Is cos(x-y)+cos(x+y)=2cos(x)cos(y) ?

The answer to whether cos(x-y)+cos(x+y)=2cos(x)cos(y) is True

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prove cos(x-y)+cos(x+y)=2cos(x)cos(y)

prove

cos (x - y) + cos (x + y) = 2 cos (x) cos (y)

True

Solution steps

cos (x - y) + cos (x + y) = 2 cos (x) cos (y)

Manipulating right side

2 cos (x) cos (y)

Rewrite using trig identities

2 cos (x) cos (y)

Use the Product to Sum identity:

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

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

Simplify

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

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

Cancel the common factor:

2

= (cos (x - y) + cos (x + y)) \cdot 1

Refine

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

p ro v e sin (\frac{π}{6} + x) + sin (\frac{π}{6} - x) = cos (x)

p ro v e (cot (x) - csc (x)) (cos (x) + 1) = - sin (x)

p ro v e csc^{4} (θ) - cot^{4} (θ) = 2 csc^{2} (θ) - 1

p ro v e \frac{cos ( θ ) cot ( θ )}{1 - sin ( θ )} - 1 = csc (θ)

p ro v e csc (θ) - sin (θ) = cos (θ) cot (θ)

Is cos(x-y)+cos(x+y)=2cos(x)cos(y) ?
The answer to whether cos(x-y)+cos(x+y)=2cos(x)cos(y) is True