Can someone please help me solve this trig identity problem: cos(x + 45) + cos(x - 45) = sq. root 2 cosx !?

Question

The x's kinda threw me off.

Answer 1

cos(x + 45) + cos(x - 45) = sq. root 2 cosx !?
cos(x+45) = cosxcos45 - sinx sin45 = 5sqrt(2)(cosx -sinx)
cos (x-45) = cosxcos45+ sin x sin45=.5sqrt(2)(cos x+sinx)
So cos(x+45) +cos(x-45)= sqrt(2)(cos x)
sqrt(2)(cos x) = sqrt(2)cos x

Answer 2

Equations involving a "variable" have this nasty habit of having x's.

Sin, cos... are functions (they are trigonometric functions) and, like any functions, they can be applied to a variable, just like powers, square roots and logarithms.

Using letters to replace numbers is a mathematician's way to see how an equation behaves for all values (at least, all values that one would allow for x).

You can use the x the same way you can use a number.

For example, there exists an identity:

cos(a+b) = cos(a)*cos(b) - sin(a)*sin(b)

It does not matter what values you use for a and b, the equation remains true.

So, we will use x and 45 instead of a and b.

cos(x+45)= cos(x)*cos(45) - sin(x)*sin(45)

it just so happens that cos(45) = sin (45) = 1/sqrt(2)

Next, use x and -45 for a and b

cos (x-45) = cos(x)*cos(-45) - sin(x)*sin(-45)

here, we have cos(-45) = cos(45) = 1/sqrt(2)
and sin(-45) = - sin(45) = -1/sqrt(2)

So far, we have:

cos(45+x) + cos(45-x) =
cos(x)*cos(45) - sin(x)*sin(45) + cos(x)*cos(-45) - sin(x)*sin(-45) =
cos(x)*( cos(45)+cos(-45) ) + sin(x)*( -sin(45) - (-sin(45)) ) =

cos(x)*(1/sqrt(2) + 1/sqrt(2)) + sin(x)*(-1/sqrt(2) + 1/sqrt(2))

The sin(x) disappear because it is multiplied by 0

You are left with cos(x)*(2/sqrt(2))

you can show that 2/sqrt(2) = sqrt(2) and you are done.

Answer 3

Cos(x+45) = Cos(x)Cos(45)+Sin(x)Sin(45)
Cos(x-45) = Cos(x)Cos(-45)+Sin(x)Sin(-45)

Sin parts cancel out since sin(45) = - sin(-45).

so 2 Cos(x)^2 (SQRT(2)/2)^2 = SQRT(2) Cos(x) - almost there

2 Cos(x)^2 = SQRT(2) Cos(x)
Cos(x) = SQRT(2)/2

Well, you can solve from there.x=45

Check

Cos(45+45)+Cos(45-45)=SQRT(2)Cos(45)
1 + 0 = SQRT(2) x SQRT(2)/2 = 1 - checked

Answer 4

I prefer radians; it's all the same though. 45 degrees = pi/4

cos(x + pi/4) + cos(x - pi/4) = sqrt(2) cos(x)

Choose the left hand side (LHS) since it's the more complex side.
Remember the cos addition identity:
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
cos(a - b) = cos(a)cos(b) + sin(a)sin(b),
and apply it here.

LHS = cos(x + pi/4) + cos(x - pi/4)
LHS = cos(x)cos(pi/4) - sin(x)sin(pi/4) + cos(x)cos(pi/4) +
sin(x)sin(pi/4)

Note that we see immediately that the terms containing the sines will cancel each other out (one is plus, one is minus).

LHS = cos(x)cos(pi/4) + cos(x)cos(pi/4)

Being two exact things, we combine them into one.

LHS = 2cos(x)cos(pi/4)

cos(pi/4) = sqrt(2)/2, so

LHS = 2cos(x) [sqrt(2)/2]

The 2 will cancel out.

LHS = cos(x)sqrt(2)

Multiplication is commutative, so

LHS = sqrt(2)cos(x) = RHS