Question

26. cos 2θ cos 2Φ + sin²(θ-Φ) - sin²(θ+ Φ) = cos(2θ + 2Φ).
prove this​

Answer 1

✪ᴘʀᴏᴏғ✪

LHS

$= cos(2θ)cos(2Φ)+sin²(θ-Φ)-sin²(θ+ Φ)$

$= cos(2θ)cos(2Φ)+$

$[sin(θ-Φ)+sin(θ+ Φ)][sin(θ-Φ) - sin(θ+ Φ)]$

$= cos(2θ)cos(2Φ) +$

$[2sin(\frac{θ-Φ+θ+ Φ}{2})cos(\frac{θ-Φ-(θ+ Φ)}{2})]</p><p>[2cos(\frac{θ-Φ+θ+ Φ}{2})sin(\frac{θ-Φ-(θ+ Φ)}{2})]$

$= cos(2θ)cos(2Φ) +$

$[2sin(θ)cos(-Φ)][2cos(θ)sin(-Φ)]$

$= cos(2θ)cos(2Φ)+$

$[2sin(θ)cos(Φ)][-2cos(θ)sin(Φ)]$

$= cos(2θ)cos(2Φ) -$

$[2sin(θ)cos(θ)][2sin(Φ)cos(Φ)]$

$= cos(2θ)cos(2Φ)-sin(2θ)sin(2Φ)$

$= cos(2θ+2Φ)$ = RHS

Hence Proved

Answer 2

Step-by-step explanation:

What is a triangle ? Only mods should answer like theRyzen and Edweena