UPSC NDA Math Model Paper 5

Let us consider B = 18°
So, 5B = 90°
⇒ 2B + 3B = 90°
⇒ 2B = 90° - 3B
By taking sine on both sides, we get
Sin 2B = sin (90° - 3B) = cos 3B
⇒ 2 sin B cos B = 4 cos‌3 B - 3 cos B
⇒ 2 sin B cos B - 4 cos‌3 B + 3 cos B = 0
⇒ cos B (2 sin B - 4 cos B + 3) = 0 Dividing both sides by cos B = cos 18˚ ≠ 0, we get
⇒ 2 sin B - 4 (1 - sin B) + 3 = 0
⇒ 4 sin B + 2 sin B - 1 = 0
This is a quadratic equation in Sine,
So, sin‌B=

−2±√4+16

2×4

=

−1±√5

4

Now since 18° is first quadrant so sin‌18° is positive
Therefore, sin‌B=

−1+√5

4

Now, cos‌36°=cos(2×18°)
⇒cos‌36°=1−2sin218°
⇒cos‌36°=1−2(

√5−1

4

)2 =(

16−2×(5+1−2√5

16

)=

4+4√5

16

cos‌36°=

√5+1

4

As we know, sin2θ+cos2θ=1
⇒sin‌θ=√1−cos2θ
Now, sin‌36°=√1−(

√5+1

4

)2
=

√10−2√5

4