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Hey guys im having a little trouble with trig and I would really appriciate it if someone would help me out, just answer any of them that you can help me with thanks!

JoeTheCat  Dec 8, 2017
 #1
avatar+92620 
+2

4. sin^4(x) + sin^2(x) [cos^2(x)]  =

 

sin^2(x)  [   sin^2(x)  + cos^2(x) ]  =

 

sin^2(x) * 1

 

sin^2 (x)

 

 

5.  (csc x + 1) (csc x - 1)  =

 

 csc^2 x  - 1  =

 

Remember that 1 + cot^2 (x) = csc^2(x)...so......subtract 1 from both sides

 

cot^2 x

 

 

 

6.    cos^2(x)  / [ 1 - sin (x) ]

 

Multiply top/bottom by 1 + sin (x)

 

cos^2(x) * (1 + sin (x)) /  [ ( 1 - sin(x))( 1 + sin (x) ]

 

cos^2(x)  (1 + sin (x)  / ( 1 - sin^2(x) ]

 

cos^2(x) (1 + sin (x) ]  / cos^2(x)

 

(1 + sin (x)

 

 

 

cool cool cool

CPhill  Dec 8, 2017
 #2
avatar+85 
0

Thank you so much Cphill! Greatly appreciated!

JoeTheCat  Dec 8, 2017
 #3
avatar+92620 
0

2.  [  sec^2x  + csc^2x]  

    __________________        =

     csc^2 x  [ 1 + tan^2 x] 

 

 

sec^2 x                                              csc^2 x

_________________       +              ______________       =

csc^2x [ sec^2x]                           csc^2 x [ sec^2 x] 

 

 

 

1                     +          1 

___                          ______        =

csc^2 x                    sec^2 x

 

 

sin^2 x  +  cos^2 x   =

 

 

1

 

 

cool cool cool

CPhill  Dec 8, 2017
 #4
avatar+92620 
0

7.      1 + cos x       +      sin x

          ________           ______                    get a common denominator

            sinx                 1 + cos x

 

 

[ 1 + cosx ] ^2 +  sin^2x

---------------------------------

  sin x  [ 1 + cos x]

 

 

 

[ 1 + 2cosx + cos^2x] + [ 1 - cos^2 (x) ]

_________________________________

  sin (x)  [1 + cos (x) ]

 

 

 

[    2  + 2cos^2(x) ] 

_______________

   sin x [ 1 + cos (x) ]

 

 

2 [  1 + cos (x) ]

______________

  sin x  [ 1 + cos(x) ]

 

 

  2

___            =       2csc x

sin x

 

I used  x instead of theta....still the same  !!!!

 

 

cool cool cool

CPhill  Dec 8, 2017
 #5
avatar+92620 
0

8.   tan^4x  + 2tan^2 x  +  1        factor as

 

[ tan^2 x + 1 ]^2   =

 

[sec^2( x) ]^2  =

 

sec^4 (x)

 

 

cool cool cool

CPhill  Dec 8, 2017
 #6
avatar+92620 
0

 

9.   sin (105)  =   

sin (60+ 45)  = 

sin60cos45 + sin45cos60  =

(√3/2) (√2/2)  + (√2/2)(1/2)  =

[ √6  +  √ 2 )  /  4

 

 

tan (15)  =   tan (45 - 30)  = 

 

[ tan (45)  -  tan (30) ]            [  1  -  √3/3 ]                   3 - √ 3

_________________     =    ___________      =        _______

1  + tan (45)tan(30)                [ 1 + 1 (√3/3) ]               3  + √3

 

 

 

 

10.    sin (42°)cos(38°)  - cos(42°)sin(38°)  =

 

sin ( 42 - 38)   =

 

sin (4°)

CPhill  Dec 8, 2017

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