(3) Trigonometric functions

Express sinx+cosx in the form Acos(x-α),where A and α are constants.

sinx + cosx = Acos(x - α)

since A = √( 1^2 + 1^2 )

A = √2

since

tan = 1/1

α = π/4

so,

sinx+cosx = √2cos(x - π/4)

please check your answer.


Sketch the graph of the function f(x)=|sinx+cosx| for 0<=x<= π.

since sinx+cosx = √2cos(x - π/4)

we draw √2cos(x - π/4)

the tips is

when comparing with the graph of f(x) = cosx

1)it transfers to the right by π/4 units along x-axis.

2)it has a range that is - √2 <= cosx <= √2,so the amplitude is √2

3)then,we may draw | √2cos(x - π/4) | by reflecting the negative side of the graph by x -

axis.


Solve the equation |sinx+cosx|=1 for 0<=x<= π

|sinx+cosx|=1

√2cos(x - π/4) = 1 or √2cos(x - π/4) = -1

cos(x - π/4) = cosπ/4 or cos(x - π/4) = 3π/4

x = π/2 or x = π

2007-03-02 17:36:07 補充：
for the last part|sinx cosx|=1after squaring both sides1 2sinxcosx = 12sinxcosx = 0sin2x = 0x = 0

a) ..Acos(x-α)
=Acos x cos α + Asinx sinα
= (Acosα) cos x + ( Asinα) sinx
=1cosx+1sinx

so, cos α = 1 / A
......sinα = 1 / A

so, A = √(1^2 + 1^2) = √2
tan α = 1
.......α =π/4

b) i don't like sketching graph so i win not show you.
c) |sinx+cosx|=1
.....sinx+cosx = +/-1
√2cos(x-π/4) = +/-1
.....cos(x-π/4) = 1 / √2 or -1 / √2 (rejected)
...........x-π/4 = π/4
.................x= π/2
# sorry may be the last part is not correct. As i am busy to leave.
### Think it yourself.