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

2 cos (x) sec^{2} (x) = \frac{2}{cos ( x )}

\frac{( sin ( t ) + cos ( t ) ) ^{2}}{sin ( t ) cos ( t )} = 2 + sec^{2} (t) (csc^{2} (t))

sec (x) + cos (x) = 1 + cos^{2} (x)

(cot (x) + tan (x)) sin (x) = sec (x)

\frac{cos ^{2} ( x )}{1 - sin ( x )} = sin (x) + 1

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