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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