Question

Prove that:
(a) sin^2(45°- theta) + sin^2(45°+ theta) = 1​

Answer 1

Required Answer:-

Given to prove:

• sin²(45° - θ) + sin²(45° + θ) = 1

Proof:

To proof this, we have to know some important formulas.

These are the formula required to prove this.

★ sin² θ + cos² θ = 1 — (i)

★ sin(90° - θ) = cos θ — (ii)

★ cos(90° - θ) = sin θ — (iii)

Now, let's prove this.

Taking LHS,

sin²(45 - θ) + sin²(45° + θ)

Can be written as,

= sin²(45 - θ) + sin²[90° - (45° - θ)] [Note that, 90° - (45° - θ) = 45° + θ only]

Now, using formula (ii), we get,

= sin²(45° - θ) + cos²(45° - θ)

Now, from formula (i), we get,

= 1

= RHS (Hence Proved)

Answer 2

Step-by-step explanation:

Given:-

sin^2(45°- theta) + sin^2(45°+ theta)

To find:-

Prove that:

sin^2(45°- theta) + sin^2(45°+ theta) = 1

Solution:-

LHS:-

sin^2(45°- theta) + sin^2(45°+ theta)

=>Sin²(45°-theta)+Sin²(90-(45°-theta)

we know that Sin(90°-A)=Cos A

=>Sin²(45°-theta)+Cos²(45°-theta)

we know that sin²A+Cos²A=1

=>1

=>RHS

LHS=RHS

sin^2(45°- theta) + sin^2(45°+ theta) = 1

Hence Proved