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!
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)
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
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 !!!!
8. tan^4x + 2tan^2 x + 1 factor as
[ tan^2 x + 1 ]^2 =
[sec^2( x) ]^2 =
sec^4 (x)
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°)