Trig problem

[mohammed_159]

the question is to show that : 2cos²x + 5sinxcosx - 3sin²x+(1/2) = (5√2)/2 sin(2x+pi/4)
I've started working from the second expression which is : (5√2)/2 sin(2x+pi/4) , and I'm stuck in: 5cosxsinx + 5cos²x - 5sin²x
thanks for helping me

 

[Deleted member 4993]

the question is to show that : 2cos²x + 5sinxcosx - 3sin²x+(1/2) = (5√2)/2 sin(2x+pi/4)
I've started working from the second expression which is : (5√2)/2 sin(2x+pi/4) , and I'm stuck in: 5cosxsinx + 5cos²x - 5sin²x
thanks for helping me

Please show us step-by-step -how did you arrive at 5cosxsinx + 5cos²x - 5sin²x starting from (5√2)/2 sin(2x+pi/4)

That will tell us what method/s you foresee to use to solve this problem.

 

[Dr.Peterson]

the question is to show that : 2cos²x + 5sinxcosx - 3sin²x+(1/2) = (5√2)/2 sin(2x+pi/4)
I've started working from the second expression which is : (5√2)/2 sin(2x+pi/4) , and I'm stuck in: 5cosxsinx + 5cos²x - 5sin²x
thanks for helping me

A couple of your coefficients are wrong. But after fixing that, you might want to move over to the left side and try to make it look like your right side.

 

[mohammed_159]

Please show us step-by-step -how did you arrive at 5cosxsinx + 5cos²x - 5sin²x starting from (5√2)/2 sin(2x+pi/4)

That will tell us what method/s you foresee to use to solve this problem.

okay : (5√2)/2(sin2x cospi/4 + sinpi/4cos2x)
=(5√2)/2(sin2X√2/2 + √2/2cos2x)
=5/2 sin2x + 5/2 cos2x
= 5/2(sin2x + cos2x)
=5/2(2cosxsinx + cos²x-sin²x)
=5cosxsinx + 5/2cos²x -5/2sin²x.
my bad I just realized the 2 wrond coefficients but yeah I'm stuck here now .

 

[mohammed_159]

A couple of your coefficients are wrong. But after fixing that, you might want to move over to the left side and try to make it look like your right side.

5√2)/2(sin2x cospi/4 + sinpi/4cos2x)
=(5√2)/2(sin2X√2/2 + √2/2cos2x)
=5/2 sin2x + 5/2 cos2x
= 5/2(sin2x + cos2x)
=5/2(2cosxsinx + cos²x-sin²x)
=5cosxsinx + 5/2cos²x -5/2sin²x.
I fixed them can you please help me solve it like that?

 

[Deleted member 4993]

okay : (5√2)/2(sin2x cospi/4 + sinpi/4cos2x)
=(5√2)/2(sin2X√2/2 + √2/2cos2x)
=5/2 sin2x + 5/2 cos2x
= 5/2(sin2x + cos2x)
=5/2(2cosxsinx + cos²x-sin²x)
=5cosxsinx + 5/2cos²x -5/2sin²x.
my bad I just realized the 2 wrong coefficients but yeah I'm stuck here now .

Hint:

1/2 = 1/2 * sin²(x) + 1/2 * cos²(x)

 

[Dr.Peterson]

I fixed them can you please help me solve it like that?

My suggestion was to look at 2cos²x + 5sinxcosx - 3sin²x+(1/2) and observe that you have to make 2cos²x - 3sin²x + (1/2) equal to 5/2cos²x - 5/2sin²x. You need to change the number of cos²x's and sin²x's, and also make a 1/2. So how can you turn some of the former into the latter?

A key to proving identities is to keep your eye on both sides, so that even if you only work on one side, you are always aiming for the other, and can adjust course as needed.

 

[mohammed_159]

I just figured it out like that I have first to show that :
2cos²x + 5sinxcosx - 3sin²x+(1/2) = 5cosxsinx + 5/2cos²x -5/2sin²x
so I did :
2cos²x + 5sinxcosx - 3sin²x+(1/2) = 2cos²x + 5sinxcosx - 3sin²x+ 1/2cos²x + 1/2 sin²x
= 4/2cos²x + 5sinxcosx - 6/2sin²x+ 1/2cos²x + 1/2 sin²x
= 4/2cos²x + 1/2cos²x + 1/2 sin²x - 6/2sin²x + 5sinxcosx
= 5/2cos²x -5/2sin²x + 5sinxcosx
then I have to show that (5√2)/2 sin(2x+pi/4) = 5cosxsinx + 5/2cos²x -5/2sin²x which I did :
(5√2)/2(sin2x cospi/4 + sinpi/4cos2x)
=(5√2)/2(sin2X√2/2 + √2/2cos2x)
=5/2 sin2x + 5/2 cos2x
= 5/2(sin2x + cos2x)
=5/2(2cosxsinx + cos²x-sin²x)
=5cosxsinx + 5/2cos²x -5/2sin²x.
so 2cos²x + 5sinxcosx - 3sin²x+(1/2) = 5cosxsinx + 5/2cos²x -5/2sin²x and (5√2)/2 sin(2x+pi/4) = 5cosxsinx + 5/2cos²x -5/2sin²x
so 2cos²x + 5sinxcosx - 3sin²x+(1/2) = (5√2)/2 sin(2x+pi/4)

is that a proper way to solve it ?

 

[Dr.Peterson]

Yes, you have shown that both sides are equal to the same thing.