# Verify Trigonometric Identities

How to verify Trigonometric Identities?

To proof and verify the identities we will make use of the basic trigonometric identities to make sure that both the sides of the equation is equal to each other.

1. If tan A = (sin θ - cos θ)/(sin θ + cos θ) then prove that,
sin
θ + cos θ  = ± √2 cos A

Solution:

We know that, sec2 A = 1 + tan2 A

⇒ sec2 A = 1 + (sin θ - cos θ)2/(sin θ + cos θ) 2

⇒ sec2 A = [(sin θ + cos θ) 2 + (sin θ - cos θ) 2]/(sin θ + cos θ) 2

⇒ sec2 A = 2(sin2 θ + cos2 θ)/ (sin θ + cos θ) 2

⇒ 1/cos2 A = 2/(sin θ + cos θ) 2

⇒ (sin θ + cos θ) 2 = 2 cos2

Now taking square root on both the sides we get,

sin θ + cos θ =  ± √2 cos A .

Proved

More examples to get the basic ideas to proof and verify Trigonometric Identities.

2. If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ – y cos θ = 0, then prove that x2 + y2 = 1, (where, sin θ ≠ 0 and cos θ ≠ 0).

Solution:

x sin θ - y cos θ = 0, (Given)

⇒ x sin θ = y cos θ

⇒ y cos θ = x sin θ

Now dividing both sides by cos θ we get,

y = x ∙ (sin θ/cos θ)

Again, x sin3 θ + y cos3 θ = sin θ cos θ

⇒ x sin3 θ + x ∙ (sin θ /cos θ) ∙ cos3 θ = sin θ cos θ [Since, y = x ∙ (sin θ/cos θ)]

⇒ x sin θ ( sin2 θ + cos2 θ) = sin θ cos θ, [since, cos θ ≠ 0]

⇒ x sin θ (1) = sin θ cos θ,[since, sin2 θ + cos2 θ = 0]

⇒ x sin θ = sin θ cos θ

Now dividing both sides by sin θ we get,

⇒ x = cos θ, [since, sin θ ≠ 0]

Therefore, y = x ∙ (sin θ/cos θ)

⇒ y = cos θ ∙ (sin θ/cos θ), [Putting x = cos θ]

⇒ y = sin θ

Now, x2 + y2

= cos2 θ + sin2 θ

= 1.

Therefore, x2 + y2 = 1.

Proved

3. If 2y cos α = x sin α and 2x sec α - y csc α = 3, then prove that x2 + 4y2 = 4

Solution:

2y cos α = x sin α, (Given)

$$\frac{cos α}{x} = \frac{sin α}{2y} = \frac{\sqrt{cos^{2} α + sin^{2} α}}{x^{2} + 4y^{2}} = \frac{1}{x^{2} + 4y^{2}}$$

$$Therefore, cos θ = \frac{x}{x^{2} + 4y^{2}} and sin θ = \frac{2y}{x^{2} + 4y^{2}}$$

Now, 2x sec α - y csc α = 3

⇒ 2x ∙ $$\frac{1}{cos α}$$ - y ∙ $$\frac{1}{sin α}$$ = 3, [Since, sec α = $$\frac{1}{cos α}$$ and csc α = $$\frac{1}{sin α}]$$

⇒ 2x ∙ $$\frac{\sqrt{x^{2} + 4y^{2}}}{x}$$ - y ∙ $$\frac{\sqrt{x^{2} + 4y^{2}}}{2y}$$ = 3, [putting the values of sin α and cos α]

⇒ $$\frac{3}{2}\sqrt{x^{2} + 4y^{2}} = 3$$

⇒ $$\sqrt{x^{2} + 4y^{2}} = 2$$

Now taking square root on both the sides we get,

⇒ x2 + 4y2 = 4.

Proved

Note: Remember there is no set method that can be applied to verify trigonometric identities. However, a few different techniques needed to follow to start verifying from one side, based on the identity which is to be verified.

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Trigonometric Functions