Proving Trigonometric Ratios
In proving trigonometric ratios we will learn how to proof the questions
step-by-step using trigonometric identities.
1. If sin
4x + sin
2 x = 1 then proof that, cot
4x + cot
2 x = 1.
Solution:
Given, sin
4x + sin
2 x = 1
⇒ sin
4x + sin
2 x - sin
2 x = 1 - sin
2 x, [subtract sin
2 x from both the sides]
⇒ sin
4x = 1 - sin
2 x
⇒ sin
4x = cos
2 x.
Now L.H.S. = cot
4x + cot
2 x
= cos
4x/sin
4x + cos
2 x/sin
2 x
= cos
4x/cos
2 x + sin
4x/sin
2 x, [since, sin
4x = cos
2 x and cos
2 x = sin
4x]
= 1 = R.H.S. [since we know, sin
2 x + cos
2 x = 1]
(Proved)
2. If sin θ -
cos θ = √2 cos θ then proof that sin θ + cos θ =
√2 sin θ, where 0 < θ < π/2
Solution:
Given, sin θ - cos θ = √2 cos θ
⇒ (sin θ - cos θ)
2 = (√2 cos θ)
2, [squaring both the sides]
⇒ sin
2 θ + cos
2 θ - 2 sin θ cos θ = 2 cos
2 θ
⇒ sin
2 θ + cos
2 θ - 2 sin θ cos θ - cos
2 θ = 2 cos
2 θ - cos
2 θ, [subtract cos
2 θ from both the sides]
⇒ sin
2 θ - 2 sin θ cos θ = cos
2 θ
⇒ sin
2 θ - 2 sin θ cos θ + 2 sin θ cos θ = cos
2 θ + 2 sin θ cos θ, [adding 2 sin θ cos θ on both the sides]
⇒ sin
2 θ = cos
2 θ + 2 sin θ cos θ
⇒ sin
2 θ + sin
2 ϴ = sin
2 θ + cos
2 θ + 2 sin θ cos θ, [adding sin
2 θ on both the sides]
⇒ 2 sin
2 θ = (sin θ + cos θ)
2
⇒ (sin θ + cos θ)
2 = 2 sin
2 θ
Now taking square root on both the sides we get,
⇒ sin
θ + cos θ = ± √2 sin θ
According to the question, 0 < θ <
π/2, hence we neglect the negative vaue.
Therefore, sin θ +
cos θ = √2 sin θ
(Proved)
The above explanation on proving trigonometric ratios will help us to
solve different types of trigonometric problems.
● Trigonometric Functions
10th Grade Math
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