2sin²(45⁰ - A) = 1 - sin2A
Answers
Step-by-step explanation:
How do I prove that [math]cos^2 (45 + x) - sin^2 (45 + x) = 0[/math]?
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This is not an identity, which can be proved.
This is an equation, which is to be solved, ie, to get the value of unknown quantity X
Let 45 + X be β
So, Cos² β - Sin²β =0
=> (Cosβ + Sinβ) (Cosβ - Sinβ)=0
On Squaring,
=> ( Cos²β +Sin²β + 2Cosβ.Sinβ)(Cos²β + Sin²β - 2Cosβ.Sinβ) =0
=> (1+ 2Cosβ.Sinβ)(1- 2Cosβ.Sinβ) =0
=> 1 - 4 Cos²β.Sin²β=0
=> 1 - 4Cos²β(1- Cos²β) =0
=> 1 - 4(Cosβ)² + 4(Cosβ)^4 =0
=> (2Cos²β —1)² =0
Now on finding square root
=> +,- ( 2Cos²β —1) = 0
=> 2Cos²β -1 =0
=> Cos²β = 1/2
=> Cosβ = +,- 1/√2
ie, cos (45 + x ) = +,- 1/√2
=> 45 + x = 0, 225 deg
=> x = 0, 180
VERIFICATION:
So, LHS= Cos²(45°+x) - Sin²(45° +x)
= Cos²45° — Sin²45°
=( 1/√2) — (1/√2)
= 0 = RHS