Find the value of (tan10 ° - Radix 3) * cos10 ° / sin50 °

Find the value of (tan10 ° - Radix 3) * cos10 ° / sin50 °

(tan10°-√3)*cos10°/sin50°=(sin10°-√3cos10°)/sin50°=2(1/2*sin10°-√3/2*cos10°)/sin50°=2(sin10°cos60°cos10°sin60°)/sin50°=2sin(10-60)°/sin50°=2sin(-50)°/sin50°=-2sin50°/sin50°=-2...

Simplify sin50 ° (1+ The result is 3 Tan 10 °)______ .

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

Coos40 + sin50 (1 + (radical 3) tan10) / sin70 radical (1 + cos40) simplified denominator detailed process formula

Molecular cos40 + sin50 (1 + √ 3tan10) = cos40 + sin50 (cos10 + √ 3sin10) / cos10 = cos40 + sin50 * 2sin40 / cos10 = cos40 + sin80 / cos10 = cos40 + 1 = 2 (cos20) ^ 2
denominator
sin70√(1+cos40)=sin70√2cos20=√2(cos20)^2
Therefore, the original formula = √ 2

Find the direct {sin50 + cos40 [1 + tan10 radical 3]} / sin70 * sin70

I calculated 4 / 1-cos140
【(2cos40*2sin40)/cos10】/1-cos140
I don't know, right

(sin50 (1 + Radix (3) * tan10) - cos20) \ \ (Radix (2) * cos80 * sin10) (sin50 (1 + Radix (3) * tan10) - cos20) divided by (Radix (2) * cos80 * sin10),

[sin50º(1+√3*tan10º)-cos20º]/(√2*cos80º*sin10º)=[sin50º(1+√3*sin10º/cos10º)-cos20º]/(√2*sin10º*sin10º)=[sin50º(cos10º+√3*sin10...

Sin 50 (1 + 3 times the root of tan10) - cos20, then divide the whole by cos80 * root sign (1-cos20) to find the value

（sin50°(1+√3*tan10°)-cos20°）/（cos80°*√（1-cos20°）)=（sin50°(1+√3*sin10°/cos10°)-cos20°）/（cos80°*√（1-cos20°））=(sin50°((cos10°+√3*sin10°)/cos10°)-cos20°）/（cos80°*√(2(si...

Cos 80 degree root 1-cos 20 degree sin 50 degree (1 + root 3 Tan 10 degree) - cos 20 degree=

Cos 80 degree root 1-cos 20 degree sin 50 degree (1 + root 3 Tan 10 degree) - cos 20 degree
=[sin50°(1+√3tan10°)-cos20°]/[cos80°√(1-cos20°)]
={[2cos40°(1/2cos10°+√3/2sin10°)]/cos10°-cos20°}/{sin10°√[2(sin10°)^2]}
=[(2cos40°sin40°/cos10°)-cos20°]/[√2(sin10°)^2]
=[sin80°/cos10°-cos20°]/[√2(sin10°)^2]
=[cos10°/cos10°-cos20°]/[√2(sin10°)^2]
=[1-cos20°]/[√2(sin10°)^2]
=2(sin10°)^2/[√2(sin10°)^2]
=√2

Evaluation: sin50 ° [1 + Radix (3) tan10 °] - cos20 °] / [cos80 ° Radix (1-cos20 °])

（sin50°(1+√3*tan10°)-cos20°）/（cos80°*√（1-cos20°）)
=（sin50°(1+√3*sin10°/cos10°)-cos20°）/（cos80°*√（1-cos20°））
=(sin50°((cos10°+√3*sin10°)/cos10°)-cos20°）/（cos80°*√(2(sin10°)^2))
=(sin50°((1/2)cos10°+(√3/2)*sin10°)/((1/2)*cos10°)-cos20°）/（cos80°*√(2(sin10°)^2))
=(sin50°(sin30°*cos10°+cos30°*sin10°)/(1/2)*cos10°-cos20°）/（(√2)cos80°*sin10°)
=(sin50°*sin40°/(1/2)*cos10°-cos20°）/（(√2)sin10°*sin10°)
=(sin50°*cos50°/(1/2)*cos10°-cos20°）/（(√2)sin10°*sin10°)
=((1/2)sin100°/(1/2)*cos10°-cos20°）/（(√2)sin10°*sin10°)
=((1/2)sin80°/(1/2)*cos10°-cos20°）/（(√2)sin10°*sin10°)
=((1/2)cos10/(1/2)*cos10°-cos20°）/（(√2)sin10°sin10°)
=(1-cos20°)/（(√2)sin10°sin10°)
=2(sin20°)^2/(√2)(sin20°)^2
=2/√2
=√2
2 respondents

Tana = - 1 / 3. CoSb = radical 5 / 5, find sin (a + b)

A. Is B the angle of a triangle?
Suppose a, B ∈ (0, π)
tanA=-1/3,
therefore
sinA=1/√10
cosA=-3/√10
and
cosB=√5/5=1/√5
be
sinB=2/√5
therefore
sin(A+B)=sinAcosB+cosAsinB
=1/√50-6/√50
=-5/√50
=-(√2)/2

It is known that Tana = radical 3, Wu

tana = √3 (π < a < 3π/2)
So a = 4 π / 3
So cos a = - 1 / 2, sin a = - √ 3 / 2
So cos a-SiN a = (√ 3 - 1) / 2