Is 4cos(x+y)+4cos(x-y)=8cos(x)cos(y) ?

The answer to whether 4cos(x+y)+4cos(x-y)=8cos(x)cos(y) is True

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prove 4cos(x+y)+4cos(x-y)=8cos(x)cos(y)

prove

4 cos (x + y) + 4 cos (x - y) = 8 cos (x) cos (y)

True

Solution steps

4 cos (x + y) + 4 cos (x - y) = 8 cos (x) cos (y)

Manipulating left side

4 cos (x + y) + 4 cos (x - y)

Factor

4 cos (x + y) + 4 cos (x - y) : 4 (cos (x + y) + cos (x - y))

4 cos (x + y) + 4 cos (x - y)

Factor out common term

4

= 4 (cos (x + y) + cos (x - y))

= (cos (x + y) + cos (x - y)) \cdot 4

Manipulating right side

8 cos (x) cos (y)

Rewrite using trig identities

8 cos (x) cos (y)

Use the Product to Sum identity:

cos (s) cos (t) = \frac{1}{2} (cos (s - t) + cos (s + t))

= 8 \cdot \frac{1}{2} (cos (x - y) + cos (x + y))

Simplify

8 \cdot \frac{1}{2} (cos (x - y) + cos (x + y)) : 4 (cos (x - y) + cos (x + y))

8 \cdot \frac{1}{2} (cos (x - y) + cos (x + y))

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 8}{2} (cos (x - y) + cos (x + y))

\frac{1 \cdot 8}{2} = 4

\frac{1 \cdot 8}{2}

Multiply the numbers:

1 \cdot 8 = 8

= \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= 4

= 4 (cos (x - y) + cos (x + y))

= 4 (cos (x - y) + cos (x + y))

= 4 (cos (x - y) + cos (x + y))

We showed that the two sides could take the same form

\Rightarrow True

Is 4cos(x+y)+4cos(x-y)=8cos(x)cos(y) ?
The answer to whether 4cos(x+y)+4cos(x-y)=8cos(x)cos(y) is True