If α is obtuse angle sin α = 1 / 3, what is tan α equal to

If α is obtuse angle sin α = 1 / 3, what is tan α equal to

α is an obtuse angle and cos α is negative, so cos α = - √ (1-sin? α) = - √ (1 - (1 / 3) 2) = - 2 √ 2 / 3
∴tanα = sinα/cosα =(1/3)/(-2√2 / 3) = - 1/2√2 = - √2/4

Let α be an acute angle and Tan α = 3, find (COS α - sin α) / cos α + sin α

（cosα-sinα）/cosα+sinα）=（1-tanα)/(1+tantanα)
=(1-3)/(1+3)=-1/2

The monotone increasing interval of the function f (x) = sin (2x + π / 4) is

If f (x) = sin (2x + π / 4), we must know that the monotone increasing interval of function y = SiNx is - π / 2 + 2K π ≤ x ≤ π / 2 + 2K π, K ∈ Z. therefore, for the function f (x) = sin (2x + π / 4), we directly replace the number in brackets with the number in ①, that is - π / 2 + 2K π ≤ 2x

Given the function f (x) = sin (2x + π / 2), let g (x) = f (x) + F (π / 4-x), find the monotone increasing interval of function g (x)

f(x)=sin(2x+π/2)=cos2x
g(x)=f(x)+f(π/4-x)
=cos2x+cos(π/2-2x)
=cos2x+sin2x
=√2sin(2x+π/4)
Single increasing interval 2x + π / 4 ∈ [2K π - π / 2, 2K π + π / 2]
x∈[kπ-3π/8, kπ+π/8] k∈Z

The monotone decreasing interval of the function y = sin (- 2x + π / 3) is

∵y=sin(-2x+π/3)
=-sin(2x-π/3)
that
2kπ-π/2≤2x-π/3≤2kπ+π/2
2kπ-π/2+π/3≤2x≤2kπ+π/2+π/3
2kπ-π/6≤2x≤2kπ+5π/6
kπ-π/12≤x≤kπ+5π/12
The monotone decreasing interval is [K π - π / 12, K π + 5 π / 12] (K ∈ z)

Find the monotone decreasing interval of the function y = sin (π / 3-2x); Why not find the monotone decreasing interval of sin (π / 3-2x) directly, but change it to y = - sin (2x - π / 3)

In fact, not necessarily. If you don't change it, sometimes it will be wrong. Even if you get it right, it is estimated that the marking teacher will deduct points. If the unknown coefficient is changed into positive, it can avoid mistakes. As for why, I am not sure why. When I encounter this kind of problem, I always turn the unknown coefficient into positive

What is the monotone decreasing interval of the function y = sin (- 2x + 6 fractions)?

According to the meaning of the title, π / 2 + 2K π - 2x + π / 6 3 π / 2 + 2K π can be simplified to - 2 π / 3 - K π and X - π / 3-K π

The function y = sin (− 2x + π) 6) The monotone decreasing interval of is () A. [−π 6+2kπ，π 3+2kπ]  k∈Z B. [π 6+2kπ，5π 6+2kπ]  k∈Z C. [−π 6+kπ，π 3+kπ]  k∈Z D. [π 6+kπ，5π 6+kπ]  k∈Z

∵y＝sin(−2x+π
6)=-sin（2x-π
6）
Let − π
2+2kπ≤2x−π
6≤π
2+2kπ
Then − π
6+kπ≤x≤π
3+kπ
The function y = sin (− 2x + π)
6) The monotone decreasing interval of [- π
6+kπ，π
3+kπ]，k∈Z
Therefore, C

The monotone decreasing interval of the function y = sin (- 2x - π / 4) is obtained

Y = sin (- 2x - π / 4) = - sin (2x + π / 4), the monotone decreasing interval of Y is also the monotone increasing interval of sin (2x + π / 4). From 2K π - π / 2 ≤ 2x + π / 4 ≤ 2K π + π / 2 (K ∈ z), X ∈ [K π - 3 π / 8, K π + π / 8] (K ∈ z) is obtained

The function y = sin (π The monotone decreasing interval of 4 − 2x) is () A. [kπ+π 8，kπ+5 8π] B. [kπ-π 8，kπ+3 8π] C. [2kπ-π 8，2kπ+3 8π] D. [2kπ-3 8π，2kπ+π 8] (K ∈ Z above)

∵y=sin（π
4-2x）=-sin（2x-π
4），
From 2K π - π
2≤2x-π
4≤2kπ+π
2 (K ∈ z), K π - π
8≤x≤kπ+3π
8（k∈Z），
∴y=sin（π
The monotone decreasing interval of 4-2x) is [K π - π
8，kπ+3π
8]（k∈Z）．
Therefore, B