Given the function y = radical 3sinx / 2 + cosx / 2 (x ∈ R), find its amplitude, period and initial phase

Given the function y = radical 3sinx / 2 + cosx / 2 (x ∈ R), find its amplitude, period and initial phase

Y = radical 3sinx / 2 + cosx / 2
=2 [radical 3 / 2sinx / 2 + 1 / 2cosx / 2]
=2sin(x/2+π/6)
Amplitude = 2
The minimum positive period T = 2 π / (1 / 2) = 4 π
Primary phase = π / 6

Given the function y = radical 3sinx cosx, find the period, amplitude and initial phase of the function

Y = 2Sin (x - π / 6) t = 2 π a = 2 initial term = - 1

If sinx-2cosx = 0, find the value of sin ^ 2x-sinxcosx-2cos ^ 2x + 2

SiNx = 2cosx, the original formula = 4cos ^ 2x-2cos ^ 2x-2cos6x + 2 = 2

Sinx-2cosx = 0. Find sin2x-cos2x / 1 + sin ^ 2x

sinx-2cosx=0
tanx = 2
sin2x-cos2x/1+sin^2x
=(2sinxcosx cos? X + sin? X) / (COS? 2 + 2Sin? X) [same as above divided by cos? X]
=(2tanx-1+tan²x)/(1+2tan²x)
=(4-1+4)/(1+8)
=7/9

SiNx = 2cosx then sin ^ 2x=

sinx=2cosx
sin²=4cos²x=4(1-sin²x)
So sin? X = 4 / 5

Sin ^ 2x + 2sinxcosx-3cos ^ 2x = 1 find SiNx

sin^2x+2sinxcosx-3cos^2x=12sinxcosx-4cos^2x=0sinxcosx=2cos^2xcosx（sinx-2cosx）=0（1）cosx=0,sinx=±1（2）sinx=2cosxtanx=2tan^x+1=51/cos^x=5cos^x=1/5sin^x=4/5sinx=±2√5/5

If SiNx = 2cosx, then sin2x + 1 = 2___ .

∵sinx=2cosx，∴tanx=2．
Then sin2x + 1 = sin2x
sin2x+cos2x+1=tan2x
tan2x+1+1=22
22+1+1=9
5．
So the answer is: 9
5．

If SiNx = 2cosx, then sin2x + 1 = 2___ .

∵sinx=2cosx，∴tanx=2．
Then sin2x + 1 = sin2x
sin2x+cos2x+1=tan2x
tan2x+1+1=22
22+1+1=9
5．
So the answer is: 9
5．

Given SiNx = 2cosx, find the value of SiNx? - 3sinxcosx

Solution:
sinx=2cosx
tanx=2
sinx²-3sinxcosx=((tanx)^2-3tanx)/((tanx)^2+1)
=(4-6)/(4+1)=-2/5

sinx-2cosx=o,sin²x-sinxcosx-2cos²+2

sinx-2cosx=0
sinx=2cosx
So TaNx = 2
So sin? X-sinxcosx-2cos? X + 2
=(sin²x-sinxcosx-2cos²x+2)/1
=(sin²x-sinxcosx-2cos²x)/1+2
=(sin? X-sinxcosx-2cos? X) / (sin? X + cos? X) + 2 [numerator and denominator divided by cos? X at the same time]
=(tan²x-tanx-2)/(tan²x+1)+2
=(4-2-2)/(4+1)+2
=2
If you don't understand, I wish you a happy study!