13 2sin2 α + 3sin α cos α - 5cos2 α=

（-13*7/2）-5