Given that SiNx = - 7 / 25 and Z belongs to (3 μ g / 2,2 μ g), find cosx, TaNx

Given that SiNx = - 7 / 25 and Z belongs to (3 μ g / 2,2 μ g), find cosx, TaNx

Cosx = square of root 1-sinx = root 1-49 / 625 = 24 / 25
TaNx = SiNx / cosx = = - 7 / 25 divided by 24 / 25 = - 7 / 24

It is known that SiNx = asiny, TaNx = btany, where x is an acute angle Proof: the square of cosx is equal to (A's Square - 1) divided by (B's Square - 1) under the root sign (the radical acts on the following large formula)

If TaNx = btany, then SiNx / cosx = bsiny / cosy and SiNx = asiny (SiNx, siny ≠ 0), then acosy = bcosx, then B? Cos? X + sin? X = a? Cos? Y + a? Sin? Y = a? 2 then (B? - 1) cos? X = a? 1, because x is an acute angle, and cosx > 0, then co

It is proved that when 0 ＜ x ＜ π / 2, SiNx ＋ TaNx ＞ 2x

Consider the auxiliary function f (x) = SiNx + tanx-2x, 0

If the acute angle X satisfies SiNx * cosx = 1 / 4, then TaNx =?

2 SiNx * cosx = 1 / 2, so sin2x = 1 / 2, tan2x = (radical 3) / 3, TaNx = 2-radical 3

X is an acute angle, (SiNx) ^ 2-sinx * cosx-2 (cosx) ^ 2 = 0. TaNx =?

If both sides of the equation are divided by the square of cosx, we can get the quadratic equation of one variable about TaNx. The solution of TaNx is 2 and - 1. Because x is an acute angle, TaNx is 2. I'm sorry to write

If the acute angle X satisfies SiNx * cosx = 1 / 4, then TaNx is

sin2x=2sinxcosx=1/2
So, 2x = 30 degrees
x=15°
tanx=tan15°=tan（60°-45°）
=（tan60°-tan45°）/（1+tan60°·tan45°）
=(root 3-1) / (root 3 + 1)
=2-radical 3

Given that x is an acute angle, cosx = 1 / 3, find the values of SiNx, TaNx, Cotx

Since x is an acute angle, SiNx, TaNx, Cotx are all greater than 0
According to (SiNx) ^ 2 + (cosx) ^ 2 = 1
We can know that SiNx = (2 root sign 2) / 3
Because: TaNx = SiNx / cosx = 2 root sign 2
Cotx = 1 / TaNx = (radical 2) / 4

Given that x is an acute angle SiNx = 7 / 8tanx = 1 / 4tanb, find X

According to SiNx = 7 / 8tanx, SiNx = 7 / 8sinx / cosx, cosx = 7 / 8, x = arccos7 / 8
Is B a constant? If so, x = arcsin (1 / 4tanb) according to SiNx = 1 / 4tanb

Verification: （1）1−2sinxcosx cos2x−sin2x=1−tanx 1+tanx； （2）（cosβ-1）2+sin2β=2-2cosβ．

(1) Left = 1 − 2sinxcosxcos2x − sin2x = cos2x + sin2x − 2sinxcosxs2x − sin2x = (cosx − SiNx) 2 (cosx + SiNx) (cosx − SiNx) = cosx − sinxcosx + SiNx = 1 − tanx1 + TaNx

If TaNx = radical 2, find the value of (1) 2 (SIN) ^ 2-sinxcosx + (cosx) ^ 2

TaNx = √ 2, from 1 + Tan ^ 2 = sec ^ 2, we get cos ^ 2x = 1 / 3
If you put cos ^ 2x in the formula you asked for, you can get: 2tan ^ 2x TaNx + 1
So the result is: 1 / 3 (2 * 2 - √ 2 + 1) = 1 - (√ 2) / 3