In this triangle, cosin C = 2?

In this triangle, cosin C = 2?

Cosine theorem: B / a = SINB / Sina = radical 3, B = a radical 3 cosine theorem: C ^ 2 = a ^ 2 + B ^ 2-2abcosc = a ^ 2 + 3A ^ 2-2 root 3 * a ^ 2 * root 3 / 2 = a ^ 2A = C = 2B = a radical 3 = 2 s △ ABC = 1 / 2absin C = 1 / 2 * 2 * 2 * 2 = radical 3 * 1 / 2 = radical 3

In the triangle ABC, we know that tanb = 3 under the root sign, COSC = 1 / 3, AC = 3 times the root sign, and then find the area of the triangle ABC

It's equal to 2 / 2

In triangle ABC, if tanb is equal to root three, COSC is equal to one third, and AC is equal to three times the root sign six, then the area of triangle ABC is given

Tanb = root 3, B = 60 degree, SINB = root 3 / 2
Sinc = (2 times root number 2) / 3 AC = b = 3 times root number 6
From C / sinc = B / SINB, we know that C = 8
According to COSC = 1 / 3 = (A2 + b2-c2) / 2Ab, a = radical 6 + 4
S = SINB * AC / 2 = 6 * Radix 2 + 8 * Radix 3

In the triangle ABC, we know that tanb = root 3, COSC = 1 / 3, AC = 3 times root 6, To find the area of the triangle ABC

As high ad on BC side, CD = AC / COSC = √ 6, ad = 4 √ 3, BD = ad / tanb = 4,
So s Δ ABC = 1 / 2x4 √ 3x (4 + √ 6) = 6 √ 2 + 8 √ 3

Why is cos (α + π / 4) equal to radical 2 / 2cos α - radical 2 / 2Sin α?

Sum and difference of two corners
cos(α+β)=cosα·cosβ-sinα·sinβ
cos(α-β)=cosα·cosβ+sinα·sinβ
cos(α+π/4)
=cosα·cosπ/4-sinα·sinπ/4
=Radical 2 / 2cos α - radical 2 / 2Sin α

If cos α + 2Sin α=- 5, then Tan α=______ .

From known
5sin（α+φ）=-
5 (where Tan φ = 1
2），
That is, sin (α + φ) = - 1,
So α + φ = 2K π - π
2，α=2kπ-π
2-φ（k∈Z），
So tan α = Tan (− π)
2−φ)=1
tanφ=2．
So the answer is: 2

Cos a + 2Sin a = negative radical 5 and Tan a =?

The simultaneous equations are established by cos a + 2Sin a = - √ 5 and cos? A + 2Sin? A = 1
The results show that sin a = (- 2 √ 5) / 5; Cos a = (- √ 5) / 5
So: Tan a = 2

If we know a point P (1, root 3) on the final edge of angle α, then cos α=

x=1,y=√3,r=√[1^2+(√3)^2]=2,
cosα= x/r=1/2

Simplification: cos10 degree tan20°+ 3sin10°•tan70°-2cos40°．

cos10°
tan20°+
3sin10°•tan70°-2cos40°
=cos10°
tan20°+
3sin10°
tan20°-2cos40°
=2sin(10°+30°)
tan20°-2cos40°
=2sin40°
tan20°-2cos40°
=4cos220°-2cos40°
=4cos220°-4cos220°+2
=2．

Cos10 ° / tan20 ° + Radix 3 × tan70 ° - 2cos40 ° =?

Cos10 ° / tan20 ° + Radix 3 × tan70 ° - 2cos40 °
=cos10°/tan20°+2cos30°/tan20°-2cos40°
=(cos10°cos20°+2cos30°cos20°)/sin20°-2cos40°
=(cos10°cos20°+2cos30°cos20°-2cos40°sin20°)/sin20°
=[(cos30°+cos10°)/2+(cos50°+cos10°)-(cos110°+cos30°)]/sin20°
=[(-cos30°+3cos10°)/2+(cos50°+cos70°)]/sin20°
=[(-cos30°+3cos10°)/2+2cos60°cos10°]/sin20°
=[(-cos30°+5cos10°)/2]/sin20°
=[(-cos30°+cos10°)/2]/sin20°+2cos10°/sin20°
=sin20°sin10°/sin20°+2cos10°/sin20°
=sin10°+2sin80°/sin20°
=sin10°+8cos20°cos40°
=sin10°+4(cos60°+cos20°)
=2+sin10°+4cos20°