化简(1+sinx)[3cosx / 2cos平方(π/4-x/2) -2tan(π/4-x/2)]_百度作业帮
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化简(1+sinx)[3cosx / 2cos平方(π/4-x/2) -2tan(π/4-x/2)]
化简(1+sinx)[3cosx / 2cos平方(π/4-x/2) -2tan(π/4-x/2)]
2.因为,2{cos[(π/4)-(x/2)]}^2 =cos2[(π/4)-(x/2)]+1 =sinx+1 xtan[(π/4)-(x/2)] =xsin[(π/4)-(x/2)]/cos[(π/4)-(x/2)] =2xsin[(π/4)-(x/2)]*cos[(π/4)-(x/2)]/2{cos[(π/4)-(x/2)]}^2 =xsin2[(π/4)-(x/2)]/(sinx+1) =xcosx/(sinx+1) 所以,原式=(sinx+1){3cosx/2{cos[(π/4)-1)]}^2-2tan[(π/4)-1]} =(sinx+1)[3cosx/(sinx+1)-2cosx/(sinx+1)] =3cosx-2cosx =cosx化简f(x)=sinx/(sinx+2sinx/2)化简到哪一步可以求周期,求详解最好是用sinx=2cos(x/2)sin(x/2)解的_百度作业帮
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化简f(x)=sinx/(sinx+2sinx/2)化简到哪一步可以求周期,求详解最好是用sinx=2cos(x/2)sin(x/2)解的
最好是用sinx=2cos(x/2)sin(x/2)解的
f(x)=sinx/(sinx+2sinx/2)=2sin（x/2）cos（x/2）/[2sin（x/2）cos（x/2）+2sinx/2]=cos（x/2）/[cos（x/2）-1]=1-1/(cosx/2-1)T=2π/(π/2)=4Solving Trigonometric Equations
students gain understanding and self-confidence in algebra
Trigonometric Equations (page
Solving trig equations use both the
you've memorized and a lot of the algebra you've learned.
Be prepared to need to think!
Solve sin(x)
+ 2 = 3 for
& x & 360°
Just as with , I'll first isolate the variable-containing term:
+ 2 = 3 &&&sin(x) = 1
Now I'll use the reference angles I've
memorized:
Solve tan2(x)
+ 3 = 0 for
& x & 360°
There's the temptation to quickly recall
that the tangent of 60°
involves the square root of 3
and slap down an answer, but this equation doesn't actually have a
How can the square of a trig
function evaluate to a negative number? It can't!
no solution
Solve &&&on
& x & 360°
To solve this, I need to do some :
Now that I've done the algebra, I can
do the trig. From the first factor, I get x
= 90° and x
= 270°. From the second factor,
I get x = 30°
and x = 330°.
= 30°, 90°, 270°, 330° & Copyright
(C) Elizabeth Stapel
All Rights Reserved
Solve sin2(x)
? sin(x) = 2
& x & 360°
This is a quadratic in sine, so I can
apply some of the same methods:
? sin(x) ? 2 = 0
? 2)(sin(x) + 1) = 0
= 2 (not possible!) or sin(x)
Only one of the factor solutions is
sensible. For sin(x)
= ?1, I get:
Solve cos2(x)
+ cos(x) = sin2(x)
& x & 360°
I can use a
to get a quadratic in cosine:
+ cos(x) = sin2(x)
+ cos(x) = 1 ? cos2(x)
+ cos(x) ? 1 = 0
? 1)(cos(x) + 1) = 0
= 1/2 &or &cos(x)
The first trig equation, cos(x)
= 1/2, gives me x
= 60° and x
= 300°. The second equation gives
me x = 180°.
So my complete solution is:
ADVERTISEMENT
= 60°, 180°, 300°
Solve sin(x)
& x & 360°
I can use a
on the right-hand side, and re then
I'll factor:
2sin(x)cos(x)
? 2sin(x)cos(x) = 0
? 2cos(x)) = 0
= 0 &or &cos(x)
I can The sine wave is zero at 0°,
and 360°.
The cosine is 1/2
and thus also at 360°
? 60° = 300°. So the complete solution
= 0°, 60°, 180°, 300°, 360°
Solve sin(x)
+ cos(x) = 1
& x & 360°
Hmm... I'm really not seeing anything
here. It sure would have been nice if one of these trig expressions
were squared...
Well, why don't I square both sides,
then, and see what happens?
cos(x))2 = (1)2
+ 2sin(x)cos(x) + cos2(x)
= 1 [sin2(x)
+ cos2(x)] + 2sin(x)cos(x)
+ 2sin(x)cos(x) = 1
2sin(x)cos(x)
= 0 sin(x)cos(x)
go figger: I squared, and got something that I could work with.
From the last line above, either sine
is zero or else cosine is zero, so my solution appears to be:
= 0°, 90°, 180°, 270°
However (and this is important!), I
squared to get this solution, so I need to check my answers in the
original equation, to make sure that I didn't accidentally
create solutions that don't actually count. Plugging back in, I see:
+ cos(0°)
= 0 + 1 = 1 &&(this
solution works) sin(90°)
+ cos(90°)
= 1 + 0 = 1 &&(this
one works, too) sin(180°)
+ cos(180°)
= 0 + (?1) = ?1 &&(okay,
so&this one does NOT work)
sin(270°)
+ cos(270°)&=
(?1) + 0 = ?1 &&(this
one doesn't work, either)
So the actual solution is:
= 0°, 90°
Note that I could have used the double-angle
identity for sine, in reverse, instead of dividing off the 2
in the next-to-last line in my computations. The answer would have been
the same, but I would have needed to account for the solution interval:
2sin(x)cos(x)
= sin(2x) = 0
= 0°, 180°, 360°, 540°, etc, and
dividing off the 2
from the x
would give me x
= 0°, 90°, 180°, 270°, which is the
same almost-solution as before. After doing the necessary check (because
of the squaring) and discarding the extraneous solutions, my final answer
would have been the same as before.
This squaring trick doesn't come up often,
but if nothing else is working, it might be worth a try.
Cite this article
Stapel, Elizabeth.
&Solving Trigonometric Equations.& Purplemath.
Available from &&&&/modules/solvtrig.htm.
Accessed [Date] [Month] 2015 &
Purplemath: && &&
Reviews ofInternet Sites:
Tutoring from Find
This lesson may be printed out for
your personal use.
&&Copyright (C)问题补充&&
7π／6]此时F（X）∈[m,2分之派]时,2X+π／6∈[π／6,m+3]所以m=1&#47你好，当属于[0，F（X）=cos2x+1+√3 sin2x+m=2sin（2x+π／6）+m+1所以最小正周期为π
梦回唐朝LX &
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Powered by数学f(x)=2sinxcosx-2cos平方x+1,怎么化简?_百度作业帮
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数学f(x)=2sinxcosx-2cos平方x+1,怎么化简?
数学f(x)=2sinxcosx-2cos平方x+1,怎么化简?
f（x）=sin（2x）-（2cos^2（x）-1）=sin（2x）-cos（2x）=根号下2x（根号下2/2sin（2x）-根号下2/2cos（2x） =根号下2xsin（2x-兀/4）