Is 3cos(x+y)+3cos(x-y)=6cos(x)cos(y) ?

The answer to whether 3cos(x+y)+3cos(x-y)=6cos(x)cos(y) is True

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prove 3cos(x+y)+3cos(x-y)=6cos(x)cos(y)

prove

3 cos (x + y) + 3 cos (x - y) = 6 cos (x) cos (y)

True

Solution steps

3 cos (x + y) + 3 cos (x - y) = 6 cos (x) cos (y)

Manipulating left side

3 cos (x + y) + 3 cos (x - y)

Factor

3 cos (x + y) + 3 cos (x - y) : 3 (cos (x + y) + cos (x - y))

3 cos (x + y) + 3 cos (x - y)

Factor out common term

3

= 3 (cos (x + y) + cos (x - y))

= (cos (x + y) + cos (x - y)) \cdot 3

Manipulating right side

6 cos (x) cos (y)

Rewrite using trig identities

6 cos (x) cos (y)

Use the Product to Sum identity:

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

= 6 \cdot \frac{1}{2} (cos (x - y) + cos (x + y))

Simplify

6 \cdot \frac{1}{2} (cos (x - y) + cos (x + y)) : 3 (cos (x - y) + cos (x + y))

6 \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 6}{2} (cos (x - y) + cos (x + y))

\frac{1 \cdot 6}{2} = 3

\frac{1 \cdot 6}{2}

Multiply the numbers:

1 \cdot 6 = 6

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

= 3 (cos (x - y) + cos (x + y))

= 3 (cos (x - y) + cos (x + y))

= 3 (cos (x - y) + cos (x + y))

We showed that the two sides could take the same form

\Rightarrow True

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

tan (x + \frac{π}{4}) = \frac{tan ( x ) + 1}{1 - tan ( x )}

csc (θ) cos (θ) tan (θ) = 1

sin (θ) cos (φ) = \frac{sin ( θ + φ ) + sin ( θ - φ )}{2}

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

Is 3cos(x+y)+3cos(x-y)=6cos(x)cos(y) ?
The answer to whether 3cos(x+y)+3cos(x-y)=6cos(x)cos(y) is True