Question

Find that value of sin²π/16 + sin² 3π/16 + sin²5π/16 + sin² 7π/16

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Answer 1

Answer:

2

Step-by-step explanation:

$\red{\boxed{\rm Solution}}$

Given,

sin²π/16 + sin² 3π/16 + sin²5π/16 + sin² 7π/16

How to find,

We are going to use the concept of allied angles and we are going the trigonometric identities

Solution,

sin²π/16 + sin² 3π/16 + sin²5π/16 + sin² 7π/16

We know that sin²θ + cos²θ = 1

So we are going to convert other two terms in the form of cos

We know that we can write 5π/16 as π/2 - 3π/16 and we can also write 7π/16 as π/2 - π/16

Therefore,

sin²π/16 + sin²3π/16 + sin²(π/2 - 3π/16) + sin²(π/2 - π/16)

We know that sin(90 - θ) = cosθ

Applying the allied angles formula,

sin²π/16 + sin²3π/16 + cos²3π/16 + cos²π/16

Rearranging the terms,

sin²π/16 + cos²π/16 + sin²3π/16 + cos²3π/16

We know that sin²θ + cos²θ = 1

Applying the above trigonometric identity,

1 + 1

2

Therefore,

sin²π/16 + sin² 3π/16 + sin²5π/16 + sin² 7π/16 = 2

Trigonometric identities:-

1. sin²θ + cos²θ = 1
2. 1 + tan²θ = sec²θ
3. 1 + cot²θ = cosec²θ