Simplified trigonometric function y = = cosx (SiNx - √ 3cosx) The trouble is that there is a detailed process and the formula on which each step is based,

Simplified trigonometric function y = = cosx (SiNx - √ 3cosx) The trouble is that there is a detailed process and the formula on which each step is based,

y=cosxsinx-√3cos^2x=1/2sin2x-√3/2(1+cos2x)
=1/2sin2x-√3/2cos2x-√3/2=sin(2x-π/3)-√3/2

A trigonometric function problem in high school mathematics: if SiNx + cosx = TaNx (0

sinX+cosX=tanX
(0

Trigonometric function Let f (SiNx) = 3-cos2x, then f (cosx)=_______ Solution ideas / directions, what will be used

Idea: the original question is f (SiNx) = 3-cos2x, the independent variable on the left is SiNx, and the independent variable on the right is cos2x, so the first idea is to transform the right into some form of independent variable, which can also be understood as setting SiNx = t, then f (SiNx) = f (T), so now what to do is to transform the right, the right is 3-cos2x, so turn it

Find the function y = sinxcosx Maximum and minimum values of 1 + cosx + SiNx

From the meaning of the question, 1 + SiNx + cosx ≠ 0, then SiNx + cosx ≠ - 1, let t = SiNx + cosx = 2Sin (x + π 4), then t ∈ [- 2, 2] and t ≠ - 1, get the square of both sides of T = SiNx + cosx, sinxcosx = t2-12, substitute y = sinxcosx1 + cosx + SiNx, y = t2-121 + T = T-12, when t = - 2, the function obtains the most

Y = maximum value of 3|sinx| + 4|cosx|

Let's take it as SiNx and cosx > 0. It's easy to do it
y=5sin(a+ x)
→ymax=5
ok
Y = 3sinx + 4cosx = 5 * (3 / 5 * SiNx + 4 / 5 * cosx) = 5 * sin (x + a), a refers to the relevant angle of a right triangle surrounded by side lengths of 3, 4 and 5 respectively

Find the maximum value of y = sin2x-3 (SiNx + cosx) ..

Let a = SiNx + cosx = √ 2Sin (x + π / 4)
So - √ 2